College Algebra

🧭 Overview

🧠 One-sentence thesis

The Cartesian Coordinate Plane unites algebra and geometry by using ordered pairs of real numbers to locate points in the plane, enabling algebraic computation of geometric properties like distance and symmetry.

📌 Key points (3–5)

• Union and intersection of sets: union collects all elements from both sets; intersection contains only shared elements.
• Interval notation: most sets of real numbers are intervals or unions of intervals, expressed using parentheses and brackets.
• Cartesian coordinates: ordered pairs (x, y) locate points by projecting to the x-axis and y-axis; order matters.
• Common confusion: (2, −4) and (−4, 2) are different points—the first coordinate is always the x-coordinate, the second is always the y-coordinate.
• Symmetry and distance: points can be reflected across axes or the origin; distance between two points is computed using the Pythagorean theorem.

🔗 Set operations and interval notation

🔗 Intersection and union

Intersection (A ∩ B): the set of elements that two sets have in common.

Union (A ∪ B): the totality of the elements in each of the sets, collected together.

• Intersection finds the overlap; union finds the combined coverage.
• Example: If A = {1, 2, 3} and B = {2, 4, 6}, then A ∩ B = {2} (only 2 is in both) and A ∪ B = {1, 2, 3, 4, 6} (all elements from both sets).
• For intervals: if A = [−5, 3) and B = (1, ∞), then A ∩ B = (1, 3) (the overlap) and A ∪ B = [−5, ∞) (the combined shaded region).

📝 Expressing sets with interval notation

• Most sets of real numbers are either intervals or unions of intervals.
• Graphing first helps: shade the region on the number line, then describe it using interval notation.
• Example: {x | x ≤ −2 or x ≥ 2} becomes (−∞, −2] ∪ [2, ∞).
• Example: {x | x ≠ 3} excludes only x = 3, so it becomes (−∞, 3) ∪ (3, ∞).
• Example: {x | x ≠ ±3} excludes both x = 3 and x = −3, breaking the line into three intervals: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞).
• A single point can be written as a set: {5} or as an interval [5, 5].

📍 The Cartesian Coordinate Plane

📍 What it is

Cartesian Coordinate Plane: the set of all possible ordered pairs (x, y) as x and y take values from the real numbers.

• Two real number lines cross at a right angle at 0.
• The horizontal line is the x-axis; the vertical line is the y-axis.
• Both axes extend indefinitely in both directions.
• This system allows us to locate points off the number lines as well as on them.

🎯 Ordered pairs and coordinates

Ordered pair (a, b): the first number is the abscissa or x-coordinate; the second is the ordinate or y-coordinate.

• The distinction between a point and its coordinates is often blurred; we speak of "the point (2, −4)."
• Think of (2, −4) as instructions: from the origin (0, 0), move 2 units right and 4 units down.
• Order matters: (2, −4) and (−4, 2) are different points.
  • (2, −4): move right 2, down 4.
  • (−4, 2): move left 4, up 2.
• Don't confuse: the first coordinate is always horizontal (x), the second is always vertical (y).

📐 Important facts about coordinates

Fact	Meaning
(a, b) and (c, d) are the same point	if and only if a = c and b = d
(x, y) lies on the x-axis	if and only if y = 0
(x, y) lies on the y-axis	if and only if x = 0
The origin	(0, 0); the only point on both axes

🗺️ Quadrants

• The axes divide the plane into four regions called quadrants, labeled with Roman numerals counterclockwise:

Quadrant	Sign of x	Sign of y	Example
I	x > 0	y > 0	(1, 2)
II	x < 0	y > 0	(−1, 2)
III	x < 0	y < 0	(−1, −2)
IV	x > 0	y < 0	(1, −2)

• Points on the axes (other than the origin) do not belong to any quadrant; they lie on the positive or negative x-axis or y-axis.
• Example: (0, 4) lies on the positive y-axis; (−117, 0) lies on the negative x-axis.

🔄 Symmetry and reflections

🔄 Three types of symmetry

Two points (a, b) and (c, d) are:

• Symmetric about the x-axis if a = c and b = −d
• Symmetric about the y-axis if a = −c and b = d
• Symmetric about the origin if a = −c and b = −d

• Symmetry is one of the most important concepts in mathematics.
• Example: starting with P(−2, 3):
  • Symmetric about the x-axis: (−2, −3) (replace y with its opposite).
  • Symmetric about the y-axis: (2, 3) (replace x with its opposite).
  • Symmetric about the origin: (2, −3) (replace both x and y with their opposites).

🪞 Reflections

• Think of the axes as mirrors.
• Reflect across the x-axis: replace y with −y.
• Reflect across the y-axis: replace x with −x.
• Reflect across the origin: replace x with −x and y with −y (equivalent to reflecting across the x-axis then the y-axis).
• Example: reflecting (−2, 3) across the x-axis gives (−2, −3); then reflecting across the y-axis gives (2, −3), which is the point symmetric about the origin.

📏 Distance in the plane

📏 The Distance Formula

Distance d between P(x₀, y₀) and Q(x₁, y₁): d = square root of [(x₁ − x₀)² + (y₁ − y₀)²]

• Distance means the length of the line segment joining P and Q.
• The formula comes from the Pythagorean theorem: imagine a right triangle with hypotenuse d and legs of length |x₁ − x₀| and |y₁ − y₀|.
• The absolute value bars can be replaced with parentheses because squaring eliminates the sign.
• Distance is never negative, so we take the positive square root.

🧮 Using the Distance Formula

• Example: distance between P(−2, 3) and Q(1, −3):
  • d = square root of [(1 − (−2))² + (−3 − 3)²]
  • d = square root of [9 + 36]
  • d = 3 times the square root of 5.
• Example: find all points with x-coordinate 1 that are 4 units from (3, 2):
  • Points have the form (1, y).
  • Set distance equal to 4: 4 = square root of [(1 − 3)² + (y − 2)²].
  • Simplify: 4 = square root of [4 + (y − 2)²].
  • Square both sides: 16 = 4 + (y − 2)².
  • (The excerpt ends here; the solution is not completed.)
• Don't confuse: the Distance Formula applies even when points are arranged vertically, horizontally, or are the same point (though the geometric triangle argument does not apply in those cases).

🧭 Overview

🧠 One-sentence thesis

The Cartesian coordinate plane unites algebra and geometry by using ordered pairs of real numbers to locate points, enabling visualization of mathematical relationships through a system of two perpendicular axes.

📌 Key points (3–5)

• What the Cartesian plane is: two perpendicular real number lines (x-axis and y-axis) that allow us to describe any point using an ordered pair (x, y).
• How coordinates work: the first number (x-coordinate) tells you horizontal movement from the origin; the second number (y-coordinate) tells you vertical movement.
• Order matters: (2, −4) and (−4, 2) are different points because the position of each number in the pair determines which axis it applies to.
• Common confusion: points on the axes vs. points in quadrants—a point lies on an axis when one coordinate is zero, and such points do not belong to any quadrant.
• Symmetry types: three fundamental symmetries (about the x-axis, y-axis, and origin) are defined by which coordinates change sign.

📐 The coordinate plane structure

📐 What the Cartesian plane is

Cartesian Coordinate Plane: the set of all possible ordered pairs (x, y) as x and y take values from the real numbers.

• It consists of two real number lines crossing at a right angle at 0.
• The horizontal line is called the x-axis; the vertical line is called the y-axis.
• Both axes extend indefinitely in both directions.
• The crossing point (0, 0) is called the origin.

🎯 How to locate a point

Ordered pair: a pair of numbers (x, y) where the first number is the abscissa or x-coordinate and the second is the ordinate or y-coordinate; together they form the Cartesian coordinates of a point.

• To locate a point from the origin:
  • Move right if x is positive, left if x is negative.
  • Move up if y is positive, down if y is negative.
• Example: (2, −4) means "move 2 units right, then 4 units down."
• The distinction between "a point" and "its coordinates" is often blurred in practice—we say "the point (2, −4)."

⚠️ Order matters

• (a, b) and (c, d) represent the same point if and only if a = c and b = d.
• Swapping the numbers gives a different point.
• Example: (2, −4) is not the same as (−4, 2)—the first is 2 units right and 4 down; the second is 4 units left and 2 up.

🗺️ Special locations and quadrants

🗺️ Points on the axes

Condition	Location	Example
y = 0	Point lies on the x-axis	(5, 0) is on the positive x-axis
x = 0	Point lies on the y-axis	(0, 5) is on the positive y-axis
x = 0 and y = 0	The origin	(0, 0) is the only point on both axes

• Points on the axes do not belong to any quadrant.
• We refer to them as lying on the "positive x-axis," "negative y-axis," etc.

🧭 The four quadrants

Quadrants: the four regions created when the axes divide the plane; labeled with Roman numerals I, II, III, IV proceeding counterclockwise.

Quadrant	Sign conditions	Example
I	x > 0, y > 0	(1, 2)
II	x < 0, y > 0	(−1, 2)
III	x < 0, y < 0	(−1, −2)
IV	x > 0, y < 0	(1, −2)

• The quadrants proceed counterclockwise starting from the upper-right.
• Don't confuse: a point on an axis (where one coordinate is zero) is not in any quadrant.

🔄 Symmetry in the plane

🔄 Three types of symmetry

Symmetry definitions: Two points (a, b) and (c, d) are symmetric about:

• the x-axis if a = c and b = −d
• the y-axis if a = −c and b = d
• the origin if a = −c and b = −d

• Symmetry is about which coordinates change sign:
  • x-axis symmetry: flip the y-coordinate (change its sign).
  • y-axis symmetry: flip the x-coordinate.
  • Origin symmetry: flip both coordinates.

🔄 How to find symmetric points

• Start with a point P (x, y).
• To find the point symmetric about the x-axis: replace y with −y → (x, −y).
• To find the point symmetric about the y-axis: replace x with −x → (−x, y).
• To find the point symmetric about the origin: replace both → (−x, −y).

Example: Given P (−2, 3):

• Symmetric about x-axis: (−2, −3)
• Symmetric about y-axis: (2, 3)
• Symmetric about origin: (2, −3)

🔄 Visualizing symmetry

• If you imagine folding the plane along the x-axis, symmetric points would overlap.
• If you imagine folding along the y-axis, symmetric points would overlap.
• If you imagine rotating 180° around the origin, symmetric points would overlap.
• Don't confuse: symmetry about the x-axis changes the y-coordinate; symmetry about the y-axis changes the x-coordinate.

🧭 Overview

🧠 One-sentence thesis

The Cartesian plane allows us to describe geometric concepts like symmetry, distance, and midpoints using algebraic formulas that connect coordinates to visual relationships.

📌 Key points (3–5)

• Symmetry types: Three kinds of symmetry (about the x-axis, y-axis, and origin) are defined by which coordinates change sign.
• Distance formula: The distance between two points comes from the Pythagorean theorem applied to a right triangle formed by the coordinate differences.
• Midpoint formula: The midpoint is found by averaging the x-coordinates and y-coordinates separately.
• Common confusion: Symmetry about the origin requires changing both coordinates' signs, not just one; don't confuse it with reflection across a single axis.
• Reflection visualization: Symmetry operations can be understood as reflections across axes, like using a mirror.

🔄 Three types of symmetry

🔄 Symmetry definitions

Two points (a, b) and (c, d) are:

• symmetric about the x-axis if a = c and b = −d
• symmetric about the y-axis if a = −c and b = d
• symmetric about the origin if a = −c and b = −d

• The key is identifying which coordinate(s) flip sign.
• x-axis symmetry: only the y-coordinate changes sign (horizontal line acts as mirror).
• y-axis symmetry: only the x-coordinate changes sign (vertical line acts as mirror).
• Origin symmetry: both coordinates change sign (equivalent to reflecting across both axes).

🪞 Reflection process

The excerpt describes symmetry as reflection:

• Across x-axis: Replace y with −y (imagine the x-axis is a mirror).
• Across y-axis: Replace x with −x (imagine the y-axis is a mirror).
• Across origin: Replace both x with −x and y with −y (reflect across both axes in sequence).

Example: Starting with (−2, 3):

• Symmetric about x-axis: (−2, −3)
• Symmetric about y-axis: (2, 3)
• Symmetric about origin: (2, −3)

Don't confuse: Reflecting across the x-axis then the y-axis gives the same result as symmetry about the origin.

📏 Distance between two points

📏 The distance formula

The distance d between points P(x₀, y₀) and Q(x₁, y₁) is: d = square root of [(x₁ − x₀)² + (y₁ − y₀)²]

• This formula comes from the Pythagorean theorem.
• Visualize a right triangle: the horizontal leg has length |x₁ − x₀|, the vertical leg has length |y₁ − y₀|, and the hypotenuse is the distance d.
• The squared terms eliminate the need for absolute value bars (since squaring makes everything positive).

🔺 Geometric derivation

The excerpt constructs a right triangle:

• The two points P and Q form the endpoints of the hypotenuse.
• Drop a perpendicular to create a right triangle.
• The legs have lengths |x₁ − x₀| and |y₁ − y₀|.
• Apply Pythagorean theorem: |x₁ − x₀|² + |y₁ − y₀|² = d²
• Simplify to (x₁ − x₀)² + (y₁ − y₀)² = d² (absolute values become parentheses when squared).
• Extract square root to get d (distance is never negative).

🧮 Application example

The excerpt shows finding distance between P(−2, 3) and Q(1, −3):

• d = square root of [(1 − (−2))² + (−3 − 3)²]
• d = square root of [9 + 36]
• d = square root of 45 = 3√5

Another example: Find all points (1, y) that are 4 units from (3, 2):

• Set up: 4 = square root of [(1 − 3)² + (y − 2)²]
• Square both sides: 16 = 4 + (y − 2)²
• Solve: (y − 2)² = 12
• Extract square root: y − 2 = ±2√3
• Two solutions: (1, 2 + 2√3) and (1, 2 − 2√3)

Why two answers? The distance condition describes points on a circle; a vertical line can intersect a circle at two points.

🎯 Midpoint of a line segment

🎯 The midpoint formula

The midpoint M of the line segment connecting P(x₀, y₀) and Q(x₁, y₁) is: M = ((x₀ + x₁)/2, (y₀ + y₁)/2)

• The midpoint is found by averaging each coordinate separately.
• Intuition: "halfway over" (average the x-values) and "halfway up" (average the y-values).
• The excerpt defines midpoint as the point whose distance from P equals its distance from Q.

✅ Verification

The excerpt notes (left as exercise):

• If d is the distance between P and Q, then the distance from P to M is d/2.
• The distance from M to Q is also d/2.
• This confirms M is equidistant from both endpoints.

Example: Midpoint of P(−2, 3) and Q(1, −3):

• M = ((−2 + 1)/2, (3 + (−3))/2)
• M = (−1/2, 0)

🔬 Abstract application

The excerpt proves that the line y = x equally divides the segment from (a, b) to (b, a) when a ≠ b:

• Find midpoint: M = ((a + b)/2, (b + a)/2) = ((a + b)/2, (a + b)/2)
• Since the x and y coordinates are equal, M lies on y = x.
• This shows the line y = x passes through the midpoint, dividing the segment equally.

📍 Special cases and axis points

📍 Points on axes

The excerpt notes:

• Points with y = 0 lie on the x-axis (positive if x > 0, negative if x < 0).
• Points with x = 0 lie on the y-axis (positive if y > 0, negative if y < 0).
• Example: (0, 4) is on the positive y-axis; (−117, 0) is on the negative x-axis.
• Such points do not belong to any of the four quadrants.

⚠️ Distance formula edge cases

The excerpt mentions:

• The geometric triangle construction doesn't work if P and Q are arranged vertically, horizontally, or are the same point.
• However, the distance formula (Equation 1.1) still applies in these cases.
• Verification is left as an exercise (Exercise 35).

🧭 Overview

🧠 One-sentence thesis

The excerpt provided does not contain substantive content on function notation; it consists only of exercises, answers, and fragments from a section on sets of real numbers and the Cartesian coordinate plane.

📌 Key points (3–5)

• The excerpt includes exercises on interval notation, set operations (intersection and union), and Cartesian coordinate geometry.
• It covers distance and midpoint formulas for points in the plane.
• It contains symmetry concepts (reflection about the x-axis, y-axis, and origin).
• The excerpt does not explain function notation, function definitions, or how to use function notation.
• Common confusion: This excerpt is labeled "1.4 Function Notation" but the actual content is from section 1.1 on sets and the Cartesian plane.

📦 What the excerpt actually contains

📦 Interval notation and set operations

The excerpt includes:

• A chart matching set-builder notation, interval notation, and number-line regions.
• Exercises on finding intersections (∩) and unions (∪) of intervals.
• Exercises on writing sets in interval notation, including sets defined by inequalities or exclusions.

Example: The set {x | -1 ≤ x < 5} is written in interval notation as [-1, 5).

📍 Cartesian coordinate plane exercises

The excerpt includes:

• Plotting and labeling points with given coordinates.
• Identifying which quadrant or axis a point lies in/on.
• Finding symmetric points about the x-axis, y-axis, and origin.

Example: The point A(-3, -7) is in Quadrant III; its reflection about the x-axis is (-3, 7).

📏 Distance and midpoint formulas

The excerpt includes:

• Exercises asking for the distance d between two points and the midpoint M of the segment connecting them.
• Verification exercises for the distance and midpoint formulas.
• Applied problems (e.g., finding all points a certain distance from a given point).

Example: One exercise asks to find all points of the form (x, -1) that are 4 units from the point (3, 2).

⚠️ Missing content

⚠️ No function notation material

The excerpt does not contain:

• A definition of function notation (e.g., f(x)).
• Explanation of how to read or evaluate function notation.
• Examples of functions written in notation form.
• Discussion of domain, range, or function properties using notation.

Don't confuse: The title "1.4 Function Notation" does not match the content, which is from section 1.1 on sets and the Cartesian plane. The excerpt appears to be a mislabeled collection of exercises and answers.

🧭 Overview

🧠 One-sentence thesis

The excerpt does not contain substantive content on function arithmetic; it presents only coordinate geometry examples (plotting points, symmetries, distances, and relations in the Cartesian plane) without discussing arithmetic operations on functions.

📌 Key points (3–5)

• What the excerpt covers: plotting points, symmetries about axes and origin, distance and midpoint formulas, and graphing relations (sets of points).
• Relations defined: a relation is a set of points in the plane, described verbally, by roster, set-builder notation, or graphically.
• Graphing techniques: plotting discrete points, horizontal/vertical line segments, open vs. closed endpoints, and infinite lines with arrows.
• Common confusion: distinguishing between inclusive (closed circle) and exclusive (open circle) endpoints when graphing inequalities.
• No function arithmetic: the excerpt does not introduce addition, subtraction, multiplication, or division of functions.

📍 Cartesian plane basics

📍 Plotting and symmetry

• The excerpt lists eight points (A through H) with their coordinates and quadrant locations.
• Symmetry rules for a point (x, y):
  • Symmetric about the x-axis: (x, −y)
  • Symmetric about the y-axis: (−x, y)
  • Symmetric about the origin: (−x, −y)
• Example: Point A at (−3, −7) in Quadrant III is symmetric about the x-axis with (−3, 7), about the y-axis with (3, −7), and about the origin with (3, 7).

📏 Distance and midpoint

• The excerpt provides results (not derivations) for distance d and midpoint M between pairs of points.
• Example: d = 5 and M = (−1, 7/2) for one pair; d = 4√10 and M = (1, −4) for another.
• Pythagorean Theorem application: For three points A, B, C, if |AB|² + |AC|² = |BC|², the triangle is a right triangle (converse of Pythagorean Theorem).

🔗 Relations as sets of points

🔗 Definition of a relation

A relation is a set of points in the plane.

• Relations can be described using:
  • Roster method: listing all points, e.g., R = {(−2, 1), (4, 3), (0, −3)}
  • Set-builder notation: e.g., {(x, 3) | −2 ≤ x ≤ 4}
  • Graphically: plotting the points
  • Algebraically: using equations or inequalities
• The excerpt emphasizes that "one method may be easier or more convenient to use than another" depending on the situation.

🖊️ Graphing discrete relations

• Example: A = {(0, 0), (−3, 1), (4, 2), (−3, 2)}
• Simply plot each point on the Cartesian plane.
• No lines or curves connect the points unless the relation specifies continuity.

📐 Graphing continuous relations

📐 Horizontal line segments

• HLS₁ = {(x, 3) | −2 ≤ x ≤ 4}: all points with y-coordinate 3 and x between −2 and 4, inclusive.
  • Includes (−2, 3), (0.829, 3), (√π, 3), (4, 3), etc.
  • Graph: a horizontal line segment from (−2, 3) to (4, 3), both endpoints filled (closed circles).
• HLS₂ = {(x, 3) | −2 ≤ x < 4}: similar, but x < 4 means (4, 3) is excluded.
  • Graph: horizontal segment from (−2, 3) with a closed circle, ending at (4, 3) with an open circle.
  • Common mistake: drawing the segment only to (3, 3) forgets points like (3.1, 3), (3.99, 3), etc.—there is no "immediately before 4."

📐 Vertical and horizontal lines

• V = {(3, y) | y is a real number}: all points with x-coordinate 3; y is unrestricted.
  • Graph: a vertical line x = 3, with arrows indicating it extends indefinitely up and down.
• H = {(x, y) | y = −2}: all points with y-coordinate −2; x is unrestricted.
  • Graph: a horizontal line y = −2, with arrows indicating it extends indefinitely left and right.

🔵 Open vs. closed endpoints

Inequality	Endpoint treatment	Graph symbol
x ≤ a or x ≥ a	Inclusive	Closed circle (filled)
x < a or x > a	Exclusive	Open circle (unfilled)

• Don't confuse: an open circle does not mean "stop one unit before"; it means the endpoint value itself is not included, but all values arbitrarily close to it are.

🔍 Regions in the plane

🔍 Unbounded regions

• R = {(x, y) | 1 < y ≤ 3}: all points where y is strictly greater than 1 and at most 3; x is unrestricted.
  • This describes a horizontal strip between the lines y = 1 (excluded) and y = 3 (included).
  • The excerpt lists this example but does not provide a detailed graph description.

🔍 Why variables are used

• The excerpt notes it is "impossible" to list all points in a continuous relation (e.g., HLS₁ contains uncountably many points).
• Variables like x and y allow compact description of infinite sets.
• Footnote: the text mentions "countable versus uncountable sets" for interested readers.

⚠️ What is missing

⚠️ No function arithmetic content

• The title "1.5 Function Arithmetic" does not match the excerpt.
• The excerpt covers:
  • Section 1.1: Sets of Real Numbers and the Cartesian Coordinate Plane (point plotting, symmetry, distance, midpoint)
  • Section 1.2: Relations (definition, graphing techniques)
• Function arithmetic (adding, subtracting, multiplying, dividing functions; composition) is not discussed.
• The excerpt does not define what a function is, nor does it explain operations on functions.

🧭 Overview

🧠 One-sentence thesis

Graphing functions connects algebraic notation with geometric visualization, allowing us to identify key features like intercepts, symmetry, increasing/decreasing behavior, and extrema both visually and analytically.

📌 Key points (3–5)

• The Fundamental Graphing Principle: the graph of a function f is the set of points (x, y) where y = f(x).
• Zeros and intercepts: zeros of f are solutions to f(x) = 0, corresponding to x-intercepts on the graph.
• Symmetry types for functions: even functions (symmetric about y-axis) satisfy f(−x) = f(x); odd functions (symmetric about origin) satisfy f(−x) = −f(x).
• Common confusion: increasing/decreasing behavior describes intervals on the x-axis, not individual points or y-values.
• Local vs absolute extrema: local extrema are "hilltops" or "valleys" in a neighborhood; absolute extrema are the overall highest or lowest function values across the entire domain.

📊 Fundamental graphing principle and basic features

📊 The core connection

The Fundamental Graphing Principle for Functions: The graph of a function f is the set of points which satisfy the equation y = f(x). That is, the point (x, y) is on the graph of f if and only if y = f(x).

• This principle unifies the algebraic process notation (Section 1.4) with the geometric point-set view (Section 1.3).
• To graph a function, graph the equation y = f(x) using techniques from earlier sections: find intercepts, test symmetry, plot additional points.

🎯 Zeros and intercepts

The zeros of a function f are the solutions to the equation f(x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f(x).

• Finding x-intercepts: set y = 0, which means solving f(x) = 0.
• Finding y-intercepts: set x = 0, which means evaluating f(0).
• Example: For f(x) = x² − x − 6, setting f(x) = 0 gives (x − 3)(x + 2) = 0, so zeros are x = −2 and x = 3, yielding x-intercepts (−2, 0) and (3, 0).

🧩 Piecewise-defined functions

• When graphing piecewise functions, apply the appropriate formula for each piece of the domain.
• Critical point: pay attention to domain boundaries where the formula changes.
• Example: For a function defined as one formula when x < 1 and another when x ≥ 1, the point at x = 1 uses the second formula, but values approaching 1 from the left use the first formula.
• Use an open circle to indicate a limiting value that is not actually attained; use a closed dot for the actual function value.

🔄 Symmetry and even/odd functions

🔄 Testing for symmetry algebraically

Testing the Graph of a Function for Symmetry:

• The graph of f is symmetric about the y-axis if and only if f(−x) = f(x) for all x in the domain.
• The graph of f is symmetric about the origin if and only if f(−x) = −f(x) for all x in the domain.

🏷️ Even and odd functions

A function is called even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin.

• Most functions are neither even nor odd.
• Testing procedure: replace x with −x and simplify; compare the result to f(x) and −f(x).
• Example: f(x) = 5/(2 − x²) is even because f(−x) = 5/(2 − (−x)²) = 5/(2 − x²) = f(x).
• Example: g(x) = 5x/(2 − x²) is odd because g(−x) = 5(−x)/(2 − x²) = −5x/(2 − x²) = −g(x).

⚠️ Don't confuse

• Symmetry about the x-axis is not relevant for functions because functions cannot have two different y-values for the same x-value.
• Testing with specific x-values can disprove evenness or oddness but cannot prove it; algebraic manipulation is required for proof.
• The calculator may suggest symmetry, but only algebra proves it.

📈 Increasing, decreasing, and constant behavior

📈 What these terms mean

Suppose f is a function defined on an interval I:

• f is increasing on I if and only if f(a) < f(b) for all real numbers a, b in I with a < b.
• f is decreasing on I if and only if f(a) > f(b) for all real numbers a, b in I with a < b.
• f is constant on I if and only if f(a) = f(b) for all real numbers a, b in I.

🗺️ Graphical interpretation

• Increasing: as you move left to right, the graph goes "uphill" (y-values get larger).
• Decreasing: as you move left to right, the graph goes "downhill" (y-values get smaller).
• Constant: the graph is "level" (y-values stay the same).

⚠️ Critical clarifications

• These behaviors describe intervals on the x-axis, not individual points.
• We report where the behavior occurs (which x-interval), not to what extent (how much the y-values change).
• A single x-value cannot be classified as increasing, decreasing, or constant by itself.
• Example: If f is increasing on [−4, −2] and decreasing on [−2, 3], the point x = −2 is in both intervals; we don't say the function is increasing or decreasing "at x = −2."

🏔️ Local and absolute extrema

🏔️ Local extrema definitions

Suppose f is a function with f(a) = b:

• f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f(a) ≥ f(x) for all x in I. The value b is called "a local maximum value of f."
• f has a local minimum at the point (a, b) if and only if there is an open interval I containing a for which f(a) ≤ f(x) for all x in I. The value b is called "a local minimum value of f."

• Local maximum: a "hilltop" where the function value is largest in some neighborhood.
• Local minimum: a "valley" where the function value is smallest in some neighborhood.
• The key requirement is an open interval around the point that stays within the domain.

🌍 Absolute extrema definitions

• The value b is the maximum of f if b ≥ f(x) for all x in the domain of f.
• The value b is the minimum of f if b ≤ f(x) for all x in the domain of f.

• Maximum: the largest function value over the entire domain.
• Minimum: the smallest function value over the entire domain.
• Not every function has local or absolute extrema.

⚠️ Boundary behavior

• Points at the endpoints of the domain typically cannot be local extrema because no open interval around them stays entirely within the domain.
• Example: If the domain is [−4, 6] and the graph has a low point at (−4, −3), this is not a local minimum because we cannot place an open interval around x = −4 that stays in the domain.
• However, such boundary points can still be absolute extrema.

🔧 Finding extrema with technology

• College Algebra techniques are limited for finding extrema analytically; Calculus provides the tools.
• Graphing calculators have "Minimum" and "Maximum" features to approximate extrema.
• These approximations should be treated as such and rounded appropriately (e.g., to two decimal places).

🧮 Working with function graphs

🧮 Reading information from graphs

When given a graph of y = f(x), you can determine:

• Domain: project the graph onto the x-axis.
• Range: project the graph onto the y-axis.
• Function values: f(a) is the y-coordinate of the point where x = a.
• Solving equations: f(x) = c means finding x-values where the y-coordinate equals c.
• Inequalities: f(x) < 0 means finding where the graph is below the x-axis.

📱 Calculator considerations

• Calculators can suggest mathematical truths by displaying graphs.
• Only algebra proves mathematical truths.
• Example: A calculator might show a graph that appears symmetric, but testing specific values and algebraic verification are needed to confirm.
• When entering piecewise functions, use logical commands available in the calculator.

🎯 Applied problems

• Functions can model real-world situations (distance, cost, population, etc.).
• Use the graphing and analysis techniques to answer questions about the model.
• Example: To minimize distance from a point to a curve, define a distance function d(x) and use the calculator's Minimum feature to approximate the solution.

🧭 Overview

🧠 One-sentence thesis

Graph transformations allow us to predict how the graph of a function changes when we systematically modify its formula by adding constants, multiplying by scalars, or changing inputs and outputs.

📌 Key points (3–5)

• Vertical vs horizontal changes: modifications to outputs (y-values) cause vertical transformations; modifications to inputs (x-values) cause horizontal transformations.
• Shifts: adding/subtracting constants moves graphs up/down (outputs) or left/right (inputs).
• Reflections: multiplying by −1 flips graphs across axes (outputs → x-axis; inputs → y-axis).
• Scalings: multiplying by positive constants stretches or compresses graphs vertically (outputs) or horizontally (inputs).
• Common confusion: horizontal transformations work "opposite" to intuition—f(x + h) shifts LEFT h units, and f(bx) with b > 1 causes horizontal SHRINKING.

📐 Vertical transformations (output changes)

⬆️ Vertical shifts

Vertical shift: To graph y = f(x) + k, add k to every y-coordinate on the graph of f.

• Adding to the output shifts the graph up (k > 0) or down (k < 0)
• The domain stays the same; the range shifts by k units
• Example: If (2, 3) is on f, then (2, 3 + k) is on f(x) + k

🪞 Reflection across x-axis

x-axis reflection: To graph y = −f(x), multiply every y-coordinate by −1.

• Multiplying the output by −1 flips the graph upside down
• The domain remains unchanged; the range is negated
• Example: If (2, 3) is on f, then (2, −3) is on −f(x)

📏 Vertical scaling

Vertical scaling: To graph y = af(x) where a > 0, multiply every y-coordinate by a.

• When a > 1: vertical stretching (graph gets taller) by factor a
• When 0 < a < 1: vertical shrinking (graph gets shorter) by factor 1/a
• The domain stays the same; the range is scaled by factor a
• Example: If (2, 3) is on f, then (2, 3a) is on af(x)

↔️ Horizontal transformations (input changes)

⬅️➡️ Horizontal shifts

Horizontal shift: To graph y = f(x + h), subtract h from every x-coordinate on the graph of f.

• Key reversal: f(x + h) shifts LEFT h units (when h > 0)
• f(x − h) shifts RIGHT h units (when h > 0)
• Why: To use f(a) = b, we solve x + h = a, giving x = a − h
• The range stays the same; the domain shifts by h units
• Example: If (2, 3) is on f, then (2 − h, 3) is on f(x + h)

🪞 Reflection across y-axis

y-axis reflection: To graph y = f(−x), multiply every x-coordinate by −1.

• Multiplying the input by −1 flips the graph left-to-right
• The range remains unchanged; the domain is negated
• Example: If (2, 3) is on f, then (−2, 3) is on f(−x)

📏 Horizontal scaling

Horizontal scaling: To graph y = f(bx) where b > 0, divide every x-coordinate by b.

• Key reversal: When b > 1, the graph undergoes horizontal shrinking by factor b (gets narrower)
• When 0 < b < 1, horizontal stretching by factor 1/b (gets wider)
• Why: To use f(a) = b, we solve bx = a, giving x = a/b
• The range stays the same; the domain is scaled by factor 1/b
• Example: If (2, 3) is on f, then (2/b, 3) is on f(bx)

Don't confuse: Horizontal scaling by f(bx) divides x-coordinates by b, not multiplies—this is opposite to vertical scaling.

🔄 Combining transformations

🎯 Order matters

The general form is g(x) = Af(Bx + H) + K

Correct order of operations:

1. Horizontal shift: Subtract H from x-coordinates (inside function first)
2. Horizontal scaling/reflection: Divide x-coordinates by B (may include reflection if B < 0)
3. Vertical scaling/reflection: Multiply y-coordinates by A (may include reflection if A < 0)
4. Vertical shift: Add K to y-coordinates (outside function last)

Why this order:

• For inputs (inside f): we solve Bx + H = a by first subtracting H, then dividing by B
• For outputs (outside f): we follow order of operations—multiply by A first, then add K

🧮 Working through transformations

To find where point (a, b) on f moves to on g(x) = Af(Bx + H) + K:

Step 1 (x-coordinate): Solve Bx + H = a

• Subtract H: Bx = a − H
• Divide by B: x = (a − H)/B

Step 2 (y-coordinate): Calculate g((a − H)/B)

• g((a − H)/B) = Af(a) + K = Ab + K

Result: Point (a, b) moves to ((a − H)/B, Ab + K)

⚠️ Common pitfalls

• Horizontal shift direction: f(x + 3) shifts LEFT 3 units, not right
• Horizontal scaling magnitude: f(2x) shrinks by factor 2 (makes graph narrower), not stretches
• Order of multiple transformations: applying transformations in wrong order produces different results
• Intermediate steps: when combining many transformations, define intermediate functions to track each step

🔍 Rigid vs non-rigid transformations

🔒 Rigid transformations

Shifts and reflections are rigid—they preserve the shape of the graph:

• Only change position (shifts) or orientation (reflections)
• Do not distort or change the fundamental shape
• Distances and angles within the graph are preserved

🔓 Non-rigid transformations

Scalings are non-rigid—they change the shape:

• Vertical scaling changes how "tall" the graph is
• Horizontal scaling changes how "wide" the graph is
• The graph is stretched or compressed, altering its proportions

Example: For f(x) = √x, both g(x) = 3√x (vertical stretch by 3) and j(x) = √(9x) (horizontal shrink by 9) can produce the same graph because √(9x) = 3√x—but this equivalence depends on algebraic properties specific to the square root function.

🧭 Overview

🧠 One-sentence thesis

Linear functions model constant-rate relationships and are characterized by their slope (rate of change) and y-intercept, making them essential tools for analyzing real-world phenomena from economics to physics.

📌 Key points (3–5)

• Slope measures steepness and direction: positive slope = increasing line, negative slope = decreasing line, zero slope = horizontal/constant, undefined slope = vertical line.
• Two key forms: point-slope form y - y₀ = m(x - x₀) and slope-intercept form y = mx + b, where m is slope and b is the y-intercept.
• Rate of change interpretation: slope represents "rise over run" or Δy/Δx, which translates to real-world rates like dollars per item or degrees per hour.
• Common confusion: don't confuse "undefined slope" (vertical lines) with "zero slope" (horizontal lines); vertical lines are not functions.
• Average rate of change: for any function over an interval [a,b], this equals the slope of the secant line connecting the endpoints.

📐 Understanding slope

📏 The slope formula

The slope m of the line containing points P(x₀, y₀) and Q(x₁, y₁) is: m = (y₁ - y₀)/(x₁ - x₀), provided x₁ ≠ x₀.

• The numerator represents vertical change (rise).
• The denominator represents horizontal change (run).
• The restriction x₁ ≠ x₀ prevents division by zero (which would indicate a vertical line).

🔍 Interpreting slope values

Slope value	Line behavior	Description
Positive	Increasing	Rises from left to right
Negative	Decreasing	Falls from left to right
Zero	Constant	Horizontal line
Undefined	Vertical	Not a function

• Magnitude matters: larger absolute value = steeper line.
• Example: slope of 1/2 means "up 1 unit for every 2 units right."

📊 Slope as rate of change

• Slope can be written as m = Δy/Δx using delta notation (Δ means "change in").
• This represents the rate of change of y with respect to x.
• Real-world interpretation: units of y per unit of x.
• Example: In a temperature scenario, if slope = 2, this means 2°F per hour—temperature increases 2 degrees for each hour that passes.

📝 Equations of lines

📍 Point-slope form

Point-slope form: y - y₀ = m(x - x₀) for a line with slope m passing through point (x₀, y₀).

• When to use: when you know one point and the slope.
• How it works: derived directly from the slope formula by algebraic manipulation.
• Any point on the line can serve as (x₀, y₀); the resulting equation will be equivalent.

📈 Slope-intercept form

Slope-intercept form: y = mx + b, where m is the slope and (0, b) is the y-intercept.

• When to use: most convenient for graphing and identifying key features quickly.
• The y-intercept: setting x = 0 gives y = b, so the line crosses the y-axis at (0, b).
• This form can represent all non-vertical lines.
• Special case: when m = 0, we get y = b (horizontal line).

🎯 Finding line equations

Process when given two points:

1. Calculate slope using m = (y₁ - y₀)/(x₁ - x₀).
2. Substitute slope and one point into point-slope form.
3. Simplify to slope-intercept form if needed.

Example: For points (-1, 3) and (2, 1):

• Slope: m = (1 - 3)/(2 - (-1)) = -2/3
• Using point (-1, 3): y - 3 = -2/3(x + 1)
• Simplified: y = -2/3·x + 7/3

🔢 Linear and constant functions

📐 Linear function definition

A linear function is a function of the form f(x) = mx + b, where m and b are real numbers with m ≠ 0. The domain is (-∞, ∞).

• The graph is a non-horizontal, non-vertical line.
• All linear functions are functions (pass the vertical line test).
• The restriction m ≠ 0 distinguishes linear from constant functions.

➡️ Constant function definition

A constant function is a function of the form f(x) = b, where b is a real number. The domain is (-∞, ∞).

• The graph is a horizontal line at height b.
• The output value never changes regardless of input.
• This is the special case where slope m = 0.

⚠️ Domain considerations

• Standard domain: all linear and constant functions have domain (-∞, ∞).
• Applied domain: real-world contexts may restrict the domain (e.g., x ≥ 0 for quantities).
• Don't confuse: a function like f(x) = (x² - 4)/(x - 2) simplifies to f(x) = x + 2, but it's NOT a linear function because its domain excludes x = 2.

💼 Real-world applications

💰 Cost functions

Structure: C(x) = (variable cost per unit)·x + (fixed cost)

• The slope represents the variable cost (cost per additional unit).
• The y-intercept C(0) represents the fixed or start-up cost.
• Example: C(x) = 80x + 150 means $80 per unit plus $150 fixed cost.

📉 Demand functions

Structure: price p as a function of quantity sold x

• Typically has negative slope (higher quantity sold at lower prices).
• Example: p(x) = -1.5x + 250 means price decreases $1.50 for each additional unit sold.
• Applied domain: both x ≥ 0 and p(x) ≥ 0 must hold (can't have negative quantities or prices).

💵 Revenue functions

Structure: R(x) = x·p(x) where p(x) is the demand function

• Revenue = (quantity sold) × (price per unit).
• When p(x) is linear, R(x) becomes quadratic (not linear).
• Example: If p(x) = -1.5x + 250, then R(x) = x(-1.5x + 250) = -1.5x² + 250x.

📊 Average rate of change

🧮 Definition for any function

The average rate of change of function f over interval [a, b] is: Δf/Δx = [f(b) - f(a)]/(b - a).

• This generalizes the concept of slope to non-linear functions.
• Geometrically: the slope of the secant line connecting (a, f(a)) and (b, f(b)).
• Some textbooks use notation m_sec for this value.

🔄 Interpretation

• Think of it as "change in outputs / change in inputs."
• Analogous to average speed on a trip: total distance divided by total time.
• The word "average" is important: the function may change at different rates within the interval.
• For linear functions: average rate of change equals the slope m (constant throughout).

📐 Connection to difference quotients

• Average rate of change over [x, x + h] equals [f(x + h) - f(x)]/h.
• This is the difference quotient from earlier sections.
• Geometric meaning: slope of the secant line through (x, f(x)) and (x + h, f(x + h)).
• This concept is foundational for calculus (leads to instantaneous rate of change/derivatives).

🌡️ Example interpretation

Temperature scenario: at 6 AM it's 24°F, at 10 AM it's 32°F.

• Average rate of change: (32 - 24)/(10 - 6) = 8/4 = 2°F per hour.
• Prediction: if this rate continues, temperature at noon (2 hours later) would be 32 + 2(2) = 36°F.
• Reality check: actual temperature may differ because the rate may not stay constant—models are approximations.

🔗 Special line relationships

↔️ Parallel lines

• Two non-vertical lines are parallel if and only if they have the same slope.
• Vertical lines are also parallel to each other (though slope is undefined).
• To find a parallel line through point P: use the same slope m with point-slope form.

⊥ Perpendicular lines

• Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals: m₁ · m₂ = -1.
• Equivalently: m₂ = -1/m₁.
• Horizontal lines are perpendicular to vertical lines.
• To find a perpendicular line through point P: use slope -1/m with point-slope form.

🧭 Overview

🧠 One-sentence thesis

Absolute value functions transform negative inputs to their positive counterparts while leaving positive inputs unchanged, producing characteristic V-shaped or piecewise-linear graphs that can be analyzed using case-by-case definitions and transformations.

📌 Key points (3–5)

• Core definition: Absolute value |x| is defined piecewise: –x when x < 0, and x when x ≥ 0.
• Solving equations: Use equality properties (e.g., |x| = c means x = c or x = –c when c > 0) or break into cases when x appears both inside and outside absolute values.
• Graphing strategy: Either rewrite using the piecewise definition or apply transformations to the base graph y = |x|, which has a characteristic ∨ shape.
• Common confusion: The expression "–x" in the definition means "the opposite of x," not "negative x"—when x is already negative, –x is positive.
• Properties: Absolute value respects multiplication and division (|ab| = |a||b|) and powers (|aⁿ| = |a|ⁿ), which simplify solving and manipulation.

📐 Definition and properties

📐 What absolute value means

Absolute value of a real number x, denoted |x|, is given by: |x| = { –x, if x < 0; x, if x ≥ 0 }

• This is a piecewise-defined function: different rules apply depending on whether x is negative or non-negative.
• Intuitive interpretations also exist: distance from x to 0 on the number line, or the square root of x squared.
• The definition ensures |x| is never negative.

Example: |5| = 5 because 5 ≥ 0, so we use the rule |x| = x. For |–5|, since –5 < 0, we use |x| = –x, giving |–5| = –(–5) = 5.

🔧 Key properties (Theorem 2.1)

Property	Statement	Notes
Product Rule	|ab| = |a||b|	Absolute value of a product equals product of absolute values
Power Rule	|aⁿ| = |a|ⁿ	Whenever aⁿ is defined
Quotient Rule	|a/b| = |a|/|b|	Provided b ≠ 0
Zero property	|x| = 0 if and only if x = 0	Only zero has absolute value zero
Positive c	|x| = c (c > 0) means x = c or x = –c	Two solutions when c is positive
Negative c	|x| = c (c < 0) has no solution	Absolute value cannot be negative

• These properties are proved by checking cases (both inputs positive, both negative, mixed signs, zero).
• The equality properties are essential for solving absolute value equations.

🧮 Solving absolute value equations

🧮 Basic equations using equality properties

When the equation has form |expression| = c:

• If c > 0: Set expression = c OR expression = –c, then solve both.
• If c = 0: Set expression = 0 (only one solution).
• If c < 0: No solution (absolute value is never negative).

Example: To solve |3x – 1| = 6:

• Since 6 > 0, write 3x – 1 = 6 or 3x – 1 = –6.
• First case: x = 7/3. Second case: x = –5/3.
• Both solutions should be checked in the original equation.

Example: To solve 3 – |x + 5| = 1:

• First isolate: |x + 5| = 2.
• Then x + 5 = 2 or x + 5 = –2.
• Solutions: x = –3 or x = –7.

Example: To solve 4 – |5x + 3| = 5:

• Isolate: |5x + 3| = –1.
• Since –1 < 0, there is no solution.

🔀 Equations with x inside and outside absolute value

When x appears both inside and outside the absolute value:

• Cannot directly use equality properties (risk losing or adding false solutions).
• Must break into cases using the piecewise definition.

Example: To solve |x| = x² – 6:

• Case 1 (x < 0): |x| = –x, so –x = x² – 6 → x² + x – 6 = 0 → (x + 3)(x – 2) = 0. Get x = –3 or x = 2. Keep only x = –3 (satisfies x < 0).
• Case 2 (x ≥ 0): |x| = x, so x = x² – 6 → x² – x – 6 = 0 → (x – 3)(x + 2) = 0. Get x = 3 or x = –2. Keep only x = 3 (satisfies x ≥ 0).
• Final solutions: x = –3 and x = 3.

Don't confuse: When isolating absolute value is possible, use equality properties. When x appears in multiple places, use case-by-case analysis.

📊 Graphing absolute value functions

📊 The base graph y = |x|

Using the piecewise definition:

f(x) = |x| = { –x, if x < 0; x, if x ≥ 0 }

• For x < 0: graph the line y = –x (slope –1).
• For x ≥ 0: graph the line y = x (slope 1).
• The two pieces meet at (0, 0), forming a ∨ shape.
• Domain: all real numbers. Range: [0, ∞).
• Decreasing on (–∞, 0], increasing on [0, ∞).
• Absolute minimum at (0, 0); no maximum.

🔄 Using transformations

To graph g(x) = |x – 3|:

• Recognize g(x) = f(x – 3) where f(x) = |x|.
• Shift the graph of y = |x| right 3 units.
• The vertex moves from (0, 0) to (3, 0).

To graph h(x) = |x| – 3:

• Recognize h(x) = f(x) – 3.
• Shift the graph of y = |x| down 3 units.
• The vertex moves from (0, 0) to (0, –3).

To graph i(x) = 4 – 2|3x + 1|:

• Rewrite as i(x) = –2f(3x + 1) + 4 where f(x) = |x|.
• Inside changes (right to left in order): Subtract 1 from x (shift left 1), then divide by 3 (horizontal shrink by factor 3).
• Outside changes (in order): Multiply by –2 (vertical stretch by 2, then reflect across x-axis), then add 4 (shift up 4).
• The vertex moves through transformations to (–1/3, 4), and the graph opens downward (∧ shape).

Key insight: All basic absolute value functions of form y = a|bx + c| + d have the characteristic V or ∧ shape.

🧩 More complex cases

When the function doesn't have simple V shape:

Example: f(x) = |x|/x

• Domain excludes x = 0 (division by zero).
• Rewrite piecewise: f(x) = { –x/x = –1 if x < 0; x/x = 1 if x > 0 }.
• Graph: horizontal line y = –1 for x < 0, horizontal line y = 1 for x > 0, with open circles at (0, –1) and (0, 1).
• Range: {–1, 1} (only two values).
• Constant on each piece; every point is both a relative max and min.

Example: g(x) = |x + 2| – |x – 3| + 1

• Two absolute value terms require breaking into multiple cases.
• Find where each expression inside absolute value changes sign: x = –2 and x = 3.
• Case 1 (x < –2): Both (x + 2) and (x – 3) are negative, so g(x) = –(x + 2) – (–(x – 3)) + 1 = –4.
• Case 2 (–2 ≤ x < 3): (x + 2) ≥ 0 and (x – 3) < 0, so g(x) = (x + 2) – (–(x – 3)) + 1 = 2x.
• Case 3 (x ≥ 3): Both positive, so g(x) = (x + 2) – (x – 3) + 1 = 6.
• Graph: horizontal line y = –4 for x < –2, line y = 2x for –2 ≤ x < 3, horizontal line y = 6 for x ≥ 3.
• The pieces connect at (–2, –4) and (3, 6).

Don't confuse: Not all functions involving absolute values have V shapes—multiple absolute value terms or ratios can produce step functions, piecewise linear graphs, or other forms.

🔍 Analysis from graphs

🔍 Finding key features

Zeros and intercepts:

• Zeros: Solve f(x) = 0 algebraically; these give x-intercepts.
• x-intercepts: Points where graph crosses x-axis, form (a, 0).
• y-intercept: Evaluate f(0) if 0 is in the domain; form (0, b).

Domain and range:

• Domain: Project graph onto x-axis; typically all real numbers unless division by zero or other restrictions.
• Range: Project graph onto y-axis; often [minimum, ∞) for V-shaped graphs opening upward, or (–∞, maximum] for graphs opening downward.

Increasing/decreasing/constant:

• Identify intervals by observing where graph rises, falls, or stays level as x increases.
• V-shaped graphs typically decrease on one side of the vertex and increase on the other.

Extrema:

• Relative (local) extrema: High or low points compared to nearby points.
• Absolute (global) extrema: Highest or lowest points on entire graph.
• Vertex of V or ∧ shape is typically an absolute extremum.
• For constant functions or piecewise constant sections, every point can be both a relative max and min.

Example: For f(x) = |x – 3|:

• Zero at x = 3; x-intercept (3, 0); y-intercept (0, 3).
• Domain: (–∞, ∞); Range: [0, ∞).
• Decreasing on (–∞, 3], increasing on [3, ∞).
• Absolute minimum 0 at (3, 0); no maximum.

Don't confuse: Relative vs absolute extrema—a relative extremum is local (best in a neighborhood), while absolute is global (best on entire domain). For simple V-shaped graphs, the vertex gives both.

🧭 Overview

🧠 One-sentence thesis

Quadratic functions—whose graphs are parabolas—can be transformed, analyzed, and optimized using standard form to find vertices, intercepts, and maximum or minimum values, making them essential for modeling real-world problems like profit maximization and projectile motion.

📌 Key points (3–5)

• Two forms of quadratic functions: general form f(x) = ax² + bx + c and standard form f(x) = a(x − h)² + k, where standard form reveals the vertex directly.
• The vertex and the coefficient a: the vertex (h, k) is the turning point; if a > 0 the parabola opens upward (vertex is a minimum), if a < 0 it opens downward (vertex is a maximum).
• Converting between forms: completing the square transforms general form into standard form, exposing the vertex and making graphing straightforward.
• Common confusion—which form to use: standard form is best for graphing and finding the vertex; general form is often easier for finding intercepts by factoring or using the quadratic formula.
• The discriminant b² − 4ac: determines the number of real solutions (x-intercepts): negative means no real solutions, zero means one, positive means two.

📐 Definition and basic shape

📐 What is a quadratic function

Quadratic function: a function of the form f(x) = ax² + bx + c, where a, b, and c are real numbers with a ≠ 0. The domain is (−∞, ∞).

• The simplest quadratic is f(x) = x², whose graph is a U-shaped curve called a parabola.
• The turning point of the parabola is called the vertex; for f(x) = x², the vertex is (0, 0).
• The parabola is symmetric about a vertical line through the vertex, called the axis of symmetry.

🔄 The role of the coefficient a

• If a > 0, the parabola opens upward and the vertex is a minimum.
• If a < 0, the parabola opens downward and the vertex is a maximum.
• The magnitude of a affects the "width" of the parabola (larger |a| means narrower).

🧩 Two forms of quadratic functions

🧩 General form vs standard form

General form: f(x) = ax² + bx + c
Standard form: f(x) = a(x − h)² + k

Form	Advantages	When to use
General	Easy to identify y-intercept (c); sometimes easier to factor	Finding intercepts, applying quadratic formula
Standard	Vertex (h, k) is immediately visible; easy to graph using transformations	Graphing, finding max/min, understanding transformations

• Don't confuse: the two forms describe the same function, just written differently.
• Example: f(x) = x² + 4x + 1 (general) is the same as f(x) = (x + 2)² − 3 (standard).

🔧 Completing the square

• Purpose: convert general form into standard form to reveal the vertex.
• Steps:
  1. Ensure the coefficient of x² is 1 (factor out a if needed).
  2. Take half the coefficient of x, square it, then add and subtract that value.
  3. Group the perfect square trinomial and simplify the constants.
• Example: f(x) = x² − 4x + 3
  • Half of −4 is −2; (−2)² = 4.
  • f(x) = (x² − 4x + 4 − 4) + 3 = (x − 2)² − 1.
• Why it works: reversing the expansion of (x − h)² = x² − 2hx + h² step by step.

🎯 Finding the vertex

🎯 Vertex formula for standard form

Theorem: For f(x) = a(x − h)² + k, the vertex is (h, k).

• This is immediate from the form: the horizontal shift is h, the vertical shift is k.
• Example: f(x) = −2(x − 3)² + 1 has vertex (3, 1).
• Don't confuse signs: f(x) = (x + 2)² − 3 is f(x) = (x − (−2))² + (−3), so h = −2, k = −3, and the vertex is (−2, −3).

🎯 Vertex formula for general form

Theorem: For f(x) = ax² + bx + c, the vertex is at x = −b/(2a), and the y-coordinate is f(−b/(2a)).

• This formula comes from completing the square in full generality.
• Example: f(x) = −1.5x² + 170x − 150
  • a = −1.5, b = 170
  • x = −170/(2(−1.5)) = 170/3 ≈ 56.67
  • y = f(170/3) = 14000/3 ≈ 4666.67
  • Vertex: (56.67, 4666.67).
• Why it works: the completed square form is a((x + b/(2a))² + (4ac − b²)/(4a), so the x-coordinate of the vertex is −b/(2a).

🔍 Intercepts and the quadratic formula

🔍 Finding x-intercepts

• Set f(x) = 0 and solve for x.
• Three methods:
  1. Factoring (if the quadratic is easily factorable): x² − 4x + 3 = (x − 3)(x − 1) = 0 gives x = 3 or x = 1.
  2. Extracting square roots (if in standard form): (x − 2)² = 3 gives x = 2 ± √3.
  3. Quadratic formula (always works): x = (−b ± √(b² − 4ac))/(2a).

🔍 The quadratic formula

Quadratic formula: If ax² + bx + c = 0 (a ≠ 0), then x = (−b ± √(b² − 4ac))/(2a).

• Derived by completing the square on the general form.
• Example: −1.5x² + 170x − 150 = 0
  • a = −1.5, b = 170, c = −150
  • x = (−170 ± √(170² − 4(−1.5)(−150)))/(2(−1.5))
  • x = (−170 ± √28000)/(−3) = (170 ± 20√70)/3.

🔍 The discriminant

Discriminant: the quantity b² − 4ac in the quadratic formula.

• What it tells you:
  • If b² − 4ac < 0: no real solutions (parabola does not cross the x-axis).
  • If b² − 4ac = 0: exactly one real solution (parabola touches the x-axis at the vertex).
  • If b² − 4ac > 0: exactly two real solutions (parabola crosses the x-axis twice).
• Don't confuse: the discriminant is not the number of solutions itself, but a test for how many real solutions exist.

🔍 Finding the y-intercept

• Set x = 0 and compute f(0).
• In general form f(x) = ax² + bx + c, the y-intercept is simply (0, c).

🚀 Applications: optimization and modeling

🚀 Profit maximization

• Setup: Profit P(x) = Revenue R(x) − Cost C(x), where x is the number of items sold.
• Often R(x) is quadratic (from a linear price-demand function) and C(x) is linear, so P(x) is quadratic.
• Goal: find the vertex of P(x) to maximize profit.
• Example: P(x) = −1.5x² + 170x − 150 for 0 ≤ x ≤ 166
  • Vertex at x = 170/3 ≈ 56.67
  • Since x must be a whole number, compare P(56) and P(57); choose x = 57 for maximum profit of $4666.50.
• Interpreting zeros (break-even points): where P(x) = 0, the business neither makes nor loses money; profit is positive between the two zeros.

🚀 Area and perimeter problems

• Typical setup: maximize area given a fixed perimeter (or vice versa).
• Example: rectangular pasture with one side along a river (no fence needed there); 200 feet of fencing available.
  • Let w = width, l = length; perimeter constraint: l + 2w = 200, so l = 200 − 2w.
  • Area A = wl = w(200 − 2w) = 200w − 2w² = −2w² + 200w.
  • Vertex at w = −200/(2(−2)) = 50; then l = 200 − 2(50) = 100.
  • Maximum area: A(50) = 5000 square feet.
• Applied domain: w > 0 and l = 200 − 2w > 0, so 0 < w < 100.

🚀 Projectile motion

• Height of an object: h(t) = −gt² + v₀t + s₀, where g is gravity (e.g., 16 or 4.9 depending on units), v₀ is initial velocity, s₀ is initial height.
• Maximum height: occurs at the vertex; t = −v₀/(2(−g)) = v₀/(2g).
• Time to hit the ground: solve h(t) = 0 using the quadratic formula.
• Example: h(t) = −5t² + 100t for 0 ≤ t ≤ 20
  • Vertex at t = 100/(2·5) = 10 seconds
  • Maximum height: h(10) = 500 feet.

🖼️ Graphing quadratics using transformations

🖼️ Starting from f(x) = x²

• The graph of f(x) = x² is a parabola with vertex (0, 0), opening upward.
• To graph g(x) = a(x − h)² + k, apply transformations in order:
  1. Horizontal shift: move h units (right if h > 0, left if h < 0).
  2. Vertical scaling and/or reflection: multiply y-values by a (stretch if |a| > 1, compress if |a| < 1; reflect across x-axis if a < 0).
  3. Vertical shift: move k units (up if k > 0, down if k < 0).
• Example: g(x) = (x + 2)² − 3
  • Shift left 2 units: vertex moves from (0, 0) to (−2, 0).
  • Shift down 3 units: vertex moves to (−2, −3).

🖼️ Key features to plot

• Vertex: the turning point; use the vertex formula.
• Axis of symmetry: the vertical line x = h (in standard form) or x = −b/(2a) (in general form).
• Intercepts: x-intercepts (if any) and the y-intercept.
• Direction: opens upward if a > 0, downward if a < 0.
• With these features, you can sketch a good graph without plotting many additional points.

🖼️ Range and increasing/decreasing intervals

• Range:
  • If a > 0 (opens upward): [k, ∞) where k is the y-coordinate of the vertex.
  • If a < 0 (opens downward): (−∞, k].
• Increasing/decreasing:
  • If a > 0: decreasing on (−∞, h], increasing on [h, ∞).
  • If a < 0: increasing on (−∞, h], decreasing on [h, ∞).

🔗 Absolute value of quadratic functions

🔗 Graphing f(x) = |g(x)| where g(x) is quadratic

• Method: graph g(x) first, then reflect any portion of the graph that is below the x-axis up above the x-axis.
• Why: |g(x)| = g(x) when g(x) ≥ 0, and |g(x)| = −g(x) when g(x) < 0.
• Example: f(x) = |x² − x − 6|
  • First graph g(x) = x² − x − 6 (a parabola with x-intercepts at x = −2 and x = 3).
  • For x between −2 and 3, g(x) is negative; reflect that portion across the x-axis.
  • The result is a "W" shape with a local maximum at the vertex of g (which becomes a local minimum of f).

🔗 Don't confuse

• This is different from f(x) = |x|² (which is just x²) or f(x) = (|x|)² (also just x²).
• The absolute value applies to the entire output of the quadratic, not to x itself.

🧭 Overview

🧠 One-sentence thesis

Inequalities involving absolute values and quadratic functions can be solved both analytically (using algebraic techniques and sign diagrams) and graphically (by comparing where one function's graph lies above or below another's).

📌 Key points (3–5)

• Graphical interpretation: Solutions to f(x) < g(x) correspond to x-values where the graph of f is below the graph of g; solutions to f(x) > g(x) correspond to where f is above g.
• Absolute value inequalities: For c > 0, |x| < c means -c < x < c (values within c units of zero); |x| > c means x < -c or x > c (values more than c units away).
• Sign diagram method: For quadratic inequalities, find zeros, divide the number line into intervals, test one value per interval to determine the sign, then choose intervals matching the desired inequality.
• Common confusion: When absolute values contain variables both inside and outside, use the definition of absolute value and break into cases rather than applying standard theorems directly.
• Practical applications: Inequalities model tolerances in manufacturing and other real-world constraints where values must stay within acceptable ranges.

📊 Graphical interpretation of inequalities

📊 How graphs reveal solutions

Graphical Interpretation Principle: The solutions to f(x) = g(x) are the x-values where the graphs of y = f(x) and y = g(x) intersect; f(x) < g(x) where f is below g; f(x) > g(x) where f is above g.

• This follows from the Fundamental Graphing Principle: point (a, b) is on the graph of f if and only if f(a) = b.
• When f(x) = g(x), both functions have the same y-value at that x, so their graphs meet.
• When f(x) < g(x), the y-coordinate on f's graph is smaller than the y-coordinate on g's graph at that x.

📈 Reading solutions from graphs

Example: If f(x) = 2x - 1 and g(x) = 5:

• The graphs intersect at x = 3 (where 2x - 1 = 5).
• For x < 3, the line y = 2x - 1 is below the horizontal line y = 5, so f(x) < g(x) on (-∞, 3).
• For x > 3, the line is above y = 5, so f(x) > g(x) on (3, ∞).

Don't confuse: The inequality symbol direction with graph position—"less than" means the graph is below, not necessarily moving downward.

📏 Absolute value inequalities

📏 Standard forms and their meanings

The excerpt presents these key patterns for c > 0:

Inequality	Equivalent form	Interpretation
|x| < c	-c < x < c	x is within c units of zero
|x| ≤ c	-c ≤ x ≤ c	x is within or at c units of zero
|x| > c	x < -c or x > c	x is more than c units from zero
|x| ≥ c	x ≤ -c or x ≥ c	x is at least c units from zero

🔍 Special cases

• For c ≤ 0: |x| < c has no solution (absolute value is never negative).
• For c < 0: |x| > c is true for all real numbers (absolute value is always non-negative).

🧮 Solving by isolating the absolute value

Example: To solve 4 - 3|2x + 1| > -2:

1. Isolate: -3|2x + 1| > -6
2. Divide (reverse inequality): |2x + 1| < 2
3. Apply theorem: -2 < 2x + 1 < 2
4. Solve: -3/2 < x < 1/2

⚠️ Variables inside and outside absolute value

When variables appear both inside and outside the absolute value (e.g., |x + 1| ≥ (x + 4)/2):

• Do not try to apply standard theorems directly.
• Do use the definition: |x + 1| = -(x + 1) if x < -1, and |x + 1| = x + 1 if x ≥ -1.
• Break into cases based on where the expression inside changes sign.
• Solve each case separately, keeping only solutions valid for that case.

🔢 Quadratic inequalities and sign diagrams

🔢 The sign diagram method

Sign diagram: A visual tool that marks zeros of a function on a number line and indicates whether the function is positive (+) or negative (-) in each interval between zeros.

Steps for solving a quadratic inequality:

1. Rewrite with zero on one side: f(x) compared-to 0
2. Find zeros of f(x) by solving f(x) = 0
3. Place zeros on a number line, dividing it into intervals
4. Choose one test value in each interval
5. Evaluate f at each test value to determine the sign
6. Select intervals matching the inequality's requirement

🧪 Why one test value per interval suffices

Key property of quadratic functions: If the function is positive at one point and negative at another, it must have a zero in between.

• A parabola cannot be above the x-axis at one point and below at another without crossing.
• Therefore, the sign stays constant within each interval between zeros.

📐 Example walkthrough

To solve 2x² ≤ 3 - x:

1. Rewrite: 2x² + x - 3 ≤ 0
2. Factor: (2x + 3)(x - 1) = 0 gives zeros x = -3/2 and x = 1
3. Test intervals: choose x = -2 (gives +), x = 0 (gives -), x = 2 (gives +)
4. Sign diagram shows (-) on (-3/2, 1)
5. Include endpoints where function equals zero: [-3/2, 1]

🔬 Non-factorable quadratics

When the quadratic doesn't factor nicely:

• Use the quadratic formula to find zeros
• Approximate the zeros to help choose test values
• Proceed with the sign diagram as usual

Example: For x² - 2x - 1 > 0, zeros are x = 1 ± √2 ≈ -0.4 and 2.4.

🎯 Special outcomes

• Single zero: If f(x) = x² - 2x + 1 = (x - 1)², the parabola touches the x-axis at exactly one point; f(x) ≥ 0 everywhere, equals zero only at x = 1.
• No real zeros: If f(x) = x² + 1, the parabola never crosses the x-axis; f(x) > 0 for all x, so f(x) < 0 has no solution.

🏭 Applications and tolerances

🏭 Tolerance problems

Tolerance: An acceptable deviation from a target value, expressed as "within d units of c" and modeled by |x - c| ≤ d.

The quantity |x - c| represents the distance from x to c.

• Solving |x - c| ≤ d finds all values within d units of the target c.
• This models manufacturing specifications, measurement accuracy, and quality control.

🔧 Manufacturing example

Problem: A particle board must have area within 0.25 square inches of 576 square inches. How close must the side length be to 24 inches?

Solution approach:

1. Express requirement: |x² - 576| ≤ 0.25
2. Rewrite: -0.25 ≤ x² - 576 ≤ 0.25
3. Add 576: 575.75 ≤ x² ≤ 576.25
4. Take square roots (valid since all values positive): √575.75 ≤ |x| ≤ √576.25
5. Since x represents length (positive), keep: [√575.75, √576.25] ≈ [23.995, 24.005]
6. Interpretation: Side length must be within approximately 0.005 inches of 24 inches.

Key technique: The increasing property of square root—if 0 ≤ a ≤ b, then √a ≤ √b—allows taking square roots across an inequality.

🗺️ Graphing regions defined by inequalities

🗺️ Shading solution regions

Relations defined by inequalities describe regions in the coordinate plane:

• y > f(x): the region above the graph of y = f(x)
• y < f(x): the region below the graph
• y ≥ f(x): the region above and including the curve
• y ≤ f(x): the region below and including the curve

🎨 Graphing conventions

• Use a dotted line for strict inequalities (< or >) to show the boundary is not included
• Use a solid line for inclusive inequalities (≤ or ≥) to show the boundary is included
• Shade the region satisfying the inequality

🔗 Compound inequalities

For relations like {(x, y) : |x| < y ≤ 2 - x²}:

• This combines two conditions: y > |x| AND y ≤ 2 - x²
• Graph both boundary curves
• Shade the region satisfying both conditions (the intersection)
• Find intersection points of the boundaries for accuracy: solve |x| = 2 - x² by cases

Example: The region between y = |x| (dotted, not included) and y = 2 - x² (solid, included) where the parabola is above the absolute value graph.

🧭 Overview

🧠 One-sentence thesis

Regression techniques allow us to find lines or parabolas that best fit real-world data by minimizing the total squared error, enabling us to model trends and make predictions even when data points don't lie perfectly on a curve.

📌 Key points (3–5)

• What regression does: finds a "line of best fit" (or parabola) that is close to all data points by minimizing the total squared error between the data and the model.
• How to measure fit quality: the correlation coefficient r (closer to ±1 means better linear fit) and the coefficient of determination R² both indicate goodness of fit.
• Linear vs quadratic models: use linear regression when data shows a steady increase or decrease; use quadratic regression when data shows a peak or valley (like temperature rising then falling).
• Common confusion: regression models are approximations—they help identify trends but don't guarantee exact predictions, especially far into the future or outside the data range.
• Why it matters: regression turns scattered real-world data into usable mathematical models for forecasting and decision-making (energy demand, population growth, temperature, etc.).

📏 How regression measures closeness

📏 Total squared error

Total squared error E: the sum of the squares of the vertical distances between each data point and the corresponding point on the line (or parabola).

• For each data point, find the point on the line with the same x-coordinate.
• Measure the vertical distance (difference in y-coordinates).
• Square each distance and add them all up.
• Example: Three points {(1, 2), (3, 1), (4, 3)} and the line y = (1/2)x + (1/2). The point (1, 2) corresponds to (1, 1) on the line; the squared difference is (2 − 1)² = 1. Do this for all points and sum.

🎯 Least squares regression line

Least squares regression line (line of best fit): the line that produces the smallest possible total squared error for the data.

• Advanced mathematics (Calculus and Linear Algebra) can find this line exactly.
• Graphing calculators have a built-in feature to compute it.
• The calculator reports the line as y = a·x + b, where a is the slope and b is the y-intercept.

📊 Measuring goodness of fit

📊 Correlation coefficient r

• The value r measures how close the data is to being on the same line.
• The closer |r| is to 1, the better the linear fit.
• Example: If r ≈ 0.327, the data points aren't close to being linear; if r ≈ 0.9696, the fit is very good.

📊 Coefficient of determination R²

• R² is also a measure of goodness of fit.
• It applies to both linear and quadratic regression.
• Closer to 1 means the model explains the data well.

⚠️ Don't confuse

• A low r or R² doesn't mean your math is wrong—it means the data doesn't follow a linear (or quadratic) pattern well.
• The excerpt emphasizes: "regression models are just that, models." They identify trends but don't replace scientific investigation into why data behaves a certain way.

🔢 Linear regression in action

🔢 Energy consumption example

• Data: US energy usage from 1950 to 2000 (in Quads).
• Process: Enter data into calculator, perform linear regression.
• Result: y = 1.287x − 2473.890 with r very close to 1 (good fit).
• Interpreting the slope: 1.287 means energy usage increases at a rate of approximately 1.287 Quads per year.

🔮 Making predictions with the line

• Predict future values: Substitute the year into the equation. For 2013, y = 1.287(2013) − 2473.890 ≈ 116.841 Quads.
• Predict when a threshold is reached: Set y = 120 and solve for x. Result: x ≈ 2015.454, meaning usage will exceed 120 Quads sometime in 2016.

⚠️ Don't confuse

• Predictions assume the trend continues unchanged—real-world factors (policy changes, technology, etc.) can break the model.
• The excerpt warns: models are approximations; investigate why data should be linear before relying on predictions.

📈 Quadratic regression in action

📈 When to use quadratic models

• Use quadratic regression when data shows a turning point: rising then falling, or falling then rising.
• Example: Hourly temperature data for a day—temperature rises in the morning, peaks in the afternoon, then falls.

📈 Temperature example

• Data: Predicted hourly temperatures from 10 AM to 4 PM.
• Observation: Data starts linear but then temperature drops—reminds us of a parabola.
• Process: Use quadratic regression to find a parabola of best fit: y = −0.321x² + 9.464x − 45.857.
• Goodness of fit: R² reasonably close to 1; graph looks like a decent fit.

🔮 Finding the maximum with the vertex

• Goal: Predict the warmest temperature and when it occurs.
• Method: The maximum of a parabola occurs at the vertex. Use the vertex formula: x = −b/(2a).
• For the temperature model: x ≈ 14.741 (roughly 2:45 PM on the 24-hour clock).
• Substitute back into the equation to find y ≈ 23.899°F.

⚠️ Don't confuse

• The model predicted 23.899°F, but the data showed 24°F—models approximate trends, they don't match every data point exactly.
• The excerpt cautions: "regression models are just that, models." Use them to observe trends, but investigate the underlying science.

🧮 The regression process summary

Step	What to do	Tool
1. Enter data	Input (x, y) pairs into calculator	Calculator data entry
2. Choose model type	Linear (steady trend) or quadratic (peak/valley)	Visual inspection of plot
3. Perform regression	Use calculator's built-in regression feature	Linear or Quadratic Regression
4. Check fit	Look at r (or R²) and the graph	Correlation coefficient, visual check
5. Interpret slope (linear) or vertex (quadratic)	Slope = rate of change; vertex = max/min	Formula or calculator output
6. Make predictions	Substitute values into the equation	Algebra

⚠️ Limitations

• Models work best within the range of the data (interpolation).
• Predictions far into the future (extrapolation) or outside the data range may not be reliable.
• The excerpt emphasizes: a more thorough investigation into why data should follow a certain pattern is usually needed—"that, most often, is the business of scientists."

🧭 Overview

🧠 One-sentence thesis

Polynomial functions are smooth, continuous curves whose behavior is determined by their degree, leading term, and the multiplicity of their zeros, allowing us to sketch their graphs by analyzing end behavior and how they cross or touch the x-axis.

📌 Key points (3–5)

• What polynomials are: functions built from powers of x with real number coefficients, having domain of all real numbers.
• End behavior matches the leading term: as x approaches positive or negative infinity, the polynomial behaves like its highest-degree term.
• Multiplicity determines crossing behavior: zeros with odd multiplicity cross through the x-axis; zeros with even multiplicity touch and rebound.
• Common confusion: not every function that looks polynomial-like actually is one—domain restrictions or non-integer exponents disqualify candidates.
• Smoothness and continuity: polynomial graphs have no breaks, holes, or sharp corners, which distinguishes them from other functions and enables the Intermediate Value Theorem.

📐 Definition and structure

📐 What makes a polynomial function

Polynomial function: a function of the form f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀, where a₀, a₁, ..., aₙ are real numbers and n ≥ 1 is a natural number. The domain is all real numbers.

• The subscripts on the coefficients indicate which power of x they belong to.
• Natural number restriction (n = 1, 2, 3, ...) keeps the functions well-behaved.
• Example: f(x) = 4x⁵ - 3x² + 2x - 5 can be rewritten as 4x⁵ + 0x⁴ + 0x³ + (-3)x² + 2x + (-5).

🔍 What disqualifies a function from being a polynomial

Not every algebraic expression is a polynomial:

Expression	Why it's NOT a polynomial
g(x) = (4 + x³)/x	Domain excludes x = 0
f(x) = ³√x = x^(1/3)	Exponent 1/3 is not a natural number
h(x) = |x|	Cannot be written as powers of x; has a sharp corner

• Key test: Can it be written in the standard form with all real numbers in the domain?
• Don't confuse: q(x) = (x³ + 4x)/(x² + 4) simplifies to x with domain of all reals, so it IS a polynomial.

🏷️ Terminology for polynomial parts

Degree: the highest power n (when aₙ ≠ 0).
Leading term: aₙxⁿ.
Leading coefficient: aₙ.
Constant term: a₀.

• Special cases: constant functions (a₀ ≠ 0) have degree 0; f(x) = 0 has no degree.
• Example: For f(x) = 4x⁵ - 3x² + 2x - 5, degree is 5, leading term is 4x⁵, leading coefficient is 4, constant term is -5.
• Tip: You can find the leading term by multiplying the leading terms of factored expressions without expanding everything.

🎢 End behavior

🎢 What end behavior means

End behavior describes what happens to y-values as x moves toward the "ends" of the x-axis:

• As x → -∞ (x becomes a large negative number)
• As x → ∞ (x becomes a large positive number)

Example: For f(x) = x², as x → -∞, f(x) → ∞, and as x → ∞, f(x) → ∞.

📊 End behavior for even-degree polynomials

For f(x) = axⁿ where n is even:

Leading coefficient	As x → -∞	As x → ∞	Graph shape
a > 0	f(x) → ∞	f(x) → ∞	Both ends up
a < 0	f(x) → -∞	f(x) → -∞	Both ends down

• The symmetry comes from the even exponent.
• As the exponent increases (x², x⁴, x⁶), the bottom becomes flatter and the sides become steeper.

📈 End behavior for odd-degree polynomials

For f(x) = axⁿ where n ≥ 3 is odd:

Leading coefficient	As x → -∞	As x → ∞	Graph shape
a > 0	f(x) → -∞	f(x) → ∞	Down on left, up on right
a < 0	f(x) → ∞	f(x) → -∞	Up on left, down on right

• These functions are odd functions (symmetric about the origin).
• Example: y = x³, y = x⁵, y = x⁷ all share this "opposite ends" behavior.

🔑 The leading term determines end behavior

Theorem: The end behavior of any polynomial f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ matches the end behavior of y = aₙxⁿ.

Why this works:

• Factor out the leading term: f(x) = aₙxⁿ(1 + (aₙ₋₁/aₙx) + ... + (a₀/aₙxⁿ))
• As x → ±∞, all terms with x in the denominator approach 0.
• What remains is approximately aₙxⁿ.

Graphically: If you zoom out far enough on a graphing calculator, the polynomial and its leading term become indistinguishable.

🎯 Zeros and multiplicity

🎯 What multiplicity means

Multiplicity: If (x - c)ᵐ is a factor of f(x) but (x - c)^(m+1) is not, then x = c is a zero of multiplicity m.

Example: For f(x) = x³(x - 3)²(x + 2), rewrite as (x - 0)³(x - 3)²(x - (-2))¹:

• x = 0 has multiplicity 3
• x = 3 has multiplicity 2
• x = -2 has multiplicity 1

🔀 How multiplicity affects the graph

Theorem: At a zero x = c with multiplicity m:

• If m is even, the graph touches and rebounds from the x-axis at (c, 0).
• If m is odd, the graph crosses through the x-axis at (c, 0).

Why this happens:

• When you test values on either side of a zero, the corresponding factor changes sign.
• If the factor is raised to an even power, squaring (or raising to any even power) makes it positive on both sides → no sign change in the function → touch and rebound.
• If the factor is raised to an odd power, the sign change is preserved → function changes sign → crosses through.

Example: For f(x) = x³(x - 3)²(x + 2):

• At x = -2 (multiplicity 1, odd): crosses through
• At x = 0 (multiplicity 3, odd): crosses through
• At x = 3 (multiplicity 2, even): touches and rebounds

🧩 Continuity and smoothness

🧩 Why polynomials are special

Polynomials are continuous and smooth everywhere:

• Continuous: no breaks or holes in the graph
• Smooth: no sharp corners or cusps

These properties are inherited from the basic power functions f(x) = xⁿ and preserved when functions are added together.

Don't confuse: f(x) = |x| has a sharp corner at x = 0, so it cannot be a polynomial even though it's continuous there.

🔗 The Intermediate Value Theorem

Intermediate Value Theorem (Zero Version): If f is continuous on an interval containing x = a and x = b (where a < b), and if f(a) and f(b) have different signs, then f has at least one zero between x = a and x = b.

What this means:

• If the graph is above the x-axis at one point and below at another, it must cross the x-axis somewhere in between.
• This seems like common sense, but it's a profound statement about real numbers.

Example application: For f(x) = x² - 2, we have f(1) = -1 and f(3) = 7. Since these have different signs, there's a real number c between 1 and 3 where f(c) = 0. That number is c = √2, proving √2 is real.

📊 Constructing sign diagrams

📊 How to build a sign diagram

Steps:

1. Find all zeros of f and place them on a number line with 0 above them.
2. Choose a test value in each interval between zeros.
3. Determine the sign of f(x) for each test value and write that sign above the interval.

Why this works: The Intermediate Value Theorem guarantees that polynomials are always positive (+) or always negative (-) on intervals that don't contain zeros.

🖼️ Using sign diagrams to sketch graphs

Once you have a sign diagram:

• Where f is (+), the graph is above the x-axis.
• Where f is (-), the graph is below the x-axis.
• The zeros give you x-intercepts.
• Smoothness and continuity let you connect the pieces.

Example: For f(x) = x³(x - 3)²(x + 2)(x² + 1):

• Zeros at x = -2, 0, 3 (note: x² + 1 = 0 has no real solutions)
• Test values at x = -3, -1, 1, 4 give signs (+), (-), (+), (+)
• Combined with end behavior (leading term is x⁸, so both ends up), you can sketch the graph without a calculator.

Note: The sign diagram doesn't tell you how high or low the graph goes—only its position relative to the x-axis.

🧭 Overview

🧠 One-sentence thesis

The Factor Theorem and Remainder Theorem establish that dividing a polynomial by (x − c) yields a remainder equal to p(c), and that c is a zero of the polynomial if and only if (x − c) is a factor, providing a systematic method to find all zeros by repeated division.

📌 Key points (3–5)

• Remainder Theorem: When polynomial p(x) is divided by (x − c), the remainder equals p(c).
• Factor Theorem: A real number c is a zero of polynomial p if and only if (x − c) is a factor of p(x).
• Synthetic division: A streamlined algorithm for dividing polynomials by divisors of the form (x − c), much faster than long division.
• Common confusion: Synthetic division works only for divisors of the form (x − c); for higher-degree divisors, you must use polynomial long division.
• Upper bound on zeros: A polynomial of degree n has at most n real zeros (counting multiplicities).

🔗 The fundamental connection between zeros and factors

🔗 What the Remainder Theorem tells us

Remainder Theorem: Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x − c, the remainder is p(c).

• This means you can evaluate p(c) by performing division instead of direct substitution.
• The proof relies on polynomial division: p(x) = (x − c)q(x) + r, where r is a constant (degree 0).
• Substituting x = c gives p(c) = (c − c)q(c) + r = 0 + r = r.
• Example: To find p(−2) for p(x) = 2x³ − 5x + 3, divide by x − (−2); the remainder in the last box of the synthetic division tableau is p(−2).

🔗 What the Factor Theorem tells us

Factor Theorem: Suppose p is a nonzero polynomial. The real number c is a zero of p if and only if (x − c) is a factor of p(x).

• This is a two-way connection:
  • If (x − c) is a factor, then p(c) = 0.
  • If p(c) = 0, then (x − c) is a factor.
• The proof uses the Remainder Theorem: if c is a zero, then p(c) = 0, so the remainder when dividing by (x − c) is 0, meaning (x − c) divides evenly.
• Example: If x = 2 is a zero of f(x) = x³ + 4x² − 5x − 14, then (x − 2) must be a factor, and we can divide to find f(x) = (x − 2)(x² + 6x + 7).

🔗 Why this matters: finding all zeros

• Once you know one zero c, you can factor out (x − c) and work with a lower-degree polynomial.
• Repeat this process (peeling off one zero at a time) until you reach a quadratic, then use the quadratic formula.
• Example: Starting with a degree-4 polynomial and one known zero, divide once to get degree 3, divide again to get degree 2, then solve the quadratic.

⚙️ Synthetic division: the efficient method

⚙️ Why synthetic division exists

• Polynomial long division works but involves writing many terms and is error-prone.
• Synthetic division is a streamlined tableau method that uses only coefficients.
• It works only for divisors of the form (x − c); for divisors like (x² + 4), you must use long division.

⚙️ How to set up the tableau

1. Write the value c (from the divisor x − c) on the left.
2. Write the coefficients of the dividend polynomial in descending order; insert 0 for any missing degree.
3. Bring down the first coefficient.
4. Multiply c by the brought-down value, write the result under the next coefficient, then add.
5. Repeat: multiply c by the new sum, write under the next coefficient, add, and so on.
6. The last number in the bottom row is the remainder; the other numbers are the coefficients of the quotient polynomial (which has degree one less than the dividend).

Example: To divide x³ + 4x² − 5x − 14 by x − 2, write:

2 | 1   4  −5  −14
  |     2  12   14
  |-----------------
    1   6   7    0

• Quotient: x² + 6x + 7; remainder: 0.

⚙️ Special cases and adjustments

• Divisor x + 2: Rewrite as x − (−2) and use c = −2.
• Divisor 2x − 3: Factor out the leading coefficient: 2x − 3 = 2(x − 3/2). First divide the dividend by 2, then use synthetic division with c = 3/2, then multiply the result by 2.
• Missing terms: Always insert 0 for missing degrees (e.g., x³ + 8 becomes 1, 0, 0, 8).

Don't confuse: Synthetic division is not a shortcut for all polynomial division—only for linear divisors of the form (x − c).

🔢 Multiplicity and repeated division

🔢 What multiplicity means

• A zero c has multiplicity m if (x − c) can be divided out of p(x) exactly m times.
• Example: If c = 1/2 is a zero of multiplicity 2, then (x − 1/2)² is a factor.

🔢 How to handle repeated zeros

• Perform synthetic division once with c to get a quotient polynomial q₁(x).
• Perform synthetic division again on q₁(x) with the same c to get q₂(x).
• Continue until the remainder is no longer 0 or you've divided the expected number of times.
• Example: Given p(x) = 4x⁴ − 4x³ − 11x² + 12x − 3 and told x = 1/2 has multiplicity 2, divide twice by 1/2 to get p(x) = (x − 1/2)²(4x² − 12), then solve 4x² − 12 = 0 to find the remaining zeros x = ±√3.

🔢 Extended tableau

• You can stack multiple synthetic division rows in one tableau.
• Each row uses the quotient coefficients from the row above.
• This is the standard technique for "peeling off" zeros one at a time until you reach a quadratic.

📐 Limits and connections

📐 Upper bound on the number of zeros

Theorem: A polynomial of degree n ≥ 1 has at most n real zeros, counting multiplicities.

• This follows from the Factor Theorem: each zero gives one factor (x − c), and a degree-n polynomial can have at most n such factors.
• Example: A cubic (degree 3) can have at most 3 real zeros; a quartic (degree 4) at most 4.

📐 Equivalent statements about zeros

The following are all equivalent for a polynomial p of degree n ≥ 1 and a real number c:

Statement	Meaning
c is a zero of p	p(c) = 0
x = c is a solution to p(x) = 0	Solving the polynomial equation
(x − c) is a factor of p(x)	p(x) = (x − c)q(x) for some polynomial q
(c, 0) is an x-intercept of y = p(x)	The graph crosses or touches the x-axis at x = c

Don't confuse: These are not separate facts—they are five ways of saying the same thing. Knowing any one tells you all the others are true.

🧭 Overview

🧠 One-sentence thesis

Finding real zeros of polynomials can be accomplished either through mathematical theorems that narrow down candidates for synthetic division or by combining limited theory with graphing technology, though some polynomials resist exact solution entirely.

📌 Key points (3–5)

• Two approaches exist: a technology-assisted method (graphing calculator + some theory) and a purely mathematical method (theorems only).
• Key theorems narrow the search: Cauchy's Bound tells us where all real zeros lie; the Rational Zeros Theorem tells us which rational numbers to test.
• Descartes' Rule of Signs estimates how many positive and negative real zeros exist (counting multiplicity).
• Common confusion: Not all polynomials have rational zeros; some have only irrational real zeros, and some techniques (like u-substitution) only work when the polynomial has exactly three terms with the first exponent exactly twice the second.
• When theorems fail: The Bisection Method can approximate irrational zeros without a calculator, though it is slow; some polynomials provably cannot be solved with algebraic symbols.

🎯 Theorems that locate zeros

📍 Cauchy's Bound – where to look

Cauchy's Bound: For a polynomial f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ of degree n ≥ 1, let M be the largest of the absolute values |a₀|/|aₙ|, |a₁|/|aₙ|, ..., |aₙ₋₁|/|aₙ|. Then all real zeros lie in the interval [−(M + 1), M + 1].

• How it works: Divide the absolute value of each coefficient (except the leading one) by the absolute value of the leading coefficient; take the largest result as M; all real zeros are guaranteed to be between −(M + 1) and (M + 1).
• Why it matters: This gives a finite search interval instead of the entire real line.
• Example: For f(x) = 2x⁴ + 4x³ − x² − 6x − 3, the largest coefficient in absolute value (other than the leading 2) is |−6| = 6. So M = 6/2 = 3, and all real zeros lie in [−4, 4].

🔢 Rational Zeros Theorem – which candidates to test

Rational Zeros Theorem: For a polynomial f(x) = aₙxⁿ + ... + a₁x + a₀ with integer coefficients, if r is a rational zero, then r = ±p/q where p is a factor of the constant term a₀ and q is a factor of the leading coefficient aₙ.

• How to use it: List all factors of the constant term; list all factors of the leading coefficient; form all fractions ±(factor of a₀)/(factor of aₙ).
• Why it works: The proof uses the fact that if p/q (in lowest terms) is a zero, then multiplying the polynomial equation by qⁿ and rearranging shows that aₙ must be a multiple of q and a₀ must be a multiple of p.
• Example: For f(x) = 2x⁴ + 4x³ − x² − 6x − 3, factors of −3 are ±1, ±3; factors of 2 are 1, 2. Possible rational zeros: ±1/2, ±1, ±3/2, ±3.
• Don't confuse: This theorem only lists possible rational zeros; if none work, the polynomial may have irrational or no real zeros.

🔍 Theorems that count zeros

🔀 Descartes' Rule of Signs

Descartes' Rule of Signs: If P is the number of sign changes in f(x) (written with descending powers), the number of positive real zeros (counting multiplicity) is P, P−2, P−4, ... If N is the number of sign changes in f(−x), the number of negative real zeros is N, N−2, N−4, ...

• What "sign change" means: Reading coefficients left to right, count how many times the sign switches from + to − or − to +; ignore zero coefficients.
• Counting multiplicity: A zero of multiplicity 2 counts as two zeros for this rule.
• Example: f(x) = 2x⁴ + 4x³ − x² − 6x − 3 has signs (+, +, −, −, −), so one sign change → exactly 1 positive real zero. For f(−x) = 2x⁴ − 4x³ − x² + 6x − 3, signs are (+, −, −, +, −), three changes → either 3 or 1 negative real zeros.
• Why the even decrements: The theorem always reduces by 2, so if there are 7 sign changes, possible counts are 7, 5, 3, or 1 positive zeros.

📏 Upper and Lower Bounds Theorem

Upper and Lower Bounds Theorem: If c > 0 is synthetically divided into f and all numbers in the final line have the same sign, then c is an upper bound (no real zero is greater than c). If c < 0 is divided in and the final line alternates signs, then c is a lower bound (no real zero is less than c).

• Treat zero flexibly: A 0 in the final line can be treated as + or − as needed to establish the pattern.
• Why it works: The theorem relies on the division algorithm f(x) = (x − c)q(x) + r; if all coefficients of q and r are nonnegative and c > 0, then for any b > c, f(b) > 0, so b cannot be a zero.
• Practical use: Once you find an upper bound, stop testing larger candidates; once you find a lower bound, stop testing smaller candidates.

🛠️ Solution techniques

🖩 Technology-assisted approach

Steps:

1. Use Cauchy's Bound to set a viewing window on the graphing calculator.
2. Use the Rational Zeros Theorem to list candidates.
3. Graph y = f(x) and visually eliminate candidates that clearly don't match the graph.
4. Use synthetic division on the remaining candidates.
5. Once the quotient is quadratic, apply the quadratic formula if needed.

Advantages: Faster; the graph provides visual confirmation and helps identify multiplicity (crossing vs. touching the x-axis).

Caution: Calculators can miss features like hidden relative extrema; mathematical analysis (finding exact zeros and multiplicities) can reveal behavior the calculator doesn't show clearly.

🧮 Pure-mathematics approach

Additional tools beyond the technology approach:

• Descartes' Rule of Signs: Tells you how many positive/negative zeros to expect, so you know when to stop searching.
• Upper and Lower Bounds Theorem: Lets you stop testing once you've bracketed all zeros.

Example workflow: For f(x) = 2x⁴ + 4x³ − x² − 6x − 3, Descartes guarantees exactly 1 positive zero and either 3 or 1 negative zeros. Testing candidates with synthetic division, you find −1 is a zero; testing −1 again succeeds (multiplicity 2). The quotient 2x² − 3 = 0 gives x = ±√(3/2) = ±√6/2. Since Descartes predicted 1 positive zero, √6/2 has multiplicity 1; since the total for negatives is 3 and we used multiplicity 2 for −1, the remaining negative zero −√6/2 has multiplicity 1.

🔄 u-substitution for "quadratics in disguise"

When it applies: Exactly three terms, and the exponent on the first term is exactly twice the exponent on the second.

Example: x⁴ + x² − 12 = 0 can be rewritten as (x²)² + (x²) − 12 = 0. Let u = x²; then u² + u − 12 = 0 factors as (u + 4)(u − 3) = 0. Back-substitute: x² = −4 (no real solutions) or x² = 3, so x = ±√3.

Don't confuse: This only works for the specific pattern; x⁵ − x − 1 has three terms but the exponents (5, 1, 0) don't fit the pattern.

🔁 When exact methods fail

🪓 The Bisection Method

Purpose: Approximate an irrational zero without a calculator.

How it works:

1. Find an interval [a, b] where f changes sign (f(a) and f(b) have opposite signs); by the Intermediate Value Theorem, a zero lies between a and b.
2. Compute the midpoint m = (a + b)/2 and evaluate f(m).
3. Determine which half-interval [a, m] or [m, b] contains the sign change.
4. Repeat, halving the interval each time, until the interval is smaller than your desired accuracy.

Characteristics: Slow and tedious, but guaranteed to work (fool-proof).

Example: For f(x) = x⁵ − x − 1, f(1) = −1 and f(2) = 29, so a zero is in [1, 2]. Midpoint 1.5: f(1.5) ≈ 5.09 > 0, so zero is in [1, 1.5]. Midpoint 1.25: f(1.25) ≈ 0.80 > 0, so zero is in [1, 1.25]. Continue until the interval is sufficiently narrow.

🚫 Polynomials with no algebraic solution

• Some polynomials, like f(x) = x⁵ − x − 1, have real zeros that cannot be expressed using radicals and standard algebraic operations.
• In such cases, approximation (Bisection Method or calculator) is the only option.

📐 Applications to equations and inequalities

⚖️ Solving polynomial equations

Key idea: Solving f(x) = g(x) is the same as finding the real zeros of f(x) − g(x).

Example: To solve 2x⁵ + 6x³ + 3 = 3x⁴ + 8x², rewrite as 2x⁵ − 3x⁴ + 6x³ − 8x² + 3 = 0 and find the zeros of p(x) = 2x⁵ − 3x⁴ + 6x³ − 8x² + 3.

📊 Solving polynomial inequalities

Steps:

1. Rewrite the inequality with 0 on one side.
2. Find all real zeros of the polynomial.
3. Construct a sign diagram: test values in each interval between zeros to determine where the polynomial is positive or negative.
4. Read off the solution from the sign diagram.

Multiplicity matters: Zeros of odd multiplicity correspond to the graph crossing the x-axis (sign change); zeros of even multiplicity correspond to the graph touching and rebounding (no sign change).

Example: To solve p(x) ≤ 0 where p(x) = 2x⁵ − 3x⁴ + 6x³ − 8x² + 3, find zeros x = −1/2 and x = 1. Test values show p(x) < 0 on (−∞, −1/2) and p(x) > 0 elsewhere. Since we want p(x) ≤ 0, the solution is (−∞, −1/2] ∪ {1}.

🎬 Graphical interpretation

• The solution to f(x) ≤ g(x) consists of x-values where the graph of f is below the graph of g, plus x-values where the graphs intersect.
• Graphing both functions can confirm the analytical solution and reveal where one function dominates the other.

---

Note: The excerpt includes extensive worked examples and exercise sets that illustrate these concepts in detail, along with calculator screenshots showing graphical verification of analytical results.

🧭 Overview

🧠 One-sentence thesis

Every polynomial of degree n ≥ 1 has exactly n complex zeros (counting multiplicity), which means polynomials can always be completely factored over the complex numbers even when they have no real zeros.

📌 Key points (3–5)

• The imaginary unit i: defined by i² = −1, allowing us to find zeros of polynomials like x² + 1 = 0.
• Complex numbers include real numbers: every number of the form a + bi (where a, b are real) is complex; when b = 0, we get the real numbers.
• Fundamental Theorem of Algebra: guarantees at least one complex zero for every polynomial of degree n ≥ 1 with complex coefficients.
• Conjugate Pairs Theorem: for polynomials with real coefficients, nonreal zeros always come in conjugate pairs (if a + bi is a zero, so is a − bi).
• Common confusion: don't apply general radical rules to even roots of negative numbers; √(−3)√(−12) ≠ √((−3)(−12)).

🔢 The imaginary unit and complex numbers

🔢 What is i?

Imaginary unit i: satisfies two properties:

1. i² = −1
2. If c is a real number with c ≥ 0, then √(−c) = i√c

• The number i acts as a square root of −1, which has no real solution.
• It behaves like other radical expressions in arithmetic: 3(2i) = 6i, 7i − 3i = 4i, etc.
• Important restriction: Property 2 only works when c ≥ 0. You cannot write √(−(−4)) = i√(−4), or you'd get 2 = −2.

🔢 Complex numbers defined

Complex number: a number of the form a + bi, where a and b are real numbers and i is the imaginary unit.

• Examples: 3 + 2i, (2/5) − i√3, 3i, 6, 0, π − √21.
• Don't forget: when a = 0 or b = 0, we still have complex numbers. Real numbers are the special case where b = 0.
• The excerpt emphasizes that complex numbers include the real numbers.

➕ Arithmetic with complex numbers

• Treat i like any other radical, but remember i² = −1.
• Addition/subtraction: combine like terms. Example: (1 − 2i) − (3 + 4i) = −2 − 6i.
• Multiplication: use distributive property, then replace i² with −1. Example: (1 − 2i)(3 + 4i) = 3 + 4i − 6i − 8i² = 3 − 2i + 8 = 11 − 2i.
• Division: multiply numerator and denominator by the conjugate of the denominator. Example: (1 − 2i)/(3 − 4i) · (3 + 4i)/(3 + 4i) = (11 − 2i)/25 = 11/25 − (2/25)i.

🔄 Complex conjugate

Conjugate of a + bi: the number a − bi, denoted with a bar: a + bi = a − bi.

• Examples: 3 + 2i = 3 − 2i, 6 = 6, 4i = −4i.
• Key properties (Theorem 3.12):
  • z = z (conjugate of conjugate is the original)
  • z + w = z + w (conjugation distributes over addition)
  • zw = z·w (conjugation distributes over multiplication)
  • (z)ⁿ = (z̄)ⁿ (conjugation works with powers)
  • z is real if and only if z = z̄
• These properties mean conjugation "plays nicely" with arithmetic operations.

⚠️ Radical rule caution

• Don't confuse: √(−3)√(−12) vs. √((−3)(−12)).
• Correct: √(−3)√(−12) = (i√3)(i√12) = i²√36 = −6.
• Also correct: √((−3)(−12)) = √36 = 6.
• The general radical property √a·√b = √(ab) does not apply to even roots of negative numbers.
• Reason: we want to solve x² = −1 without breaking established algebraic rules.

🎯 The Fundamental Theorem of Algebra

🎯 Existence of zeros

Fundamental Theorem of Algebra: If f is a polynomial function with complex coefficients of degree n ≥ 1, then f has at least one complex zero.

• This is an existence theorem: it guarantees a zero exists but doesn't tell you how to find it.
• Similar to the Intermediate Value Theorem: it assures you something is there, but provides no algorithm.
• Applies to polynomials with any complex coefficients, not just real ones.

🎯 Why it matters

• Before this theorem, we had polynomials with no real zeros (like x² + 1).
• Now we know: every polynomial has some zero in the complex numbers.
• This gives a sense of "completeness" or "closure" to polynomial theory.
• Historical note: it took mathematicians hundreds of years to prove this theorem in full generality.

🏗️ Complete factorization over complex numbers

🏗️ Complex Factorization Theorem

Complex Factorization Theorem: If f is a polynomial with complex coefficients of degree n ≥ 1, then f has exactly n complex zeros (counting multiplicity). If z₁, z₂, …, zₖ are the distinct zeros with multiplicities m₁, m₂, …, mₖ, then
f(x) = a(x − z₁)^(m₁)(x − z₂)^(m₂)···(x − zₖ)^(mₖ).

• The constant a is the leading coefficient of f(x).
• How we get this: apply the Fundamental Theorem repeatedly. Each zero gives a factor by the Factor Theorem; repeat n times until the quotient is a constant.
• A polynomial is completely determined by its zeros, their multiplicities, and its leading coefficient.

🏗️ Factoring example

• Example from excerpt: f(x) = 12x⁵ − 20x⁴ + 19x³ − 6x² − 2x + 1.
• After synthetic division, zeros are: x = 1/2 (multiplicity 2), x = −1/3, x = (1 + i√3)/2, x = (1 − i√3)/2.
• Factored form: f(x) = 12(x − 1/2)²(x + 1/3)(x − [1 + i√3)/2])(x − [1 − i√3)/2]).
• This is the complete factorization over the complex numbers.

🏗️ Factoring over the reals

• If we want to factor over the real numbers only, we stop at irreducible quadratics.
• Same example: f(x) = 12(x − 1/2)²(x + 1/3)(x² − x + 1).
• The quadratic x² − x + 1 has nonreal zeros, so it's irreducible over the reals (cannot be factored further using real numbers).

🔗 Conjugate Pairs and real polynomials

🔗 Conjugate Pairs Theorem

Conjugate Pairs Theorem: If f is a polynomial with real coefficients and z is a zero of f, then z̄ (the conjugate of z) is also a zero of f.

• What this means: nonreal zeros of real-coefficient polynomials come in conjugate pairs.
• Example: if 2 + 3i is a zero, then 2 − 3i is also a zero.
• Why it's true: the proof uses properties of conjugation. If f(z) = 0, then f(z̄) = f(z) = 0̄ = 0 (because coefficients are real, so they equal their own conjugates).

🔗 Irreducible quadratic factors

• When a + bi and a − bi are both zeros, the Factor Theorem gives us (x − [a + bi]) and (x − [a − bi]) as factors.
• Multiplying these: (x − [a + bi])(x − [a − bi]) = x² − 2ax + (a² + b²), which is a quadratic with real coefficients.
• This quadratic is irreducible over the reals (cannot be factored further with real numbers).

🔗 Real Factorization Theorem

Real Factorization Theorem: A polynomial f with real coefficients can be factored into a product of linear factors (corresponding to real zeros) and irreducible quadratic factors (giving the nonreal zeros).

• This tells us that even if we can't factor completely over the reals, we can always get down to linear and quadratic factors.
• Example: f(x) = x⁴ + 64 factors as (x² − 4x + 8)(x² + 4x + 8) over the reals, or as (x − [2 − 2i])(x − [2 + 2i])(x − [−2 + 2i])(x − [−2 − 2i]) over the complex numbers.

🛠️ Finding and using complex zeros

🛠️ Using the Quadratic Formula

• For quadratics that don't factor nicely, use the Quadratic Formula.
• Example: x² − 2x + 5 = 0 gives x = (2 ± √(4 − 20))/2 = (2 ± √(−16))/2 = (2 ± 4i)/2 = 1 ± 2i.
• The zeros 1 + 2i and 1 − 2i are conjugates, as expected for a real-coefficient polynomial.

🛠️ Synthetic division with complex zeros

• Synthetic division works the same way for complex numbers as for real numbers.
• The Factor Theorem and Remainder Theorem still hold.
• Example: to verify x = 2 + 2i is a zero of f(x) = x⁴ + 64, perform synthetic division and check that the remainder is 0.

🛠️ Strategy for finding all zeros

1. Use techniques from Section 3.3 (Rational Root Theorem, synthetic division) to find some zeros.
2. If you find a nonreal zero z of a real-coefficient polynomial, you automatically know z̄ is also a zero.
3. Divide by both (x − z) and (x − z̄) to reduce the degree.
4. Repeat until the quotient is quadratic, then use the Quadratic Formula if needed.

🛠️ Constructing polynomials from zeros

• Given zeros and other conditions, use the Complex Factorization Theorem in reverse.
• Example: if x = 1/3 is a zero of even multiplicity (lowest degree → multiplicity 2), and x = 3i is a zero, then x = −3i is also a zero (conjugate pair).
• Factor form: f(x) = a(x − 1/3)²(x − 3i)(x + 3i).
• Expand and choose a to satisfy other conditions (like leading coefficient or end behavior).

📐 Practical considerations and scope

📐 When to use complex vs. real factorization

Context	Factorization type	Example
Theoretical completeness	Over complex numbers	f(x) = (x − [1+2i])(x − [1−2i])
Applied problems, Calculus prep	Over real numbers	f(x) = x² − 2x + 5 (irreducible quadratic)

• For most applications in early Calculus, we work with real variables, so irreducible quadratics are acceptable.
• Example: f(x) = 1/(x² + 1) has domain "all real numbers" because x² + 1 > 0 for all real x, even though x² + 1 = 0 has complex solutions x = ±i.

📐 Applications of complex numbers

• The excerpt notes that complex numbers are used in fluid dynamics, electromagnetism, quantum mechanics, and electronics.
• However, these applications require mathematics beyond College Algebra.
• The text restricts attention to real numbers for the remainder (except Section 11.7), to prepare students for the standard Calculus sequence.
• Don't confuse: "imaginary" doesn't mean "useless in reality"; it's a historical name, and complex numbers have real-world applications.

📐 Multiplicity and graph behavior

• Even multiplicity at a zero: the graph touches but doesn't cross the x-axis.
• Odd multiplicity: the graph crosses the x-axis.
• Example: if (1/3, 0) is a local minimum and the graph touches the x-axis there, then x = 1/3 has even multiplicity (lowest degree → multiplicity 2).

🧭 Overview

🧠 One-sentence thesis

Rational functions—ratios of polynomials—exhibit distinctive behaviors including vertical asymptotes (where the function becomes unbounded), horizontal or slant asymptotes (describing end behavior), and holes (removable discontinuities), all determined by analyzing the degrees and factorizations of numerator and denominator.

📌 Key points (3–5)

• What rational functions are: functions formed by dividing one polynomial by another, written as r(x) = p(x)/q(x).
• Domain restrictions: values that make the denominator zero are excluded from the domain.
• Three types of asymptotes: vertical (where function becomes unbounded), horizontal (constant end behavior), and slant (linear end behavior).
• Common confusion—asymptote vs hole: both occur where the denominator is zero, but a hole appears when that factor cancels with the numerator; a vertical asymptote appears when it doesn't cancel.
• Degree comparison determines end behavior: comparing numerator and denominator degrees reveals whether the graph has horizontal asymptotes, slant asymptotes, or neither.

📐 Definition and domain

📐 What is a rational function

Rational function: a function that is the ratio of polynomial functions; r(x) = p(x)/q(x) where p and q are polynomials.

• All polynomial functions are also rational functions (set q(x) = 1).
• The key feature: division of polynomials, which may produce non-polynomial results.

🚫 Finding the domain

• The rule: exclude any x-values that make the denominator equal to zero.
• Method: solve q(x) = 0 and exclude those solutions from the domain.
• Example: for f(x) = (2x - 1)/(x + 1), solve x + 1 = 0 to get x = -1, so domain is (-∞, -1) ∪ (-1, ∞).

✏️ Simplifying rational expressions

• Combine terms over common denominators when adding or subtracting.
• Factor numerator and denominator completely.
• Cancel common factors—but remember: the domain is determined before canceling.
• Don't confuse: two functions with the same simplified formula may be different functions if their domains differ.

📏 Vertical asymptotes and holes

📏 Vertical asymptotes defined

Vertical asymptote: the line x = c is a vertical asymptote if as x approaches c from left or right, f(x) approaches positive or negative infinity.

• Graphically: the graph shoots up or down near x = c, resembling a vertical line.
• Notation examples:
  • As x → c⁻, f(x) → ∞ (approaching from left, function goes to positive infinity)
  • As x → c⁺, f(x) → -∞ (approaching from right, function goes to negative infinity)

🕳️ Holes in the graph

• A hole (or removable discontinuity) appears at x = c when:
  • x = c is not in the domain
  • But after simplifying, x = c no longer makes the denominator zero
• Graphically: represented by an open circle at the point where the function "should" be.
• Example: h(x) = (2x² - 1)/(x² - 1) - (3x - 2)/(x² - 1) simplifies to (2x - 1)/(x + 1), creating a hole at x = 1.

🔍 How to distinguish asymptotes from holes

Theorem (Location of Vertical Asymptotes and Holes): Suppose r(x) = p(x)/q(x) is in lowest terms (no common factors), and c is not in the domain:

Condition	Result
q(c) ≠ 0	Hole at (c, p(c)/q(c))
q(c) = 0	Vertical asymptote at x = c

• In plain language: if a factor in the denominator cancels with the numerator, you get a hole; if it doesn't cancel, you get a vertical asymptote.
• Process:
  1. Factor completely
  2. Cancel common factors
  3. Check what still makes the denominator zero → those are vertical asymptotes
  4. Check what made the original denominator zero but now doesn't → those are holes

🧮 Analyzing behavior near vertical asymptotes

• Make tables of values approaching from left (x → c⁻) and right (x → c⁺).
• Determine whether f(x) → ∞ or f(x) → -∞ for each direction.
• Example: for f(x) = (2x - 1)/(x + 1) near x = -1:
  • From left: as x → -1⁻, f(x) → ∞
  • From right: as x → -1⁺, f(x) → -∞

↔️ Horizontal asymptotes

↔️ Horizontal asymptotes defined

Horizontal asymptote: the line y = c is a horizontal asymptote if as x approaches positive or negative infinity, f(x) approaches c.

• Describes the "leveling off" behavior as x becomes very large (positive or negative).
• A rational function can have at most one horizontal asymptote.
• The graph may actually cross a horizontal asymptote (unlike the informal idea that asymptotes are never crossed).

📊 Determining horizontal asymptotes by degree

Theorem (Location of Horizontal Asymptotes): For r(x) = p(x)/q(x) where p has degree n with leading coefficient a, and q has degree m with leading coefficient b:

Degree comparison	Horizontal asymptote
n = m (same degree)	y = a/b
n < m (numerator smaller)	y = 0
n > m (numerator larger)	None

🔬 Why this works

• Intuition: as x becomes very large, only the highest-degree terms matter.
• For same degrees: r(x) ≈ (ax^n)/(bx^n) = a/b
• For numerator smaller: r(x) ≈ (ax^n)/(bx^m) where m > n, so the denominator dominates and r(x) → 0
• For numerator larger: the numerator dominates and the function becomes unbounded

📈 Behavior notation

• As x → -∞, f(x) → c⁺ means approaching c from above
• As x → ∞, f(x) → c⁻ means approaching c from below
• Example: f(x) = (2x - 1)/(x + 1) has horizontal asymptote y = 2 because degrees are equal (both 1) and leading coefficients give 2/1 = 2

⤴️ Slant asymptotes

⤴️ Slant asymptotes defined

Slant (oblique) asymptote: the line y = mx + b (where m ≠ 0) is a slant asymptote if as x approaches positive or negative infinity, the graph resembles this non-horizontal line.

• Occurs when the graph "levels off" to follow a slanted line rather than a horizontal line.
• Notation: as x → ±∞, f(x) → mx + b
• Alternative interpretation: f(x) - (mx + b) → 0 as x → ±∞

📐 When slant asymptotes occur

Theorem (Determination of Slant Asymptotes): If r(x) = p(x)/q(x) and the degree of p is exactly one more than the degree of q, then the graph has a slant asymptote y = L(x), where L(x) is the quotient from dividing p(x) by q(x).

Degree of numerator vs denominator	Type of asymptote
Exactly one more	Slant asymptote
Same	Horizontal asymptote
Less than	Horizontal asymptote y = 0
Two or more greater	No asymptote

🧮 Finding slant asymptotes

• Method: perform polynomial long division of p(x) ÷ q(x)
• The quotient (ignoring remainder) is the slant asymptote
• Example: g(x) = (x² - 4)/(x + 1)
  • Degree of numerator (2) is one more than degree of denominator (1)
  • Long division gives quotient x - 1
  • Slant asymptote: y = x - 1
• Why it works: g(x) = (x - 1) - 3/(x + 1), and as x → ±∞, the term 3/(x + 1) → 0

🎯 Behavior near slant asymptotes

• The graph approaches the slant asymptote from above or below
• Make tables or use a calculator to determine which
• Example: for h(x) = (x³ + 1)/(x² - 4) with slant asymptote y = x:
  • As x → -∞, graph approaches from below
  • As x → ∞, graph approaches from above

🌍 Real-world applications

🦠 Doomsday equations

• Model populations that grow without bound at a finite time
• Example: P(t) = 100/(5 - t)² for bacteria population (in thousands) where 0 ≤ t < 5
  • As t → 5⁻, P(t) → ∞ (vertical asymptote at t = 5)
  • Interpretation: population becomes infinite at t = 5 days, which is impossible—the environment collapses before then
  • Called "doomsday" because the model predicts catastrophic failure at a specific time

📚 Learning curves

• Rational functions can model skill acquisition over time
• Example: typing speed Y as a function of practice pages X
• Horizontal asymptotes represent the maximum achievable skill level
• As practice → ∞, performance → maximum limit

💰 Cost models

• Average cost functions: C̄(x) = C(x)/x where C(x) is total cost
• Horizontal asymptotes represent the variable cost per unit as production increases
• Vertical asymptotes at x = 0 relate to fixed costs
• Example: if C(x) = 100x + 2000, then C̄(x) = 100 + 2000/x
  • As x → ∞, C̄(x) → 100 (the variable cost per unit)
  • As x → 0⁺, C̄(x) → ∞ (fixed costs dominate when few units are made)

🐟 Removal/remediation problems

• Cost to remove p% of invasive species: C(p) = 1770p/(100 - p)
• Vertical asymptote at p = 100: cost → ∞ as you try to remove 100%
• Interpretation: impossible to remove every last individual; cost becomes prohibitive

🔧 Analysis techniques

🔧 Complete analysis procedure

For a rational function r(x) = p(x)/q(x):

1. Domain: solve q(x) = 0 and exclude those values
2. Simplify: factor completely and cancel common factors (but remember original domain)
3. Vertical asymptotes and holes:
   • Values that made original denominator zero but cancel → holes
   • Values that still make simplified denominator zero → vertical asymptotes
4. Horizontal or slant asymptotes:
   • Compare degrees of numerator and denominator
   • Apply appropriate theorem
5. Behavior analysis:
   • Near vertical asymptotes: make tables or analyze signs
   • Near horizontal/slant asymptotes: make tables for large |x| values
6. Verify with graphing calculator

📊 Using tables effectively

• For vertical asymptotes: approach from both sides (x → c⁻ and x → c⁺)
• For horizontal asymptotes: use increasingly large positive and negative x values
• Look for patterns: values increasing without bound, approaching a constant, etc.
• Example table structure for x → c⁻:

x	f(x)	Interpretation
c - 0.1	value
c - 0.01	value	Pattern emerging
c - 0.001	value
c - 0.0001	value	Conclusion

⚠️ Common pitfalls

• Don't confuse: simplified formula vs. original domain—always determine domain before simplifying
• Don't assume: graphs never cross horizontal asymptotes—they can, just not infinitely often
• Remember: a rational function has at most one horizontal asymptote, but can have multiple vertical asymptotes
• Check carefully: whether a denominator factor cancels (hole) or doesn't (asymptote)

🧭 Overview

🧠 One-sentence thesis

Graphing rational functions requires a systematic six-step procedure that identifies domain restrictions, asymptotes, holes, intercepts, and end behavior to reveal how these functions can change sign at excluded domain values without crossing the x-axis.

📌 Key points (3–5)

• Sign diagrams for rational functions: Unlike polynomials, rational functions can change sign at values excluded from their domain (vertical asymptotes and holes) without crossing the x-axis, because they are not continuous at those points.
• Six-step graphing procedure: Find domain → reduce to lowest terms → find intercepts → locate vertical asymptotes/holes → analyze end behavior (horizontal/slant asymptotes) → construct sign diagram and sketch.
• Detailed asymptote analysis: Use "number sense" to determine exactly how the graph approaches vertical and horizontal/slant asymptotes (from above/below, toward ±∞).
• Common confusion: Graphs can cross horizontal asymptotes (unlike the myth); reduced formulas reveal when this happens and help analyze end behavior more precisely.
• The interrobang symbol (‽): Used on sign diagrams to mark domain exclusions (vertical asymptotes and holes), distinct from zeros marked with "0".

🔍 Understanding sign changes in rational functions

🔍 Why rational functions are different

Rational functions are continuous on their domains, but vertical asymptotes and holes occur at values excluded from their domains—where the functions are not continuous.

• The Intermediate Value Theorem applies only to continuous functions.
• Polynomials are continuous everywhere, so they can only change sign by crossing the x-axis.
• Rational functions can "jump" across the x-axis at vertical asymptotes or holes without actually crossing it.
• Example: A graph might be positive on one side of a vertical asymptote and negative on the other, even though it never touches the x-axis at that point.

🔍 Constructing sign diagrams with domain exclusions

The sign diagram must include both zeros and excluded values:

Mark on diagram	Meaning	Symbol
Zeros of the function	Where f(x) = 0	0
Excluded domain values	Vertical asymptotes or holes	‽ (interrobang)

• Place ‽ above vertical asymptotes and holes to signal "caution and wonderment."
• Choose test values in each interval created by these marks.
• Determine the sign of the function in each interval.

📋 The six-step graphing procedure

📋 Step 1: Find the domain

• Set the denominator equal to zero and solve.
• Domain excludes all values that make the denominator zero.
• Express as a union of intervals.

📋 Step 2: Reduce to lowest terms

• Factor both numerator and denominator completely.
• Cancel common factors if any exist.
• Important: Write the reduced formula with the restriction "x ≠ [canceled value]" to remember the hole location.
• Example: If (x + 1) cancels, write the reduced formula with "x ≠ -1."

📋 Step 3: Find intercepts

• x-intercepts: Set the reduced numerator equal to zero; only include solutions in the domain.
• y-intercept: Evaluate f(0) if 0 is in the domain.
• Don't confuse: A zero that was canceled during reduction creates a hole, not an x-intercept.

📋 Step 4: Vertical asymptotes and holes

• Vertical asymptotes: Values excluded from the domain that remain in the reduced denominator.
• Holes: Values excluded from the domain that were canceled during reduction; find the y-coordinate by substituting into the reduced formula.
• Analyze behavior near each vertical asymptote using "number sense."

📋 Step 5: End behavior and horizontal/slant asymptotes

Compare degrees of numerator (n) and denominator (d):

Condition	Asymptote type	How to find
n < d	Horizontal: y = 0	The x-axis
n = d	Horizontal: y = ratio of leading coefficients	Divide leading terms
n = d + 1	Slant (oblique)	Perform long division; quotient is the asymptote
n > d + 1	None of the above	Analyze using long division remainder

• Use long division to rewrite the function and analyze the remainder term for detailed end behavior.
• Determine whether the graph approaches the asymptote from above or below as x → ±∞.

📋 Step 6: Sign diagram and final sketch

• Mark all zeros (0) and excluded values (‽) on a number line.
• Test the sign in each interval.
• Combine with asymptote analysis to sketch the complete graph.

🧮 "Number sense" for asymptote analysis

🧮 Analyzing behavior near vertical asymptotes

Use mental estimation rather than tables:

• Imagine substituting a value very close to the asymptote (e.g., x = -2.000001 for asymptote at x = -2).
• Evaluate each factor:
  • Factors not involving the asymptote value: approximate with their actual value at the asymptote.
  • The factor causing the asymptote: becomes "very small (+)" or "very small (-)."
• Simplify mentally: "very small" in the denominator creates "very big" in the result.
• Example: If f(x) ≈ 6 / (very small (-)), then f(x) ≈ very big (-), meaning f(x) → -∞.

🧮 Analyzing end behavior

For x → ±∞, use similar mental estimation:

• Substitute a very large number (e.g., x = 1 billion).
• In polynomials, the highest-degree term dominates; lower-degree terms become insignificant.
• Example: x² - 4 ≈ x² when x is very large.
• Determine whether the function approaches the asymptote from above (slightly positive) or below (slightly negative).

🧮 Using long division for detailed analysis

When the function can be rewritten as f(x) = quotient + remainder/denominator:

• The quotient gives the asymptote (horizontal if constant, slant if linear).
• The remainder term shows how the graph approaches the asymptote.
• As x → ±∞, the remainder term → 0, but its sign tells whether the graph is above or below the asymptote.
• Don't confuse: The graph can cross a horizontal or slant asymptote at finite x-values; asymptotes describe only end behavior.

🎯 Special cases and subtleties

🎯 Graphs crossing horizontal asymptotes

Myth: Graphs of rational functions never cross their horizontal asymptotes.

Reality: Graphs can and do cross horizontal asymptotes at finite x-values.

• Horizontal asymptotes describe behavior as x → ±∞ only.
• To find crossing points: Set f(x) equal to the asymptote value and solve.
• Example: If f(x) = 2 - (x-7)/(x²-x-6), the graph crosses y = 2 when (x-7)/(x²-x-6) = 0, i.e., at x = 7.

🎯 Holes in the graph

• Occur when a factor cancels from both numerator and denominator.
• Mark the x-value with ‽ on the sign diagram.
• Find the y-coordinate by substituting into the reduced formula.
• The graph approaches the hole from both sides but has a "missing point" there.
• Analyze behavior near the hole using the same "number sense" technique as for vertical asymptotes.

🎯 When standard theorems don't apply

If the numerator degree exceeds the denominator degree by more than 1:

• No horizontal or slant asymptote exists.
• Use long division to rewrite as polynomial + remainder/denominator.
• The graph approaches the polynomial curve (not a line) as x → ±∞.
• Example: f(x) = (x⁴ + 1)/(x² + 1) = x² - 1 + 2/(x² + 1); the graph approaches the parabola y = x² - 1.

🎯 Functions with no domain restrictions

• If the denominator has no real zeros (e.g., x² + 1), the domain is all real numbers.
• No vertical asymptotes or holes exist.
• The graphing procedure simplifies, but calculator help may be needed to find relative extrema.
• Check for symmetry: even functions (f(-x) = f(x)) are symmetric about the y-axis; odd functions (f(-x) = -f(x)) are symmetric about the origin.

⚠️ Common pitfalls and reminders

⚠️ Don't skip reduction

• Always factor and reduce to lowest terms before analyzing.
• Failing to reduce means you'll miss holes and misidentify vertical asymptotes.

⚠️ Remember domain restrictions after reduction

• Even after canceling a factor, that x-value is still excluded from the domain.
• Write the reduced formula with an explicit restriction (e.g., "x ≠ -1").

⚠️ Use the correct formula at each step

• For intercepts and asymptote analysis: use the reduced formula.
• For finding hole coordinates: substitute into the reduced formula.
• For sign diagrams: the reduced formula determines the sign (except at excluded values).

⚠️ Asymptote analysis is not optional

• The six-step procedure alone may miss important features like relative extrema or where the graph crosses its horizontal asymptote.
• Detailed analysis (or a calculator) reveals these subtleties.
• Calculus provides the ultimate tools for complete analysis.

🧭 Overview

🧠 One-sentence thesis

Rational inequalities require sign-diagram analysis rather than clearing denominators (which works for equations), and rational functions model real-world problems involving rates, costs, and variation relationships.

📌 Key points (3–5)

• Equation vs inequality: clearing denominators works for rational equations but not for inequalities, because multiplying by a variable expression may reverse the inequality sign depending on whether the expression is positive or negative.
• Sign-diagram technique: to solve rational inequalities, move all terms to one side, combine into a single rational expression, then analyze where the expression is positive or negative using test intervals.
• Rate problems: rates are additive—when traveling downstream, add the river speed to the canoe speed; when traveling upstream, subtract the river speed.
• Common confusion: don't forget the applied domain—even if algebra gives you an answer, context (e.g., "you can't produce half a system") may restrict the final solution.
• Variation language: "varies directly" means multiply by a constant; "varies inversely" means divide by the variable; "varies jointly" means multiply by multiple variables.

🔀 Solving rational equations vs inequalities

🔀 How to solve a rational equation

• Clearing denominators is safe: multiply both sides by the least common denominator to eliminate fractions.
• Check for extraneous solutions: any value that makes a denominator zero must be discarded.
• Example: To solve (x³ − 2x + 1)/(x − 1) = (1/2)x − 1, multiply both sides by 2(x − 1) to get 2x³ − 4x + 2 = x² − 3x + 2, then solve the polynomial equation. The solution x = 1 is extraneous because it makes the original denominator zero.

🚫 Why you cannot clear denominators in inequalities

• The sign of the multiplier matters: if you multiply both sides by (x − 1), the inequality direction depends on whether (x − 1) is positive or negative.
• Instead, use a sign diagram: collect all terms on one side so that one side is zero, combine into a single rational expression, then analyze the sign of that expression.
• Example: To solve (x³ − 2x + 1)/(x − 1) ≥ (1/2)x − 1, rewrite as (2x³ − x² − x)/(2x − 2) ≥ 0, then build a sign diagram.

📊 Sign-diagram method for rational inequalities

1. Move everything to one side: get 0 on the right-hand side.
2. Combine into a single fraction: find a common denominator.
3. Identify critical values: zeros of the numerator (where the expression equals zero) and zeros of the denominator (where the expression is undefined).
4. Test intervals: pick a test value in each interval between critical values and check the sign of the expression.
5. Read the solution from the diagram: include intervals where the sign matches the inequality, and include zeros if the inequality is ≥ or ≤ (but never include values that make the denominator zero).

Example: For (2x³ − x² − x)/(2x − 2) ≥ 0, the numerator factors as x(2x + 1)(x − 1) and the denominator is 2(x − 1). The critical values are x = −1/2, x = 0, and x = 1 (undefined). Testing intervals gives (+) on (−∞, −1/2), (−) on (−1/2, 0), (+) on (0, 1), and (+) on (1, ∞). The solution is (−∞, −1/2] ∪ [0, 1) ∪ (1, ∞). Note that x = 1 is excluded because it makes the denominator zero.

🖼️ Graphical interpretation

• Equation f(x) = g(x): the solutions are the x-coordinates where the graphs intersect.
• Inequality f(x) ≥ g(x): the solution includes not only intersection points but also intervals where the graph of f is above the graph of g.
• Don't confuse: the calculator may not show holes in the graph (removable discontinuities), so always check algebraically for values excluded from the domain.

🚤 Rate problems and work problems

🚤 Distance-rate-time relationship

distance = rate · time

• Rates are additive: when traveling downstream, the effective speed is (canoe speed in still water) + (river speed); when traveling upstream, the effective speed is (canoe speed in still water) − (river speed).
• Example: Carl canoes 5 miles downstream and 5 miles upstream in 3 hours total. His canoe speed in still water is 6 mph. Let R = river speed. Then:
  • Time downstream: 5/(6 + R)
  • Time upstream: 5/(6 − R)
  • Total time: 5/(6 + R) + 5/(6 − R) = 3
  • Solving gives R = 4 mph.

🛠️ Work-rate-time relationship

amount of work done = rate of work · time spent working

• Rates are additive: if Taylor weeds the garden in 4 hours alone, Taylor's rate is 1/4 garden per hour. If Carl and Taylor together weed the garden in 3 hours, their combined rate is 1/3 garden per hour.
• Example: Taylor's rate is 1/4 garden/hour. Together, Taylor and Carl's rate is 1/3 garden/hour. So Carl's rate is 1/3 − 1/4 = 1/12 garden/hour. Therefore, Carl alone takes 12 hours to weed the garden.

📏 Tracking units

• Why it matters: units help you check whether an equation makes sense—don't try to add apples to oranges.
• Example: In the equation 5/(6 + R) + 5/(6 − R) = 3, the left side has units of hours (distance in miles divided by speed in miles/hour), and the right side is 3 hours, so the units match.

💰 Average cost and applied domain

💰 Average cost function

Average cost function: C̄(x) = C(x)/x, where C(x) is the total cost to produce x items.

• What it measures: the cost per item when x items are produced.
• Example: If C(x) = 80x + 150, then C̄(x) = (80x + 150)/x = 80 + 150/x.

📉 Solving cost inequalities

• Applied domain matters: even if algebra gives you x > 7.5, the context (number of items produced) may require x to be a whole number, so the final answer is x ≥ 8.
• Example: Solve C̄(x) < 100 where C̄(x) = (80x + 150)/x. Rewrite as (150 − 20x)/x < 0. The sign diagram shows the solution is (7.5, ∞), but since x must be a whole number, the final answer is [8, ∞).

📈 Behavior as x → ∞

• Horizontal asymptote interpretation: if C̄(x) = 80 + 150/x, then as x → ∞, 150/x → 0⁺, so C̄(x) → 80⁺. This means the average cost approaches $80 per item but is always slightly above $80.
• Why: the $80 is the variable cost per item; the fixed cost of $150 is spread over more and more items, so its per-item contribution shrinks toward zero.
• Don't confuse: C̄(x) never equals 80 (because 150/x = 0 has no solution), but it gets arbitrarily close to 80 as production increases.

📦 Optimization with rational functions

📦 Box with no top example

• Setup: a box with square base and no top has volume 1000 cm³. Let x = width (and depth) in cm, h = height in cm.
• Volume constraint: x²h = 1000, so h = 1000/x².
• Applied domain: x > 0 (width must be positive), and every positive x gives a positive h, so the domain is (0, ∞).

🔍 Solving h(x) ≥ x

• What it means: find widths x such that the height is at least as large as the width.
• Method: rewrite 1000/x² ≥ x as (1000 − x³)/x² ≥ 0. The numerator is zero when x = 10. Sign diagram shows (+) on (0, 10) and (−) on (10, ∞). Solution: (0, 10].
• Interpretation: the width can be at most 10 cm for the height to be at least as large as the width.

📐 Surface area minimization

• Surface area formula: S = x² + 4xh (base plus four sides; no top).
• Substitute h: S(x) = x² + 4x(1000/x²) = x² + 4000/x.
• Behavior as x → ∞: S(x) ≈ x², so S(x) → ∞ (not a horizontal asymptote at y = 0).
• Calculator: the minimum occurs at x ≈ 12.60 cm, giving h ≈ 6.30 cm.

🔗 Variation relationships

🔗 Direct, inverse, and joint variation

Type	Definition	Formula
Direct	y varies directly with x	y = kx
Inverse	y varies inversely with x	y = k/x
Joint	z varies jointly with x and y	z = kxy

• Constant of proportionality: the constant k in each formula.
• Example: "Force F is directly proportional to extension x" means F = kx (Hooke's Law).
• Example: "Pressure P is inversely proportional to volume V" means P = k/V (Boyle's Law).

🧮 Translating variation statements

• Direct and inverse together: "Current I is directly proportional to voltage V and inversely proportional to resistance R" means I = kV/R (Ohm's Law).
• Joint variation: "Volume V varies jointly with height h and the square of radius r" means V = khr² (volume of a cone).
• Multiple relationships: "Force F varies directly with the product of masses mM and inversely with the square of distance r²" means F = kmM/r² (Newton's Law of Universal Gravitation).

📊 Using data to find the constant

• Method: if P = k/V, then k = PV. Multiply each pressure value by the corresponding volume value; if the product is approximately constant, that constant is k.
• Example: Boyle's data gives PV ≈ 1400 for all data points, so P = 1400/V.
• Power regression: fitting data to y = ax^b with a ≈ 1400 and b ≈ −1 confirms the inverse relationship y = 1400/x.

⚠️ Ambiguity in wording

• Algebraic operations: if the problem says "varies jointly with height and the square of radius," the squaring is part of the formula: V = khr², not V = k(hr)².
• Don't confuse: "the square of the radius" means r², not (radius)² as a separate variable.

🧭 Overview

🧠 One-sentence thesis

Function composition creates new functions by feeding the output of one function into another, enabling us to build complex algebraic functions from simpler building blocks like linear and quadratic functions.

📌 Key points (3–5)

• What composition is: A two-step process where the output of function f becomes the input to function g, written as (g ∘ f)(x) = g(f(x)).
• Order matters: In general, g ∘ f and f ∘ g produce different results (composition is not commutative).
• Domain before simplification: Always determine the domain of a composite function before algebraic simplification to capture all restrictions.
• Common confusion: Don't confuse "inside out" vs "outside in" evaluation methods—both yield the same result but approach the substitution differently.
• Why it matters: Composition lets us decompose complicated functions into simpler components and model multi-step real-world processes.

🔗 What is function composition

🔗 The definition

Composite of g with f, denoted g ∘ f, is defined by the formula (g ∘ f)(x) = g(f(x)), provided x is in the domain of f and f(x) is in the domain of g.

• Read as "g composed with f" or "g of f"
• At the formula level: replace every x in g(x) with the entire formula for f(x)
• From a process perspective: take input x → apply f → get f(x) → apply g → get g(f(x))

🎯 Inside vs outside functions

• In the expression g(f(x)), function f is the inside function and g is the outside function
• Example: For g(f(x)) where f(x) = x² - 4x and g(x) = 2 - √(x + 3), the inside is the polynomial and the outside is the expression with the square root

🔄 Two evaluation methods

🔄 Inside out approach

• Insert the expression for f(x) into g first
• Example: To find (g ∘ f)(x) where f(x) = x² - 4x and g(x) = 2 - √(x + 3):
  • Start with g(f(x)) = g(x² - 4x)
  • Substitute into g: 2 - √((x² - 4x) + 3) = 2 - √(x² - 4x + 3)

🔄 Outside in approach

• Use the formula for g(x) first, leaving f(x) as a placeholder
• Same example:
  • Start with g(f(x)) = 2 - √(f(x) + 3)
  • Then substitute: 2 - √((x² - 4x) + 3) = 2 - √(x² - 4x + 3)
• Both methods yield identical results

⚠️ Domain determination

⚠️ The critical rule

Always find the domain BEFORE simplifying the composite function.

• Look at the unsimplified form to identify all restrictions
• Need elements in the domain of f whose outputs f(x) are also in the domain of g

⚠️ Example walkthrough

For (g ∘ f)(x) = 2 - √((x² - 4x) + 3):

• Before simplifying, examine 2 - √((x² - 4x) + 3)
• Square root requires: x² - 4x + 3 ≥ 0
• Factor and solve: zeros at x = 1 and x = 3
• Using a sign diagram: domain is (-∞, 1] ∪ [3, ∞)
• Don't confuse: If you simplify first, you might miss domain restrictions from the original composition

⚠️ Multiple restrictions

When composing with rational functions or radicals:

• Check for division by zero in denominators
• Check for negative values under square roots
• Combine all restrictions using set intersection/union

Example: For (g ∘ h)(x) = 2 - √((2x/(x+1)) + 3), need both x ≠ -1 (denominator) and (2x/(x+1)) + 3 ≥ 0 (radical)

🔢 Key properties

🔢 Associativity

Property: h ∘ (g ∘ f) = (h ∘ g) ∘ f, provided the composites are defined

• When composing three or more functions, the order matters but grouping doesn't
• Can write h ∘ g ∘ f without parentheses
• Both expressions equal h(g(f(x)))
• Don't confuse: This is different from commutativity—g ∘ f ≠ f ∘ g in general

🔢 Identity function

Property: If I(x) = x for all real x, then I ∘ f = f ∘ I = f

• The identity function I acts like the number 1 in multiplication
• Composing any function with I returns the original function
• Just as 1 · x = x · 1 = x for numbers, I ∘ f = f ∘ I = f for functions

🔢 Non-commutativity

• In general, g ∘ f and f ∘ g are different functions
• Think of processes: putting on socks then shoes ≠ putting on shoes then socks
• Example from excerpt: (f ∘ g)(x) = x - 1 but (g ∘ f)(x) = 2 - √(x² - 4x + 3) are completely different

🔁 Special cases

🔁 Composing a function with itself

• Written as (f ∘ f)(x) = f(f(x))
• Called iterating the function
• Example: If h(x) = 2x/(x+1), then (h ∘ h)(x) = 4x/(3x+1)
• Real-world analogy: setting a washing machine to "double rinse"

🔁 Multiple compositions

• Can compose more than two functions: (h ∘ g ∘ f)(x) = h(g(f(x)))
• Apply functions in order from right to left
• Example: (h ∘ (g ∘ f))(x) means first compose g with f, then compose h with the result

🌍 Real-world applications

🌍 Relating quantities through a common variable

Composition connects quantities that aren't directly related but share an intermediate variable.

Example from excerpt: Surface area of an inflating sphere

• Surface area depends on radius: S(r) = 4πr²
• Radius depends on time: r(t) = 3t²
• Composition gives surface area as a function of time directly: (S ∘ r)(t) = S(r(t)) = 4π(3t²)² = 36πt⁴
• Eliminates the "middle man" variable r

🌍 Multi-step processes

• Any process with sequential steps can be viewed as function composition
• Each step is a function; the overall process is the composition
• Example: Computing revenue per hour from number of items sold, where items sold depends on time

🧩 Decomposing functions

🧩 Breaking down complex functions

A useful skill: express a complicated function as a composition of simpler functions.

Method 1 - Inside/outside perspective:

• For F(x) = |3x - 1|, identify 3x - 1 as "inside" the absolute value
• Let f(x) = 3x - 1 (inside) and *g(x) = |x| (outside)
• Then F = g ∘ f

Method 2 - Operational approach:

• For G(x) = 2/(x² + 1), list operations: square → add 1 → divide into 2
• Let f(x) = x², g(x) = x + 1, h(x) = 2/x
• Then G = h ∘ g ∘ f

Method 3 - Identify simple pieces:

• For H(x) = (√x + 1)/(√x - 1), recognize √x as a building block
• Let f(x) = √x and g(x) = (x + 1)/(x - 1)
• Then H = g ∘ f

🧩 Why decompose

• Simplifies analysis of complex functions
• Useful in calculus for differentiation (chain rule)
• Helps understand function structure and behavior
• Multiple correct decompositions exist for any given function

🧭 Overview

🧠 One-sentence thesis

A function is invertible if and only if it is one-to-one (different inputs produce different outputs), and the inverse function reverses the original process by swapping inputs and outputs.

📌 Key points (3–5)

• Reversibility concept: Some functions can be "undone" (like putting on and removing shoes), while others cannot (like cooking a steak).
• One-to-one requirement: For a function to have an inverse, different inputs must produce different outputs; graphically, this means passing the Horizontal Line Test.
• Finding inverses: Switch x and y in the equation, then solve for y; the result is the inverse function.
• Common confusion: The notation f⁻¹(x) does NOT mean 1/f(x); it means the inverse function that undoes f.
• Domain restrictions: Non-one-to-one functions (like x²) can be made invertible by restricting their domain to a portion where they pass the Horizontal Line Test.

🔄 What makes a function reversible

🔄 The reversibility concept

The excerpt introduces inverse functions through a process analogy:

• A function f(x) = 3x + 4 applies two steps: multiply by 3, then add 4.
• To reverse this, a function g must undo these steps in reverse order: subtract 4, then divide by 3.
• This gives g(x) = (x - 4)/3.

Example: If f(5) = 19, then g(19) = 5, returning the original input.

🔗 Composition verification

Two functions f and g are inverses when:

• (g ∘ f)(x) = x for all x in the domain of f
• (f ∘ g)(x) = x for all x in the domain of g

This means each function undoes the other completely.

📐 Graphical relationship

Definition 5.2: Suppose f and g are two functions such that (g ∘ f)(x) = x for all x in the domain of f and (f ∘ g)(x) = x for all x in the domain of g, then f and g are inverses of each other and the functions f and g are said to be invertible.

The graphs of inverse functions are reflections across the line y = x.

🎯 The one-to-one property

🎯 What one-to-one means

Definition 5.3: A function f is said to be one-to-one if f matches different inputs to different outputs. Equivalently, f is one-to-one if and only if whenever f(c) = f(d), then c = d.

• Different inputs must produce different outputs.
• If two inputs produce the same output, the function cannot be inverted.

Example problem: f(x) = x² is not one-to-one because f(-2) = 4 and f(2) = 4; two different inputs produce the same output.

📏 Horizontal Line Test

Theorem 5.4: A function f is one-to-one if and only if no horizontal line intersects the graph of f more than once.

• If any horizontal line crosses the graph more than once, the function fails the test and is not one-to-one.
• This is the graphical method to check invertibility.

⚠️ Why f(x) = x² fails

When you try to find an inverse for f(x) = x²:

• Both x = -2 and x = 2 produce y = 4.
• An inverse would need to take 4 back to both -2 and 2, which violates the definition of a function (one input cannot map to two outputs).
• Graphically, reflecting y = x² across y = x produces a sideways parabola that fails the Vertical Line Test.

🔧 Finding inverse functions

🔧 The algorithm

Steps for finding the inverse of a one-to-one function:

1. Write y = f(x)
2. Interchange x and y
3. Solve x = f(y) for y to obtain y = f⁻¹(x)

The interchange step reminds us we are switching inputs and outputs.

📝 Linear example

For f(x) = (1 - 2x)/5:

• Write y = (1 - 2x)/5
• Switch: x = (1 - 2y)/5
• Solve: 5x = 1 - 2y → 2y = 1 - 5x → y = -5x/2 + 1/2
• Result: f⁻¹(x) = -5x/2 + 1/2

📝 Rational example

For g(x) = 2x/(1 - x):

• Write y = 2x/(1 - x)
• Switch: x = 2y/(1 - y)
• Solve: x(1 - y) = 2y → x = xy + 2y → x = y(x + 2) → y = x/(x + 2)
• Result: g⁻¹(x) = x/(x + 2)

✅ Verification methods

Check both compositions:

• (f⁻¹ ∘ f)(x) = x for all x in the domain of f
• (f ∘ f⁻¹)(x) = x for all x in the domain of f⁻¹

Also verify graphically that the functions are reflections across y = x.

🔒 Domain restrictions

🔒 Making non-invertible functions invertible

The function f(x) = x² is not one-to-one on its natural domain (-∞, ∞), but:

• Restrict the domain to x ≥ 0 to create g(x) = x², x ≥ 0.
• Now g passes the Horizontal Line Test (only the right half of the parabola).
• The inverse is g⁻¹(x) = √x.

🔒 Why restrictions work

When finding the inverse of y = x², x ≥ 0:

• Switch: x = y², y ≥ 0
• Solve: y = ±√x
• The restriction y ≥ 0 tells us to choose y = √x (not the negative root).

Don't confuse: Without the domain restriction, we'd get y = |x| when composing, not y = x.

🔒 Quadratic with domain restriction

For j(x) = x² - 2x + 4, x ≤ 1:

• This selects only the left half of the parabola.
• Using the quadratic formula to solve for y gives two solutions.
• The restriction y ≤ 1 determines which solution to keep: j⁻¹(x) = 1 - √(x - 3).

📊 Key theorems and properties

📊 Properties of inverse functions

Theorem 5.2: Suppose f and g are inverse functions. The range of f is the domain of g and the domain of f is the range of g; f(a) = b if and only if g(b) = a; (a, b) is on the graph of f if and only if (b, a) is on the graph of g.

These properties follow from the definition and show how inverses exchange the roles of inputs and outputs.

📊 Uniqueness

Theorem 5.3: Suppose f is an invertible function. There is exactly one inverse function for f, denoted f⁻¹ (read f-inverse). The graph of y = f⁻¹(x) is the reflection of the graph of y = f(x) across the line y = x.

Every invertible function has exactly one inverse.

📊 Equivalent conditions

Theorem 5.5: Suppose f is a function. The following statements are equivalent: f is invertible; f is one-to-one; the graph of f passes the Horizontal Line Test.

All three conditions mean the same thing—if one is true, all are true.

⚠️ Notation warning

The notation f⁻¹ does NOT mean 1/f(x).

• f⁻¹ is the inverse function that undoes f.
• 1/f(x) is the reciprocal of f(x).
• For f(x) = 3x + 4: f⁻¹(x) = (x - 4)/3, but 1/f(x) = 1/(3x + 4).

The notation f⁻¹ alludes to the property that f⁻¹ ∘ f = I (identity function), similar to how 3⁻¹ · 3 = 1 in arithmetic.

🧮 Application example

🧮 Price-demand context

Given p(x) = -1.5x + 250 for 0 ≤ x ≤ 166, where x is weekly sales and p is price per system:

• p is one-to-one (non-horizontal line segment).
• p⁻¹(x) = (500 - 2x)/3 with domain [1, 250].
• Interpretation: p⁻¹(220) = 20 means 20 systems will be sold weekly if the price is $220.

🧮 Profit composition

Given profit P(x) = -1.5x² + 170x - 150:

• (P ∘ p⁻¹)(x) gives weekly profit as a function of price.
• The composition chains: price → weekly sales (via p⁻¹) → weekly profit (via P).
• After simplification: (P ∘ p⁻¹)(x) = -(2/3)x² + 220x - 40450/3.

🧮 Optimization

To maximize profit:

• The graph of (P ∘ p⁻¹)(x) is a downward-opening parabola.
• The vertex x-coordinate is x = -b/(2a) = 165.
• Maximum profit occurs when the price is set at $165 per system.

🧭 Overview

🧠 One-sentence thesis

Algebraic functions—combinations of polynomials, rational functions, and radicals—require careful handling of domains, exponents, and radicals, and their graphs can exhibit new features like cusps and unusual steepness that distinguish them from simpler function families.

📌 Key points (3–5)

• What algebraic functions are: combinations of polynomial/rational functions with radical operations (roots).
• Domain concerns: even-indexed radicals require non-negative radicands; negative exponents create denominators that must not be zero.
• Fractional exponents vs radicals: converting to radical notation avoids errors, especially when canceling exponents (e.g., √(x²) = |x|, not x).
• Common confusion: when simplifying fractional exponents like (x^(2/3))^(3/2), directly multiplying exponents can give wrong answers because √(x²) = |x|; rewrite as radicals first.
• New graphical features: algebraic functions can have cusps, unusual steepness (vertical tangents), and multiple horizontal asymptotes—features not seen in polynomial or simple rational functions.

📐 Radicals and their properties

📐 Definition of principal nth root

Definition 5.4: Let x be a real number and n a natural number.

• If n is odd, the principal nth root of x, denoted ⁿ√x, is the unique real number satisfying (ⁿ√x)ⁿ = x.
• If n is even, ⁿ√x is defined similarly provided x ≥ 0 and ⁿ√x ≥ 0.
  The index is n and the radicand is x. For n = 2, write √x instead of ²√x.

• Why the restrictions for even n: both x = −2 and x = 2 satisfy x⁴ = 16, but ⁴√16 = 2, not −2.
• Functional interpretation: f(x) = ⁿ√x is the inverse of g(x) = xⁿ (with domain [0, ∞) when n is even).
• Example: the graph of y = √x is the reflection of y = x² (restricted to x ≥ 0) across the line y = x.

🧮 Properties of radicals (Theorem 5.6)

Let x and y be real numbers and m and n be natural numbers. If ⁿ√x and ⁿ√y are real numbers, then:

Property	Formula	Notes
Product Rule	ⁿ√(xy) = ⁿ√x · ⁿ√y	Multiply inside or outside the radical
Powers of Radicals	ⁿ√(xᵐ) = (ⁿ√x)ᵐ	Raise the root to a power
Quotient Rule	ⁿ√(x/y) = ⁿ√x / ⁿ√y	y ≠ 0
Simplifying powers	ⁿ√(xⁿ) = x if n is odd; ⁿ√(xⁿ) = |x| if n is even	The absolute value appears for even n

• Why the absolute value matters: ⁴√((-2)⁴) = ⁴√16 = 2 = |-2|, not -2.
• Proof sketch: the product rule follows from the definition of roots and properties of exponents; the quotient rule is similar.
• Don't confuse: ⁿ√(xⁿ) = x only when n is odd; for even n, you must write |x|.

📊 Graph behavior of radical functions

• Even-indexed radicals (√x, ⁴√x, ⁶√x):
  • Domain: [0, ∞)
  • Vertical steepening near x = 0; horizontal flattening as x → ∞
• Odd-indexed radicals (³√x, ⁵√x, ⁷√x):
  • Domain: (−∞, ∞)
  • Steepening near x = 0; flattening as x → ±∞
• Example: the excerpt shows graphs of y = √x, y = ⁴√x, y = ⁶√x all pass through (0, 0) and exhibit similar steepening/flattening.

🔢 Rational exponents and their pitfalls

🔢 Definition of rational exponents (Definition 5.5)

Let x be a real number, m an integer, and n a natural number.

• x^(1/n) = ⁿ√x, defined whenever ⁿ√x is defined.
• x^(m/n) = (ⁿ√x)ᵐ = ⁿ√(xᵐ), defined whenever (ⁿ√x)ᵐ is defined.

• Rational exponents behave like integer exponents with one critical exception.

⚠️ The critical exception: canceling exponents

• Tempting but wrong: (x^(2/3))^(3/2) = x^(2/3 · 3/2) = x^1 = x.
• Counterexample: substitute x = −1:
  • (−1)^(2/3) = (³√(−1))² = (−1)² = 1
  • ((−1)^(2/3))^(3/2) = 1^(3/2) = 1 ≠ −1
• Why it fails: when you cancel the 2's in (2/3) · (3/2), you are canceling a square with a square root, but √(x²) = |x|, not x.
• Correct simplification: (x^(2/3))^(3/2) = ((³√x)²)^(3/2) = (√((³√x)²))³ = (|³√x|)³ = |(³√x)³| = |x|.

✅ Best practice

• Moral of the story: when simplifying fractional exponents, rewrite them as radicals to avoid errors.
• Example: the excerpt notes that (x^(3/2))^(2/3) = x does work (verification left as exercise).

🔍 Domains and sign diagrams for algebraic functions

🔍 Steps for constructing a sign diagram

Suppose f is an algebraic function:

1. Place any values excluded from the domain on the number line with an '⊗' above them.
2. Find the zeros of f and place them on the number line with 0 above them.
3. Choose a test value in each interval determined in steps 1 and 2.
4. Determine the sign of f(x) for each test value and write that sign above the corresponding interval.

• Why this works: algebraic functions are continuous on their domains (by the Intermediate Value Theorem, Theorem 3.1), so sign changes only at zeros or domain boundaries.

🔍 Domain concerns for algebraic functions

• Even-indexed radicals: require radicand ≥ 0.
• Negative exponents: indicate denominators; set denominator ≠ 0.
• Example: for g(x) = √(2 − ⁴√(x + 3)), you need:
  • x + 3 ≥ 0 (for the fourth root)
  • 2 − ⁴√(x + 3) ≥ 0 (for the square root)
  • Solve the second inequality using a sign diagram: 2 − ⁴√(x + 3) ≥ 0 gives [−3, 13].

🔍 Finding zeros

• Set f(x) = 0 and solve.
• Caution: if you raise both sides to an even power, check for extraneous solutions.
• Example: solving √(2 − ⁴√(x + 3)) = 0:
  • Square both sides: 2 − ⁴√(x + 3) = 0
  • Solve: ⁴√(x + 3) = 2 → x + 3 = 16 → x = 13
  • Check: g(13) = √(2 − ⁴√16) = √(2 − 2) = 0 ✓

📈 New graphical features of algebraic functions

📈 Cusps

• What it is: a sharp turn or point where the graph is not smooth.
• Example: f(x) = x^(2/3) has a cusp at x = 0 (the graph of y = ³√(x²) turns sharply at the origin).
• Don't confuse with a corner in an absolute value graph; cusps arise from fractional exponents with even numerators.

📈 Unusual steepness (vertical tangents)

• What it is: the graph becomes nearly vertical at a point.
• Example: near x = 2 in f(x) = 3x · ³√(2 − x), the graph steepens dramatically.
• The proper Calculus term is "vertical tangent," but the excerpt calls it "unusual steepness."
• Example: even-indexed radicals like √x exhibit unusual steepness at x = 0.

📈 Multiple horizontal asymptotes

• New behavior: unlike rational functions (which have at most one horizontal asymptote), algebraic functions can have different horizontal asymptotes as x → ∞ and x → −∞.
• Example: k(x) = 2x / √(x² − 1)
  • As x → ∞, |x| = x, so k(x) ≈ 2x / x = 2 → horizontal asymptote y = 2.
  • As x → −∞, |x| = −x, so k(x) ≈ 2x / (−x) = −2 → horizontal asymptote y = −2.

📈 Vertical asymptotes

• Occur where the denominator is zero (and not canceled by the numerator).
• Example: h(x) = ³√(8x / (x + 1)) has a vertical asymptote at x = −1.

🧩 Solving inequalities with algebraic functions

🧩 General approach

• Get 0 on one side: rewrite the inequality so one side is zero.
• Build a sign diagram: for the nonzero side, treating it as a function r(x).
• Read the solution: from the sign diagram, identify intervals where r(x) has the desired sign.
• Check graphically: graph both sides and confirm where one is above/below the other.

🧩 Example: solving x^(2/3) < x^(4/3) − 6

1. Rewrite: x^(4/3) − x^(2/3) − 6 > 0.
2. Let r(x) = x^(4/3) − x^(2/3) − 6.
3. Find zeros: this is a "quadratic in disguise." Let u = x^(2/3), so u² − u − 6 = 0 → u = −2 or u = 3.
   • x^(2/3) = −2 has no real solution (since ³√(x²) ≥ 0).
   • x^(2/3) = 3 → ³√(x²) = 3 → x² = 27 → x = ±3√3.
4. Sign diagram: r(x) > 0 on (−∞, −3√3) ∪ (3√3, ∞).
5. Graphical check: graph f(x) = x^(2/3) and g(x) = x^(4/3) − 6; f is below g on the solution intervals.

🧩 Factoring vs common denominator

• Factoring approach: factor out the term with the smallest exponent (even if negative).
  • Example: 3(2 − x)^(1/3) − x(2 − x)^(−2/3) = (2 − x)^(−2/3) [3(2 − x)^1 − x] = (2 − x)^(−2/3) (6 − 4x).
• Common denominator approach: rewrite all terms with a common denominator.
  • Example: 3(2 − x)^(1/3) − x / (2 − x)^(2/3) = [3(2 − x)^1 − x] / (2 − x)^(2/3) = (6 − 4x) / (2 − x)^(2/3).
• Both methods give the same result; choose whichever is clearer.

🌍 Applications of algebraic functions

🌍 Distance and cost optimization

• Scenario: Carl wants to run cable from a junction box 50 miles down Route 117 to an outpost 30 miles off the road. Cable costs $15/mile on-road and $20/mile off-road.
• Setup: let x = miles run along the road before turning off. Then:
  • On-road cost: 15x
  • Off-road distance: z = √((50 − x)² + 900) (by Pythagorean Theorem)
  • Off-road cost: 20z
  • Total cost: C(x) = 15x + 20√((50 − x)² + 900), 0 ≤ x ≤ 50.
• Solution: graph C(x) and use the calculator's "Minimum" feature to find the minimum cost ≈ $1146.86 at x ≈ 15.98 miles.

🌍 Other applications

• Wind chill formula: W = 35.74 + 0.6215T_a − 35.75V^(0.16) + 0.4275T_a V^(0.16), where W is wind chill (°F), T_a is air temperature (°F), and V is wind speed (mph).
  • Defined only for T_a ≤ 50°F and V > 3 mph.
• Pendulum period: T = 2π√(L/g), where L is length (meters) and g = 9.8 m/s².
• Relativity: observed mass m(x) = m_r / √(1 − x²/c²), where x is speed and c is speed of light.
  • As x → c⁻, m(x) → ∞ (mass increases without bound as speed approaches light speed).

🧭 Overview

🧠 One-sentence thesis

Logarithmic functions are defined as inverses of exponential functions, and their primary use is to "undo" exponentiation by transforming products into sums, quotients into differences, and powers into multiples.

📌 Key points (3–5)

• Logarithms undo exponentials: The logarithm base b of x answers "what power of b gives x?" and reverses the exponential operation.
• Inverse relationship: If b raised to a equals c, then log base b of c equals a; these are two ways of writing the same relationship.
• Transformations work predictably: Exponential and logarithmic graphs can be shifted, reflected, and stretched using the same transformation rules as other functions.
• Common confusion: When finding inverses procedurally, remember that logarithms specifically undo the "put as an exponent" step—add/subtract operations are undone in reverse order.
• Algebraic properties mirror exponents: Logarithms convert multiplication into addition, division into subtraction, and exponentiation into multiplication (the Product, Quotient, and Power rules).

🔄 The inverse relationship between exponentials and logarithms

🔄 Definition and equivalence

b raised to the power a equals c if and only if log base b of c equals a.

• This is the fundamental definition: logarithms and exponentials are two sides of the same coin.
• Example: 2 cubed equals 8 can be rewritten as log base 2 of 8 equals 3.
• The excerpt emphasizes this equivalence is bidirectional—you can always rewrite an exponential equation as a logarithmic one and vice versa.

🔄 Inverse function properties

The excerpt states two key inverse properties:

• Logarithm of an exponential: log base b of (b raised to x) equals x for all real x.
• Exponential of a logarithm: b raised to (log base b of x) equals x for all x greater than 0.

These properties confirm that applying a logarithm and then the corresponding exponential (or vice versa) returns the original input.

Don't confuse: The domain restrictions matter—logarithms require positive inputs, so the second property only works when x is greater than 0.

📊 Graphing exponentials and logarithms using transformations

📊 Exponential transformations

The excerpt demonstrates graphing f(x) = 2 raised to (x minus 1) minus 3 by starting with g(x) = 2 raised to x and applying transformations:

• Horizontal shift: Adding 1 to x-coordinates shifts the graph right 1 unit.
• Vertical shift: Subtracting 3 from y-coordinates shifts the graph down 3 units.
• Asymptote moves: The horizontal asymptote y = 0 becomes y = negative 3 after the vertical shift.
• The domain remains all real numbers; the range becomes (negative 3, infinity).

📊 Logarithmic transformations

For the inverse f inverse of x = log base 2 of (x plus 3) plus 1, the excerpt starts with j(x) = log base 2 of x:

• Horizontal shift: Subtracting 3 from x-values (including the vertical asymptote) shifts left 3 units.
• Vertical shift: Adding 1 to y-values shifts up 1 unit.
• Asymptote moves: The vertical asymptote x = 0 becomes x = negative 3.
• The domain is (negative 3, infinity); the range is all real numbers.

Key observation: The excerpt notes the reader may experience "déjà vu"—the transformed points on the logarithmic graph match the transformed points on the exponential graph, reflecting the inverse relationship and symmetry about the line y = x.

🔧 Finding inverses procedurally

🔧 The procedural perspective

The excerpt introduces a step-by-step method for finding inverses without algebraically solving for y:

1. Break the function into steps: For f(x) = 2 raised to (x minus 1) minus 3, the steps are (a) subtract 1, (b) put as an exponent on 2, (c) subtract 3.
2. Reverse each step: To undo, (a) add 3, (b) take the logarithm base 2, (c) add 1.
3. Write the inverse: f inverse of x = log base 2 of (x plus 3) plus 1.

🔧 Why logarithms undo exponentiation

The excerpt explicitly states: "How do we undo the second step? The answer is we use the logarithm. By definition, log base 2 of x undoes exponentiation by 2."

This is the core operational role of logarithms—they are the tool for reversing the "put as an exponent" operation.

Don't confuse: The order matters. Undo operations in reverse order: if the original function subtracts 1 first, then exponentiates, then subtracts 3, the inverse must add 3 first, then take the log, then add 1.

✅ Verifying inverses through composition

✅ Composition in both directions

The excerpt verifies that f and f inverse are true inverses by checking two compositions:

• (f inverse composed with f)(x) = x for all x in the domain of f.
• (f composed with f inverse)(x) = x for all x in the domain of f inverse.

✅ Example calculation

For f(x) = 2 raised to (x minus 1) minus 3 and f inverse of x = log base 2 of (x plus 3) plus 1:

• First composition: f inverse of (f(x)) = log base 2 of ((2 raised to (x minus 1) minus 3) plus 3) plus 1 = log base 2 of (2 raised to (x minus 1)) plus 1 = (x minus 1) plus 1 = x.
  • The key step uses: log base 2 of (2 raised to u) equals u for all real u.
• Second composition: f of (f inverse of x) = 2 raised to ((log base 2 of (x plus 3) plus 1) minus 1) minus 3 = 2 raised to (log base 2 of (x plus 3)) minus 3 = (x plus 3) minus 3 = x.
  • The key step uses: 2 raised to (log base 2 of u) equals u for all u greater than 0.
  • The excerpt notes "pay attention" because this step requires x greater than negative 3 (so that x plus 3 is positive).

✅ Graphical symmetry

The excerpt confirms that graphing f and f inverse on the same axes shows symmetry about the line y = x, which is the visual signature of inverse functions.

🧮 Algebraic properties of logarithms

🧮 Product Rule

log base b of (u times w) = log base b of u plus log base b of w

• Logarithms convert multiplication into addition.
• The excerpt explains this from the inverse perspective: exponential functions add inputs to multiply outputs (b raised to (u plus w) = b raised to u times b raised to w), so logarithms must multiply inputs to add outputs.

🧮 Quotient Rule

log base b of (u divided by w) = log base b of u minus log base b of w

• Logarithms convert division into subtraction.
• This mirrors the exponential quotient rule: b raised to (u minus w) = (b raised to u) divided by (b raised to w).

🧮 Power Rule

log base b of (u raised to w) = w times log base b of u

• Logarithms convert exponentiation into multiplication.
• This corresponds to the exponential power rule: (b raised to u) raised to w = b raised to (u times w).

🧮 Why these properties hold

The excerpt offers two explanations:

1. Definitional proof: Let a = log base b of (uw), c = log base b of u, d = log base b of w. Then b raised to a = uw, b raised to c = u, b raised to d = w. So b raised to a = b raised to c times b raised to d = b raised to (c plus d). By the one-to-one property of exponentials, a = c plus d.
2. Functional perspective: Inverse functions interchange inputs and outputs. Since exponentials add inputs to multiply outputs, logarithms must multiply inputs to add outputs.

Don't confuse: These properties only apply when the arguments are positive (u greater than 0, w greater than 0), because logarithms are only defined for positive inputs.

📐 Applications: Logarithmic scales

📐 Richter scale for earthquakes

M(x) = log of (x divided by 0.001)

• M is the magnitude; x is the seismograph reading in millimeters at 100 km from the epicenter.
• The baseline is 0.001 mm (a "magnitude 0 event").
• Example: M(0.001) = log(1) = 0.
• Multiplicative interpretation: An earthquake measuring 6.7 has a seismograph reading ten times larger than one measuring 5.7 (because log of (10x divided by 0.001) minus log of (x divided by 0.001) = log(10) = 1).

📐 Decibel scale for sound intensity

L(I) = 10 times log of (I divided by 10 raised to negative 12)

• L is the sound intensity level in decibels; I is intensity in watts per square meter.
• The baseline is 10 raised to negative 12 watts per square meter (threshold of human hearing).
• Example: L(10 raised to negative 6) = 10 times log(10 raised to 6) = 10 times 6 = 60 decibels.
• Damage starts around 115 decibels; the threshold of pain is around 140 decibels.

📐 pH scale for acidity

pH = negative log of [H plus]

• [H plus] is the hydrogen ion concentration in moles per liter.
• pH less than 7 is acidic; pH greater than 7 is alkaline (basic); pH equals 7 is neutral.
• Example: Pure water has [H plus] = 10 raised to negative 7, so pH = negative log(10 raised to negative 7) = 7.
• Gastric acid has pH about 0.7, corresponding to [H plus] = 10 raised to negative 0.7, approximately 0.1995 moles per liter.

Common pattern: All three scales compare a measured quantity to a baseline using a common logarithm, converting wide-ranging multiplicative differences into manageable additive scales.

🧭 Overview

🧠 One-sentence thesis

The algebraic properties of logarithms—Product, Quotient, and Power Rules—allow us to transform complex logarithmic expressions by converting multiplication into addition, division into subtraction, and exponents into coefficients, all stemming from the inverse relationship between logarithms and exponentials.

📌 Key points (3–5)

• Three core properties: Product Rule turns log of a product into a sum; Quotient Rule turns log of a quotient into a difference; Power Rule turns log of a power into a coefficient times the log.
• Why they work: These properties arise because logarithms are inverses of exponentials—adding inputs in exponentials produces multiplying outputs, so the inverse (logarithm) must turn products of inputs into sums of outputs.
• Expanding vs. condensing: Reading properties left-to-right expands a single log into multiple terms; reading right-to-left condenses multiple logs into one.
• Common confusion: Expanding or condensing logs can change the domain of the resulting function, so the original and simplified expressions may not be equivalent everywhere.
• Change of Base formula: Any logarithm or exponential can be rewritten in a different base by multiplying (for exponentials) or dividing (for logarithms) by an appropriate conversion factor.

📐 The three algebraic properties

🔢 Product Rule

Product Rule: log base b of (u times w) equals log base b of u plus log base b of w.

• In symbols: log_b(uw) = log_b(u) + log_b(w).
• Why it works: If a = log_b(uw), c = log_b(u), and d = log_b(w), then by definition b^a = uw, b^c = u, and b^d = w. Since uw = b^c · b^d = b^(c+d), we have b^a = b^(c+d), so a = c + d by the one-to-one property of exponentials.
• Inverse perspective: Exponential functions add inputs to multiply outputs (b^(u+w) = b^u · b^w), so the inverse function (logarithm) must take products of inputs and return sums of outputs.
• Example: log_2(8x) = log_2(8) + log_2(x) = 3 + log_2(x).

➗ Quotient Rule

Quotient Rule: log base b of (u divided by w) equals log base b of u minus log base b of w.

• In symbols: log_b(u/w) = log_b(u) − log_b(w).
• This follows from the same reasoning as the Product Rule, using the exponential property that division corresponds to subtraction of exponents.
• Example: log_2(8/x) = log_2(8) − log_2(x) = 3 − log_2(x).

🔺 Power Rule

Power Rule: log base b of (u to the power w) equals w times log base b of u.

• In symbols: log_b(u^w) = w · log_b(u).
• The exponent inside the log becomes a coefficient outside the log.
• Example: log_0.1(x²) = 2 · log_0.1(x).
• Don't confuse: The power must apply to the entire quantity inside the log; if only part of the expression has an exponent, use Product or Quotient Rule first to isolate it.

🔧 Expanding logarithms

🧩 What "expanding" means

• Expanding = rewriting a single complicated logarithm as a sum, difference, or multiple of simpler logarithms.
• Read the properties left to right: products inside → sums outside; quotients inside → differences outside; powers inside → coefficients outside.

🎯 Order of operations: reverse priority

• Rule of thumb: Apply log properties in reverse order of operations.
• If you would compute the inside of the log by first squaring, then multiplying, then dividing, you expand by first applying the Quotient Rule (last operation), then Product Rule, then Power Rule.
• Example: For log_0.1(10x²), you would compute x² first, then multiply by 10. So expand by Product Rule first (multiplication is last), then Power Rule (exponent is first).

📝 Handling radicals and special cases

• Radicals: Rewrite cube root or nth root as a fractional exponent (e.g., cube root of (100x²yz⁵) = (100x²yz⁵)^(1/3)), then apply Power Rule.
• Factoring: If you see a difference inside the log (e.g., log_117(x² − 4)), try factoring it into a product (x² − 4 = (x+2)(x−2)) so you can use the Product Rule.
• Example: log_117(x² − 4) = log_117((x+2)(x−2)) = log_117(x+2) + log_117(x−2).

⚠️ Domain warning

• Expanding a logarithm can restrict the domain.
• Example: f(x) = log_117(x² − 4) has domain (−∞, −2) ∪ (2, ∞), but g(x) = log_117(x+2) + log_117(x−2) has domain (2, ∞) only, because each individual log requires its argument to be positive.
• The expanded form is valid only where all individual logs are defined.

🔄 Condensing logarithms

🧩 What "condensing" means

• Condensing = rewriting a sum, difference, or multiple of logarithms as a single logarithm.
• Read the properties right to left: sums outside → products inside; differences outside → quotients inside; coefficients outside → powers inside.

🎯 Order of operations: normal priority

• Apply log properties in normal order of operations: Power Rule first (deal with coefficients), then Product and Quotient Rules (deal with addition and subtraction).
• Example: log(x) + 2log(y) − log(z) becomes log(x) + log(y²) − log(z) (Power Rule), then log(xy²) − log(z) (Product Rule), then log(xy²/z) (Quotient Rule).

🔢 Converting constants to logs

• To combine a constant with a logarithm, rewrite the constant as a log in the same base.
• Example: 4log_2(x) + 3 becomes log_2(x⁴) + log_2(2³) = log_2(x⁴) + log_2(8) = log_2(8x⁴), because 3 = log_2(2³).
• For natural log: 1/2 = ln(e^(1/2)) = ln(√e).

⚠️ Domain warning (reverse)

• Condensing a logarithm can expand the domain.
• Example: f(x) = log_3(x−1) − log_3(x+1) has domain (1, ∞), but g(x) = log_3((x−1)/(x+1)) has domain (−∞, −1) ∪ (1, ∞).
• This matters when solving equations: you may get extraneous solutions that satisfy the condensed form but not the original expanded form.

🔀 Change of Base formulas

🔄 Two forms of the formula

The Change of Base formulas let you rewrite any logarithm or exponential in a different base.

Form	Formula	What it does
Exponential	a^x = b^(x · log_b(a))	Converts base-a exponential to base-b by multiplying the input by log_b(a)
Logarithmic	log_a(x) = log_b(x) / log_b(a)	Converts base-a logarithm to base-b by dividing the output by log_b(a)

• Both formulas say the same thing from different perspectives: exponentials and logs in different bases are just scalings of one another.
• Why they work: Start with b^(x · log_b(a)). By Power Rule, the exponent x · log_b(a) = log_b(a^x). By the inverse property, b^(log_b(a^x)) = a^x. For the log form, note that log_a(x) · log_b(a) = log_b(a^(log_a(x))) = log_b(x), then divide both sides by log_b(a).

🧮 Using Change of Base with calculators

• Calculators typically have only LOG (base 10) and LN (base e) buttons.
• To compute log_2(7), use the formula: log_2(7) = ln(7) / ln(2) or log(7) / log(2).
• Example: 3² to base 10 becomes 10^(2·log(3)). Typing this into a calculator gives 9, as expected.
• Example: 2^x to base e becomes e^(x·ln(2)). Graphing both shows they are identical.

🌐 All bases are equivalent

• The Change of Base formulas show that all exponential and logarithmic functions are scalings of one another.
• You could do all mathematics with a single base (10, e, 42, or any other valid base)—the graphs have similar shapes because they differ only by a scaling factor.
• This is why the natural log (base e) is preferred in calculus: it simplifies certain formulas, but any base would work in principle.

⚠️ Common pitfalls

🚫 Invented properties that don't exist

• Do NOT invent your own log rules. If it's not written in a textbook, it's probably not true.
• Wrong: log_117(x² − 4) = log_117(x²) − log_117(4). (You cannot split a difference inside a log into a difference of logs.)
• Wrong: log_b(x + y) = log_b(x) + log_b(y). (Addition inside a log does not become addition outside.)
• Wrong: log_b(x − y) = log_b(x) − log_b(y). (Subtraction inside a log does not become subtraction outside.)
• Wrong: log_b(x/y) = log_b(x) / log_b(y). (Division inside a log becomes subtraction outside, not division outside.)

🔍 Domain changes

• Expanding or condensing logs changes the domain of the function.
• Always check: the original and simplified expressions may not be equal for all x.
• This is why you must check for extraneous solutions when solving logarithmic equations (covered in Section 6.4).

🧭 Overview

🧠 One-sentence thesis

Exponential equations can be solved either by expressing both sides with a common base and equating exponents or by taking the natural logarithm of both sides, and these same principles extend to solving exponential inequalities using sign diagrams.

📌 Key points (3–5)

• Two main solution strategies: rewrite with a common base and equate exponents, or take the natural log of both sides and use the Power Rule.
• When to use each method: common base is convenient when both sides are easily expressed as powers of the same base; otherwise, use natural logarithm.
• Quadratics in disguise: some exponential equations have three terms where one exponent is twice another, allowing substitution to convert to a quadratic equation.
• Common confusion: when taking ln of both sides, divide by ln(2) as a number, not by "ln" as a symbol—just as you wouldn't divide by the square root symbol.
• Inequalities use sign diagrams: find zeros and undefined points, test intervals, and apply the Intermediate Value Theorem since exponential functions are continuous.

🔧 Core solution techniques

🔧 Isolate the exponential first

• Before applying any technique, get the exponential function by itself on one side.
• Example: solving 2000 = 1000 · 3^(−0.1t) requires dividing both sides by 1000 first to get 3^(−0.1t) = 2.

🔧 Method 1: Common base

• If both sides can be written as powers of the same base, use the one-to-one property of exponential functions.
• The one-to-one property says: if b^u = b^w, then u = w.
• Example: 2^(3x) = 16^(1−x) becomes 2^(3x) = 2^(4(1−x)), so 3x = 4(1−x), giving x = 4/7.
• This method works cleanly when the numbers involved are powers of a common base.

🔧 Method 2: Take natural logarithm

• When a common base is inconvenient, take ln of both sides.
• Apply the Power Rule: ln(b^u) = u·ln(b).
• Example: from 3^(−0.1t) = 2, take ln of both sides to get −0.1t·ln(3) = ln(2), then solve for t = −10·ln(2)/ln(3).
• Don't confuse: divide by ln(2) as a number, not by the symbol "ln"—just as you wouldn't divide by √ when solving x√2 = 5.

🎯 Special equation types

🎯 Quadratics in disguise

• Some equations have three terms where one exponent is exactly twice another.
• Substitution converts them to quadratic form.
• Example: 25^x = 5^x + 6 becomes (5^2)^x = 5^x + 6, or 5^(2x) = 5^x + 6.
• Let u = 5^x, so u² = 5^(2x), giving u² = u + 6, which factors as u² − u − 6 = 0.
• Solutions: u = −2 or u = 3. Since 5^x = −2 has no real solution (exponentials are always positive), only 5^x = 3 works, giving x = ln(3)/ln(5).

🎯 Equations with e^x and e^(−x)

• Rewrite e^(−x) = 1/e^x to clear the negative exponent.
• This often creates a fraction that leads to another quadratic in disguise.
• Example: (e^x − e^(−x))/2 = 5 becomes e^x − 1/e^x = 10, then e^(2x) − 1 = 10e^x.
• Let u = e^x, so u² − 10u − 1 = 0, giving u = 5 ± √26 by the quadratic formula.
• Since 5 − √26 < 0, only e^x = 5 + √26 works, so x = ln(5 + √26).

🎯 Rational expressions with exponentials

• Clear denominators first to isolate the exponential.
• Example: 75 = 100/(1 + 3e^(−2t)) becomes 75(1 + 3e^(−2t)) = 100, then 225e^(−2t) = 25, so e^(−2t) = 1/9.
• Taking ln: ln(e^(−2t)) = ln(1/9), which simplifies to −2t = −ln(9), giving t = ln(9)/2 = ln(3).

📊 Solving inequalities

📊 Sign diagram method

• Get zero on one side of the inequality.
• Set r(x) equal to the non-zero side.
• Find the domain (exclude where undefined).
• Find zeros by solving r(x) = 0.
• Build a sign diagram: mark zeros and undefined points on a number line, test intervals.
• The Intermediate Value Theorem applies because exponential functions are continuous.

📊 Example: rational exponential inequality

• To solve e^x/(e^x − 4) ≤ 3, first get zero on one side: e^x/(e^x − 4) − 3 ≤ 0.
• Common denominator: (e^x − 3(e^x − 4))/(e^x − 4) = (12 − 2e^x)/(e^x − 4) ≤ 0.
• Domain: e^x − 4 ≠ 0, so x ≠ ln(4).
• Zeros: 12 − 2e^x = 0 gives e^x = 6, so x = ln(6).
• Test intervals around ln(4) and ln(6) using the fact that ln is increasing: ln(3) < ln(4) < ln(5) < ln(6) < ln(7).
• Evaluate at test points like ln(3), ln(5), ln(7) using e^(ln(a)) = a.
• Solution: (−∞, ln(4)) ∪ [ln(6), ∞).

📊 Choosing test values

• When testing around values like ln(4), use properties: e^(ln(3)) = 3, so you can evaluate without a calculator.
• Use the fact that ln is increasing: if a < b, then ln(a) < ln(b).
• Example: to test near ln(6), use ln(5) or ln(7) as test points.

🔄 Finding inverses

🔄 Rational-exponential compositions

• For functions like f(x) = 5e^x/(e^x + 1), write y = f(x) and swap x and y.
• Solve for y by clearing denominators and isolating the exponential.
• Example: x = 5e^y/(e^y + 1) becomes x(e^y + 1) = 5e^y, then xe^y + x = 5e^y, so x = e^y(5 − x).
• Therefore e^y = x/(5 − x), and taking ln: y = ln(x/(5 − x)).
• The inverse is f^(−1)(x) = ln(x/(5 − x)).
• Verify graphically: f and f^(−1) should be symmetric about the line y = x, with domain of f equal to range of f^(−1) and vice versa.

✅ Verification strategies

✅ Analytical verification

• Substitute the solution back into the original equation.
• Use log properties and inverse properties to simplify.
• Example: to verify t = −10·ln(2)/ln(3) solves 2000 = 1000·3^(−0.1t), substitute and use change of base: 3^(ln(2)/ln(3)) = 3^(log₃(2)) = 2 by the inverse property.

✅ Graphical verification

• Graph both sides of the equation as separate functions.
• The x-coordinate of intersection points are the solutions.
• For inequalities, observe where one graph is above/below the other.
• Example: for 25^x = 5^x + 6, graph f(x) = 25^x and g(x) = 5^x + 6; they intersect at x = ln(3)/ln(5) ≈ 0.6826.

🧭 Overview

🧠 One-sentence thesis

Logarithmic equations and inequalities can be solved either by equating arguments when bases match or by converting to exponential form, but solutions must always be checked because logarithms are undefined for non-positive arguments.

📌 Key points (3–5)

• Two main strategies: equate arguments when logs have the same base, or rewrite the log equation as an exponential equation.
• Domain restrictions matter: any solution that produces a negative number (or zero) inside a logarithm must be discarded as extraneous.
• For inequalities: use sign diagrams because logarithmic functions are continuous on their domains.
• Common confusion: algebraically correct solutions may violate domain restrictions—always verify that arguments of logarithms remain positive.
• Change of base: when logs have different bases, convert to a common base before combining.

🔧 Two core strategies for equations

🔧 Strategy A: Equate arguments

• When to use: both sides of the equation are logarithms with the same base.
• If log base b of (expression 1) equals log base b of (expression 2), then expression 1 equals expression 2 (by Theorem 6.4).
• Example: log₂(x) = log₂(5) immediately gives x = 5.

🔧 Strategy B: Convert to exponential form

• When to use: one side is a logarithm, the other is a number (or when Strategy A is inconvenient).
• Rewrite using the definition: log base b of (x) = c means b raised to the power c equals x.
• Example: log₂(x) = 3 becomes 2³ = x, so x = 8.

🔄 Combining the strategies

• Sometimes you can rewrite a number as a logarithm (using Theorem 6.3): 3 = log₂(2³) = log₂(8).
• Then log₂(x) = 3 becomes log₂(x) = log₂(8), so x = 8.
• The exponential approach is usually faster.

⚠️ Domain restrictions and extraneous solutions

⚠️ Why extraneous solutions appear

• Algebraic manipulation (like squaring, combining logs) can introduce solutions that violate the original domain.
• Key rule: the argument of every logarithm must be strictly positive.
• Example: solving log₁₁₇(1 − 3x) = log₁₁₇(x² − 3) gives x = −4 and x = 1. Substituting x = 1 yields log₁₁₇(−2) = log₁₁₇(−2), which is undefined in real numbers, so x = 1 is extraneous.

⚠️ How to check

• Graphically: plot both sides as functions and find intersections.
• Analytically: substitute back into the original equation and verify that all logarithm arguments are positive.
• Don't confuse: two undefined expressions that "look identical" (like log(−2) = log(−2)) do not make a valid solution.

⚠️ Common domain violations

• Negative arguments: log(−2) is not a real number.
• Zero arguments: log(0) is undefined.
• Example: x = 1 − √3 < 0 causes 2 log₂(1 − √3) to be undefined, so it is discarded.

🧮 Techniques for solving equations

🧮 Isolate the logarithm first

• Move all non-logarithmic terms to one side.
• Example: 2 − ln(x − 3) = 1 becomes ln(x − 3) = 1, then e¹ = x − 3, so x = e + 3.

🧮 Use logarithm properties to combine

• Product Rule: log(A) + log(B) = log(A·B).
• Quotient Rule: log(A) − log(B) = log(A/B).
• Power Rule: k·log(A) = log(A^k).
• Example: log₆(x + 4) + log₆(3 − x) = 1 becomes log₆[(x + 4)(3 − x)] = 1, then 6¹ = (x + 4)(3 − x), giving x = −3 and x = 2 (both valid).

🧮 Change of base when needed

• If logs have different bases, convert to a common base using the change-of-base formula: log base b of (x) = (log base c of (x)) / (log base c of (b)).
• Example: 1 + 2 log₄(x + 1) = 2 log₂(x). Since 4 = 2², log₄(x + 1) = (1/2) log₂(x + 1). Rewrite the equation with base 2 throughout, then combine.

🧮 Quadratics in disguise

• If the equation has (log(x))² or similar, substitute u = log(x) to get a standard quadratic.
• Example: (log₂(x))² − 2 log₂(x) − 3 = 0. Let u = log₂(x), so u² − 2u − 3 = 0, giving u = −1 or u = 3. Then x = 2⁻¹ = 1/2 or x = 2³ = 8.

📊 Solving inequalities with sign diagrams

📊 Why sign diagrams work

• Logarithmic functions are continuous on their domains, so they change sign only at zeros or domain boundaries.
• Move all terms to one side to get r(x) ≥ 0 (or < 0, etc.), then analyze where r(x) is positive or negative.

📊 Steps for sign diagrams

1. Find the domain: identify where all logarithms are defined (arguments > 0).
2. Find zeros: solve r(x) = 0.
3. Choose test values: pick values in each interval (preferably powers of the base for easy evaluation).
4. Determine signs: evaluate r at test values to see if r is (+) or (−) in each interval.
5. Read the solution: select intervals where the inequality holds.

📊 Example walkthrough

• Solve (1 / (ln(x) + 1)) ≤ 1.
• Rearrange: (1 / (ln(x) + 1)) − 1 ≤ 0, which simplifies to (ln(x) / (ln(x) + 1)) ≥ 0.
• Domain: x > 0 and ln(x) + 1 ≠ 0, so x ≠ 1/e. Domain is (0, 1/e) ∪ (1/e, ∞).
• Zeros: ln(x) = 0 gives x = 1.
• Test values: use powers of e: 0 < 1/e² < 1/e < 1/√e < 1 < e.
• Sign analysis: r(1/e²) = (−2)/(−2 + 1) = 2 > 0; test other intervals similarly.
• Solution: (0, 1/e) ∪ [1, ∞).

📊 Compound inequalities

• For a ≤ log(x) ≤ b, solve each piece separately and take the intersection.
• Example: 7.8 ≤ −log[H⁺] ≤ 8.5 becomes −7.8 ≥ log[H⁺] ≥ −8.5. Solve each: [H⁺] ≤ 10⁻⁷·⁸ and [H⁺] ≥ 10⁻⁸·⁵, so 10⁻⁸·⁵ ≤ [H⁺] ≤ 10⁻⁷·⁸.

🔄 Finding inverses of logarithmic functions

🔄 Standard procedure

1. Write y = f(x).
2. Interchange x and y.
3. Solve for y (this is f⁻¹(x)).

🔄 Example: f(x) = log(x) / (1 − log(x))

• Write y = log(x) / (1 − log(x)).
• Interchange: x = log(y) / (1 − log(y)).
• Solve: x(1 − log(y)) = log(y), so x − x·log(y) = log(y), hence x = (x + 1)·log(y), giving log(y) = x/(x + 1).
• Rewrite as exponential: y = 10^(x/(x+1)).
• Therefore f⁻¹(x) = 10^(x/(x+1)).

🔄 Verifying the inverse

• Check that (f⁻¹ ∘ f)(x) = x for all x in the domain of f.
• Check that (f ∘ f⁻¹)(x) = x for all x in the domain of f⁻¹.
• The range of f equals the domain of f⁻¹, and vice versa.

🔄 Composition of inverses

• If f = g ∘ h, then f⁻¹ = h⁻¹ ∘ g⁻¹ (the order reverses).
• Example: f(x) = log(x) / (1 − log(x)) can be written as g(h(x)) where h(x) = log(x) and g(x) = x/(1 − x). Then f⁻¹ = h⁻¹ ∘ g⁻¹.

🧪 Applied examples

🧪 pH and hydrogen ion concentration

• Recall: pH = −log[H⁺], where [H⁺] is hydrogen ion concentration in moles per liter.
• Example: to breed Ippizuti fish, pH must satisfy 7.8 ≤ pH ≤ 8.5.
• This means 7.8 ≤ −log[H⁺] ≤ 8.5, or −7.8 ≥ log[H⁺] ≥ −8.5.
• Solving: 10⁻⁸·⁵ ≤ [H⁺] ≤ 10⁻⁷·⁸ (approximately 3.16 × 10⁻⁹ ≤ [H⁺] ≤ 1.58 × 10⁻⁸).

🧪 Graphical verification

• Plot f(x) = −log(x) and the horizontal lines y = 7.8 and y = 8.5.
• The graph of f lies between the two lines on the interval [10⁻⁸·⁵, 10⁻⁷·⁸].

📋 Summary table: equation-solving strategies

Situation	Strategy	Example
Same base on both sides	Equate arguments	log₂(x) = log₂(5) → x = 5
Log equals a number	Convert to exponential	log₂(x) = 3 → x = 2³ = 8
Multiple logs, same base	Combine using properties, then solve	log(A) + log(B) = c → log(A·B) = c
Different bases	Change to common base first	log₄(x) and log₂(x) → express log₄ in terms of log₂
(log(x))² or similar	Substitute u = log(x), solve quadratic	u² − 2u − 3 = 0

Don't confuse: algebraic solutions vs. valid solutions—always check domain restrictions.

🧭 Overview

🧠 One-sentence thesis

Exponential and logarithmic functions model diverse real-world phenomena—from compound interest and population growth to radioactive decay and cooling—using mathematically similar equations that differ primarily in their rate constants and physical interpretations.

📌 Key points (3–5)

• Compound interest models: Both discrete compounding (Equation 6.2) and continuous compounding (Equation 6.3) use exponential functions; continuous compounding involves the natural base e.
• Growth and decay share structure: Uninhibited growth (populations, cells) and radioactive decay follow the same exponential form A(t) = A₀e^(kt), differing only in the sign of the rate constant k.
• Limited growth models: Newton's Law of Cooling and logistic growth describe situations where a quantity approaches a limiting value rather than growing indefinitely.
• Common confusion: The same mathematical form (exponential function) applies to seemingly unrelated contexts—financial, biological, physical—but the constants and variables have different units and meanings.
• Logarithms linearize data: Taking logarithms of data can transform exponential or power relationships into linear ones, making it easier to fit models and estimate parameters.

💰 Financial applications of exponential functions

💵 Simple vs. compound interest

Simple interest: I = Prt, where P is principal, r is annual rate, t is time in years; total amount A = P(1 + rt).

• Simple interest pays only on the original principal.
• Compound interest pays "interest on interest."

Example: $100 at 5% annual rate for one year yields $105 with simple interest. If compounded semiannually, the same investment yields $105.06 because interest earned in the first half-year itself earns interest in the second half.

🔄 Compound interest formula

Equation 6.2: If principal P is invested at annual rate r, compounded n times per year, the amount after t years is A(t) = P(1 + r/n)^(nt).

• The more frequently interest is compounded (larger n), the more money accumulates.
• The effect of increasing compounding frequency diminishes as n grows large.

Example: $2000 at 7.125% compounded monthly (n = 12) gives A(t) = 2000(1.0059375)^(12t). After 5 years, A(5) ≈ $2852.92.

♾️ Continuous compounding

Equation 6.3: If interest is compounded continuously, A(t) = Pe^(rt).

• As the number of compoundings per year approaches infinity, the formula converges to the continuous model.
• The natural base e arises from the limit: as n → ∞, (1 + 1/n)^n → e.
• Don't confuse: Monthly vs. continuous compounding yields similar results for typical interest rates; the difference is often less than 1% over decades.

Example: Comparing monthly and continuous compounding over 35 years at 7.125% on $2000 yields about $24,035 vs. $24,213—a difference under 1%.

🌱 Biological growth models

📈 Uninhibited growth

Equation 6.4 (Law of Uninhibited Growth): N(t) = N₀e^(kt), where N₀ is initial population, k > 0 is the growth rate constant, and the instantaneous rate of change at time t equals kN(t).

• The premise: the rate at which a population increases is directly proportional to the current population size.
• The more organisms present, the faster reproduction occurs.
• This model applies to early-stage populations with abundant resources.

Example: A culture of 12,000 cells grows to 5 million in one week. Using N(t) = 12e^(kt) (in thousands), solving 12e^(7k) = 5000 gives k ≈ 0.8618, an 86.18% daily growth rate.

🧬 Radioactive decay

Equation 6.5 (Radioactive Decay): A(t) = A₀e^(kt), where A₀ is initial amount, k < 0 is the decay constant, and the instantaneous rate of change equals kA(t).

• The rate of decay is directly proportional to the amount remaining.
• The model is identical to uninhibited growth except k is negative.
• Half-life: the time it takes for half the substance to decay; used to determine k.

Example: Iodine-131 has a half-life of 8 days. Starting with 5 grams, A(8) = 2.5 gives k = −ln(2)/8 ≈ −0.08664, meaning an 8.664% daily loss.

Don't confuse: The decay constant k is always negative, but half-life is always positive; k = −ln(2)/h where h is half-life.

🌡️ Limited growth and cooling models

🥤 Newton's Law of Cooling (Warming)

Equation 6.6: T(t) = T_a + (T₀ − T_a)e^(−kt), where T₀ is initial temperature, T_a is ambient (surrounding) temperature, and k > 0.

• The rate of temperature change is proportional to the temperature difference between the object and its surroundings, not the object's temperature itself.
• The object approaches ambient temperature asymptotically.
• The same equation applies to both cooling (T₀ > T_a) and warming (T₀ < T_a).

Example: A 40°F roast in a 350°F oven reaches 125°F after 2 hours. Using T(t) = 350 − 310e^(−kt) and T(2) = 125 gives k ≈ 0.1602. The roast reaches 165°F (done) in about 3.22 hours.

Don't confuse: The constant k is always positive; whether the function increases or decreases depends on whether T₀ is below or above T_a.

📊 Logistic growth

Equation 6.7 (Logistic Growth): N(t) = L / (1 + Ce^(−kLt)), where L is the limiting population, N₀ is initial population, C = L/N₀ − 1, and k > 0.

• Combines uninhibited growth with a limiting factor: growth rate depends on both current population and available room to grow.
• As t → ∞, N(t) → L (the population approaches but never exceeds the limit).
• The inflection point (point of diminishing returns) occurs at half the limiting population; before this point, growth resembles uninhibited growth; after, it resembles limited growth.

Example: Rumor spread modeled by N(t) = 84 / (1 + 2799e^(−t)) (in hundreds). Initially N(0) = 3 people; as t → ∞, N(t) → 84 (8400 people). Half the limit (4200 people) is reached at t ≈ 7.937 days.

📐 Applications of logarithms

🔐 Password entropy

Information entropy: H = L log₂(N), where L is password length in characters and N is the number of possible symbols per character.

• Higher entropy indicates a stronger password.
• Entropy is measured in bits.

Example: A 7-character case-sensitive password using letters and numbers has N = 62 symbols, so H = 7 log₂(62) ≈ 41.68 bits. To achieve 50 bits with 7 characters requires N = 2^(50/7) ≈ 142 symbols.

🩸 Buffer solutions and pH

The excerpt mentions the Henderson-Hasselbalch equation applied to blood pH:

pH = 6.1 + log(800/x)

where x is the partial pressure of CO₂ in arterial blood (in torr).

Example: If pH = 7.4, then log(800/x) = 1.3, so x = 800/10^(1.3) ≈ 40.09 torr.

📉 Linearizing data with logarithms

Scientists use logarithms to transform nonlinear relationships into linear ones for easier analysis.

Assumed model	Take logs	Result	What to plot
N = Bt^A (power)	ln(N) = A ln(t) + ln(B)	Y = AX + ln(B)	ln(N) vs. ln(t)
N = Be^(At) (exponential)	ln(N) = At + ln(B)	Y = AX + ln(B)	ln(N) vs. t

Example: WHO data on H1N1 cases were fit to multiple models (quadratic, power, exponential, logistic). The logistic model N = 10739 / (1 + 42.416e^(−0.268t)) fit well and made scientific sense (finite limiting population), whereas the quadratic model predicted unlimited growth.

Don't confuse: A good statistical fit (high R²) does not guarantee the model makes physical or biological sense; always consider the underlying science.

🧭 Overview

🧠 One-sentence thesis

Conic sections are curves formed by slicing a double-napped cone with a plane, and the angle of the slice determines which type of curve results.

📌 Key points (3–5)

• What conics are: curves created by intersecting a plane with a double-napped cone.
• How they are formed: the orientation of the slicing plane determines the resulting shape.
• Circle formation: a horizontal plane slicing the cone produces a circle.
• Ellipse formation: tilting the plane slightly from horizontal produces an ellipse.
• Common confusion: the difference between shapes depends on the angle of the plane—horizontal gives a circle, slight tilt gives an ellipse.

📐 What are conic sections

📐 Definition and origin

Conic Sections: curves formed by slicing a double-napped cone with a plane.

• The name literally means "sections of a cone."
• A double-napped cone is a three-dimensional cone shape that extends in both directions from a central vertex (like two ice cream cones joined at their tips).
• The resulting curve depends entirely on how the plane intersects the cone.

✂️ The slicing method

• Visualize a plane (a flat surface) cutting through the cone at different angles.
• The intersection between the plane and the cone's surface creates the conic curve.
• Different angles and positions of the plane produce different types of conics.

🔵 Types of conics introduced

🔵 Circle

• How it forms: slice the cone with a horizontal plane (perpendicular to the cone's axis).
• The plane cuts through the cone parallel to the base.
• This produces a closed, perfectly round curve.

🥚 Ellipse

• How it forms: tilt the horizontal plane "ever so slightly."
• The plane is no longer perpendicular to the cone's axis but still intersects only one nappe (one half of the double cone).
• This produces a closed, oval-shaped curve.

🔍 Distinguishing circle from ellipse

• Circle: plane is exactly horizontal (no tilt).
• Ellipse: plane has a slight tilt from horizontal.
• Both are closed curves, but the tilt changes the symmetry from perfectly round to elongated.

🧭 Overview

🧠 One-sentence thesis

A circle is fully determined by its center point and radius, and its equation can be derived from the distance formula and manipulated through completing the square.

📌 Key points (3–5)

• Definition: A circle is the set of all points at a fixed distance (radius) from a center point.
• Standard equation: The equation (x − h)² + (y − k)² = r² describes a circle with center (h, k) and radius r.
• Completing the square: Expanded circle equations can be converted to standard form by grouping variables and completing the square.
• Common confusion: Not every equation of the form (x − h)² + (y − k)² = c represents a circle—c must be positive.
• Unit circle: The circle centered at the origin with radius 1 has the special equation x² + y² = 1.

📐 Geometric definition and derivation

📍 What defines a circle

Definition 7.1: A circle with center (h, k) and radius r > 0 is the set of all points (x, y) in the plane whose distance to (h, k) is r.

• The definition is based purely on distance: every point on the circle is exactly r units away from the center.
• This is not about the area inside or the circumference; it's about the collection of points satisfying the distance condition.

🧮 From distance to equation

• A point (x, y) is on the circle if and only if its distance to (h, k) equals r.
• Using the distance formula: r = square root of [(x − h)² + (y − k)²]
• Squaring both sides (valid since r > 0) gives the standard form.

Standard equation:

(x − h)² + (y − k)² = r²

• Example: Center (−2, 3) and radius 5 gives (x − (−2))² + (y − 3)² = 5², which simplifies to (x + 2)² + (y − 3)² = 25.

🔄 Working with circle equations

📖 Reading the standard form

• From (x + 2)² + (y − 1)² = 4, identify:
  • x + 2 is x − h, so h = −2
  • y − 1 is y − k, so k = 1
  • r² = 4, so r = 2
• Center: (−2, 1); radius: 2

🔧 Completing the square

When given an expanded equation like x² + 4x + y² − 2y + 1 = 0, convert to standard form:

Steps:

1. Group same variables together; move constant to the other side
2. Complete the square on both variables
3. Divide by the coefficient of the squared terms (must be the same for circles)

• Example: 3x² − 6x + 3y² + 4y − 4 = 0
  • Group: 3(x² − 2x) + 3(y² + 4/3 y) = 4
  • Complete: 3(x² − 2x + 1) + 3(y² + 4/3 y + 4/9) = 4 + 3(1) + 3(4/9)
  • Factor: 3(x − 1)² + 3(y + 2/3)² = 25/3
  • Divide by 3: (x − 1)² + (y + 2/3)² = 25/9
  • Center: (1, −2/3); radius: 5/3

⚠️ When it's not a circle

• Equations like (x − 3)² + (y + 1)² = 0 or (x − 3)² + (y + 1)² = −1 do not represent circles.
• The right side must be positive for a circle to exist.
• Zero right side: only the single point (3, −1) satisfies the equation.
• Negative right side: no real points satisfy the equation (sum of squares cannot be negative).

🎯 Special cases and applications

⭕ The unit circle

Definition 7.2: The Unit Circle is the circle centered at (0, 0) with radius 1. Its equation is x² + y² = 1.

• This is the most important circle in mathematics.
• It's the standard form with h = 0, k = 0, and r = 1.
• Example: Find points with y-coordinate √3/2:
  • Substitute: x² + (√3/2)² = 1
  • Solve: x² + 3/4 = 1, so x² = 1/4, giving x = ±1/2
  • Points: (1/2, √3/2) and (−1/2, √3/2)

📏 Using diameter endpoints

• If two points are endpoints of a diameter, the center is their midpoint.
• The radius is half the distance between the endpoints.
• Example: Endpoints (−1, 3) and (2, 4):
  • Midpoint (center): ((−1+2)/2, (3+4)/2) = (1/2, 7/2)
  • Distance: square root of [(2−(−1))² + (4−3)²] = √10
  • Radius: √10/2
  • Equation: (x − 1/2)² + (y − 7/2)² = 10/4

Markdown notes complete.

🧭 Overview

🧠 One-sentence thesis

Parabolas can be defined geometrically as the set of all points equidistant from a fixed point (focus) and a fixed line (directrix), and this distance-based definition leads directly to the algebraic equations we use to graph and analyze them.

📌 Key points (3–5)

• Geometric definition: A parabola is the set of all points equidistant from a focus point F and a directrix line D.
• Two orientations: Parabolas can be vertical (opening up/down) or horizontal (opening left/right), determined by which variable is squared.
• Key components: Every parabola has a vertex (closest point to focus), focus, directrix, and latus rectum (line segment through focus parallel to directrix).
• Common confusion: The sign of p determines direction—positive p means up (vertical) or right (horizontal); negative p means down or left; don't confuse the focal length |p| with the directed distance p.
• Standard forms: Vertical parabolas use (x - h)² = 4p(y - k); horizontal parabolas use (y - k)² = 4p(x - h).

📐 Geometric foundation

📏 Distance-based definition

A parabola is the set of all points equidistant from F (focus) and D (directrix).

• The focus F is a fixed point in the plane.
• The directrix D is a fixed line that does not contain F.
• Every point on the parabola has the same distance to F as it does to the nearest point on D.
• Example: If a point is 5 units from the focus, it must also be 5 units from the directrix (measured perpendicularly).

🎯 The vertex

• The vertex is the point on the parabola closest to the focus.
• It lies exactly halfway between the focus and the directrix.
• The vertex serves as the reference point for the standard equation forms.

📊 Directed distance p

• p represents the directed distance from vertex to focus (same as vertex to directrix).
• "Directed" means p carries sign information about orientation:
  • p > 0: focus is above vertex (vertical) or right of vertex (horizontal)
  • p < 0: focus is below vertex (vertical) or left of vertex (horizontal)
• The focal length is |p|, the absolute value (always positive).

🔢 Vertical parabolas

📝 Standard equation form

The equation of a vertical parabola with vertex (h, k) and focal length |p| is: (x - h)² = 4p(y - k)

• Only the x variable is squared.
• If p > 0, the parabola opens upwards.
• If p < 0, the parabola opens downwards.
• This form comes from applying the distance formula to the geometric definition.

🧮 Deriving the equation

Starting with vertex at (0, 0), focus at (0, p), and directrix y = -p:

• Distance from (x, y) to focus (0, p) equals distance from (x, y) to directrix point (x, -p).
• Using distance formula: square root of (x² + (y - p)²) = square root of ((y + p)²)
• Squaring both sides and simplifying: x² = 4py
• This matches the quadratic form y = x²/(4p) with coefficient a = 1/(4p).

🔍 Key features to identify

Feature	How to find it	Example for (x + 1)² = -8(y - 3)
Vertex	(h, k) from standard form	(-1, 3)
Value of p	Solve 4p from equation	4p = -8, so p = -2
Focus	Move	p
Directrix	Move	p

🔄 Horizontal parabolas

📝 Standard equation form

The equation of a horizontal parabola with vertex (h, k) and focal length |p| is: (y - k)² = 4p(x - h)

• Only the y variable is squared (this is the key difference).
• If p > 0, the parabola opens to the right.
• If p < 0, the parabola opens to the left.
• The directrix is now a vertical line instead of horizontal.

↔️ Orientation rules

• Vertical parabolas: x is squared → opens up or down
• Horizontal parabolas: y is squared → opens left or right
• Don't confuse: The variable that is NOT squared determines whether the parabola is vertical or horizontal.

🛠️ Working with parabola equations

📐 The latus rectum and focal diameter

• The latus rectum is a line segment through the focus, parallel to the directrix, with endpoints on the parabola.
• The focal diameter (length of latus rectum) is |4p|.
• This measurement helps graph the parabola accurately by showing its width at the focus.
• Example: If 4p = -8, the focal diameter is 8, meaning the parabola is 8 units wide at the focus (4 units on each side).

🔧 Converting to standard form

When given an equation not in standard form:

1. Group variables: Put the squared variable on one side, the non-squared variable and constant on the other.
2. Complete the square: If needed, complete the square for the squared variable.
3. Factor: Factor out the coefficient of the non-squared variable.

Example: y² + 4y + 8x = 4

• Rearrange: y² + 4y = -8x + 4
• Complete square: y² + 4y + 4 = -8x + 8
• Factor: (y + 2)² = -8(x - 1)

🔎 Quick identification

• Circle vs parabola: In a circle equation, both x and y are squared; in a parabola, only one variable is squared.
• Which way it opens: Check the sign of p and which variable is squared.

🌟 Applications

📡 Reflective property

• Parabolic mirrors and dishes use the focus as a concentration point.
• Incoming parallel rays (like satellite signals) reflect off the parabola and concentrate at the focus—optimal receiver placement.
• Outgoing rays from a source at the focus (like a light bulb) reflect off the parabola as parallel beams—directional light in flashlights.

🛰️ Paraboloid of revolution

• A three-dimensional shape formed by rotating a parabola around its axis of symmetry.
• Every cross-section through the vertex is a parabola with the same focus.
• Used in satellite dishes, telescope mirrors, and flashlight reflectors.

📏 Solving design problems

Example approach: A satellite dish 12 feet wide with focus 2 feet above vertex—how deep?

• Set up equation with vertex at (0, 0): x² = 4py with p = 2, so x² = 8y
• The edge is 6 feet from center (half of 12 feet width)
• Substitute x = 6: 36 = 8y, so y = 4.5 feet deep

⚠️ Common pitfalls

🚫 Sign confusion

• Don't confuse p (directed distance, can be negative) with |p| (focal length, always positive).
• The sign of p tells direction; the magnitude |p| tells distance.

🚫 Vertex location

• The vertex is always between the focus and directrix, equidistant from both.
• To find vertex from focus and directrix: locate the midpoint of the perpendicular distance between them.

🚫 Standard form identification

• (x - h)² = 4p(y - k): vertical parabola
• (y - k)² = 4p(x - h): horizontal parabola
• The squared variable determines orientation, not the sign of p.

🧭 Overview

🧠 One-sentence thesis

An ellipse is defined by the sum of distances from any point on the curve to two fixed foci being constant, and this geometric property determines its equation, shape, and reflective behavior.

📌 Key points (3–5)

• Definition: An ellipse is the set of all points where the sum of distances to two foci equals a fixed distance d (which equals the length of the major axis).
• Standard equation: The form (x − h)²/a² + (y − k)²/b² = 1 describes an ellipse centered at (h, k), where a and b control horizontal and vertical stretching.
• Major vs minor axis: The longer axis is the major axis (containing the foci and vertices); the shorter is the minor axis (perpendicular to the major axis through the center).
• Common confusion: Which axis is major depends on which denominator is bigger—if a² > b², the major axis is horizontal; if b² > a², it's vertical.
• Eccentricity: The ratio e = c/a (distance center-to-focus over distance center-to-vertex) measures "roundness"—closer to 0 is more circular, closer to 1 is more elongated.

📐 Defining an ellipse

📐 The two-foci definition

Ellipse: Given two distinct points F₁ and F₂ (the foci) and a fixed distance d, an ellipse is the set of all points (x, y) such that the sum of the distances from F₁ to (x, y) and from F₂ to (x, y) equals d.

• Think of anchoring a string at two points and tracing a curve with a pencil keeping the string taut—the result is an ellipse.
• The fixed distance d is not arbitrary; it equals the length of the major axis (2a).
• Example: If the foci are 6 units apart and the sum of distances is 10, every point on the ellipse satisfies distance₁ + distance₂ = 10.

🎯 Key parts of an ellipse

Part	Definition
Center	Midpoint of the line segment connecting the two foci
Major axis	Line segment through the center and foci, connecting two opposite ends of the ellipse (the longer axis)
Minor axis	Line segment through the center, perpendicular to the major axis, connecting two opposite ends (the shorter axis)
Vertices	The two points where the ellipse intersects the major axis
Foci	The two fixed points used in the definition

• The center is also the midpoint of the vertices.
• The major axis is always longer than the minor axis.

🧮 Standard equation and its derivation

🧮 The standard form

Standard equation of an ellipse: For positive unequal numbers a and b, the equation of an ellipse with center (h, k) is
(x − h)²/a² + (y − k)²/b² = 1

• The values a and b tell you how far to move horizontally and vertically from the center to reach points on the ellipse.
• If a > b, the major axis is horizontal; if b > a, the major axis is vertical.
• Don't confuse with a circle: A circle has the same denominator for both squared terms; an ellipse has different denominators.

🔗 Relationship between a, b, and c

• The distance from the center to each focus is c.
• The relationship is: c = √(bigger denominator − smaller denominator).
• Specifically:
  • If a² > b², then c = √(a² − b²) and the foci lie left/right of the center.
  • If b² > a², then c = √(b² − a²) and the foci lie above/below the center.
• This comes from the derivation: using the definition and distance formula, the algebra yields b² = a² − c² (when a > b).

🔄 Connection to circles

• A circle's alternate standard form is (x − h)²/r² + (y − k)²/r² = 1 (same denominators).
• An ellipse can be thought of as a stretched circle: replace x with x/a and y with y/b in the unit circle x² + y² = 1.
• Both involve a sum of squares equal to 1; the difference is whether the denominators are equal (circle) or unequal (ellipse).

📊 Graphing and identifying ellipse features

📊 From standard form to graph

Steps:

1. Identify the center (h, k) from (x − h) and (y − k).
2. Identify a² and b²; compute a and b.
3. Determine which axis is major: compare a² and b²—the larger denominator corresponds to the major axis direction.
4. Draw a "guide rectangle" centered at (h, k) with width 2a and height 2b; sketch the ellipse inside.
5. Mark the vertices (on the major axis, a units from center) and endpoints of the minor axis (b units from center).
6. Compute c = √(larger − smaller) and mark the foci (c units from center along the major axis).

Example: For (x + 1)²/9 + (y − 2)²/25 = 1:

• Center: (−1, 2)
• a² = 9 → a = 3; b² = 25 → b = 5
• Since b² > a², the major axis is vertical (along x = −1).
• Vertices: (−1, 7) and (−1, −3) (5 units up/down from center)
• Endpoints of minor axis: (−4, 2) and (2, 2) (3 units left/right)
• c = √(25 − 9) = 4 → foci at (−1, 6) and (−1, −2)

🔧 Converting to standard form

Procedure (when the equation is not in standard form):

1. Group x terms together and y terms together; move the constant to the other side.
2. Complete the square for both x and y.
3. Divide both sides by the constant so the right side equals 1.

Example: x² + 4y² − 2x + 24y + 33 = 0

• Group: x² − 2x + 4y² + 24y = −33
• Factor out coefficient of y²: x² − 2x + 4(y² + 6y) = −33
• Complete squares: (x² − 2x + 1) + 4(y² + 6y + 9) = −33 + 1 + 36
• Simplify: (x − 1)² + 4(y + 3)² = 4
• Divide by 4: (x − 1)²/4 + (y + 3)²/1 = 1
• Now in standard form with center (1, −3), a = 2, b = 1.

🎯 Eccentricity and shape

🎯 What eccentricity measures

Eccentricity (e): the ratio of the distance from the center to a focus over the distance from the center to a vertex.
e = c/a

• For an ellipse, the foci are closer to the center than the vertices, so 0 < e < 1.
• The closer e is to 0, the more circular (round) the ellipse.
• The closer e is to 1, the more elongated (eccentric) the ellipse.

Example: An ellipse with e ≈ 0.66 is rounder than one with e ≈ 0.98.

🧩 Using eccentricity to find the equation

• Given vertices and eccentricity, you can find c from e = c/a, then use c = √(a² − b²) to solve for b².

Example: Vertices at (±5, 0) and e = 1/4:

• a = 5 (distance from center to vertex)
• e = c/a → 1/4 = c/5 → c = 5/4
• c = √(a² − b²) → 5/4 = √(25 − b²) → b² = 375/16
• Equation: x²/25 + 16y²/375 = 1

🔊 Reflective property and applications

🔊 The whispering gallery effect

• Sound waves (or light) emanating from one focus reflect off the ellipse and converge at the other focus.
• This is exploited in whispering galleries: a person whispering at one focus can be heard clearly by someone at the other focus, even across a large room.
• Famous examples: St. Paul's Cathedral (London), National Statuary Hall (Washington, D.C.), Cincinnati Museum Center.

🏛️ Practical example

Example: A whispering gallery is 100 feet wide and 40 feet tall (half-ellipse cross section). How far from the outer wall should two people stand to be at the foci?

• Model the ellipse centered at (0, 0): a = 50 (half the width), b = 40 (the height).
• Equation: x²/50² + y²/40² = 1
• Compute c = √(50² − 40²) = √900 = 30
• The foci are 30 units from the center, so 50 − 30 = 20 feet from the outer wall.

💡 Other applications

• Extracorporeal shock-wave lithotripsy: Uses the reflective property of an ellipsoid to focus shock waves on kidney stones (one focus is the stone, the other is the shock-wave source).
• Planetary orbits: Earth's orbit around the Sun is an ellipse with the Sun at one focus and eccentricity e ≈ 0.0167 (nearly circular).

🧭 Overview

🧠 One-sentence thesis

Hyperbolas arise when we measure points where the absolute difference of distances to two foci equals a constant, and they are distinguished from ellipses by a subtraction (rather than addition) in their defining equation and by branches that open outward along asymptotes.

📌 Key points (3–5)

• What a hyperbola is: the set of all points where the absolute value of the difference of distances to two foci equals a fixed distance d.
• Key structural features: two branches, a center (midpoint of the foci), a transverse axis (through the vertices), a conjugate axis (perpendicular), and asymptotes that guide the branches.
• Standard forms: horizontal hyperbolas have the form (x − h)²/a² − (y − k)²/b² = 1; vertical hyperbolas swap the roles of x and y.
• Common confusion—hyperbola vs ellipse: ellipses add distances to foci (sum of squares in the equation); hyperbolas subtract distances (difference of squares); also, AC < 0 signals a hyperbola, while AC > 0 signals an ellipse or circle.
• Why it matters: hyperbolas model positioning problems (e.g., LORAN navigation, earthquake epicenter location) and appear in engineering structures like cooling towers and telescopes.

📐 Definition and core geometry

📐 What defines a hyperbola

Hyperbola: Given two distinct points F₁ and F₂ (the foci) and a fixed distance d, a hyperbola is the set of all points (x, y) such that the absolute value of the difference of the distances from F₁ and F₂ to (x, y) equals d.

• The key operation is subtraction of distances, not addition (which gives an ellipse).
• The hyperbola has two branches (two separate curves).
• Example: if the distance from F₁ to a point minus the distance from F₂ to that point equals d, the point lies on one branch; if the reverse difference equals d, it lies on the other branch.

🎯 Key geometric elements

Element	Definition
Center	Midpoint of the line segment joining the two foci
Transverse axis	Line segment connecting the two vertices (opposite ends of the hyperbola); contains the center and foci
Vertices	Points where the hyperbola intersects the transverse axis
Conjugate axis	Line segment through the center, perpendicular to the transverse axis, with length equal to the segment through a vertex that connects the asymptotes
Asymptotes	Lines that the branches approach as x and y grow large; they pass through the center and the corners of the guide rectangle

• The guide rectangle is constructed using the transverse and conjugate axes; its diagonals lie along the asymptotes.
• Don't confuse: the conjugate axis does not intersect the hyperbola itself (unlike the transverse axis, which does at the vertices).

🧮 Standard equations and parameters

🧮 Horizontal hyperbola (opens left and right)

Standard form (horizontal): (x − h)²/a² − (y − k)²/b² = 1

• Center: (h, k)
• Transverse axis: horizontal line y = k
• Conjugate axis: vertical line x = h
• Vertices: (h ± a, k)
• The x² term is positive; the y² term is subtracted.

🧮 Vertical hyperbola (opens up and down)

Standard form (vertical): (y − k)²/b² − (x − h)²/a² = 1

• Center: (h, k)
• Transverse axis: vertical line x = h
• Conjugate axis: horizontal line y = k
• Vertices: (h, k ± b)
• The y² term is positive; the x² term is subtracted.

🧮 Relationship among a, b, and c

• c: distance from the center to each focus.
• a: distance from the center to each vertex (half the transverse axis length).
• b: half the length of the conjugate axis.
• Formula: c² = a² + b² (note the plus sign, unlike the ellipse where c² = a² − b²).
• The fixed distance d from the definition equals 2a (the length of the transverse axis).

🧮 Asymptotes

• Slopes: ± b/a (for horizontal hyperbolas) or ± b/a (for vertical hyperbolas, but roles of a and b are swapped in the formula).
• Equations (horizontal hyperbola centered at (h, k)): y − k = ± (b/a)(x − h).
• Equations (vertical hyperbola centered at (h, k)): y − k = ± (b/a)(x − h).
• The asymptotes pass through the center and the corners of the guide rectangle.
• Example: for a horizontal hyperbola with center (2, 0), a = 2, b = 5, the asymptotes are y = ± (5/2)(x − 2), which simplify to y = (5/2)x − 5 and y = −(5/2)x + 5.

🔄 Converting to standard form

🔄 Procedure for completing the square

1. Group variables: move all x terms together, all y terms together, and the constant to the other side.
2. Complete the square in both x and y as needed.
3. Divide both sides by the constant so the right-hand side becomes 1.

🔄 Example walkthrough

• Given: 9y² − x² − 6x = 10
• Group: 9y² − (x² + 6x) = 10
• Complete the square in x: 9y² − (x² + 6x + 9) = 10 − 9 → 9y² − (x + 3)² = 1
• Divide by 1 (already done): y²/(1/9) − (x + 3)²/1 = 1
• Standard form: (y − 0)²/(1/9) − (x − (−3))²/1 = 1
• This is a vertical hyperbola with center (−3, 0), b² = 1/9 (so b = 1/3), a² = 1 (so a = 1).
• Vertices: (−3, ±1/3); foci: c = √(a² + b²) = √(1 + 1/9) = √(10/9) = √10/3, so foci are at (−3, ±√10/3).
• Asymptotes: slopes ± b/a = ± (1/3)/1 = ±1/3, so y = (1/3)(x + 3) and y = −(1/3)(x + 3).

🔍 Distinguishing conic sections

🔍 General form and the role of A and C

• General equation: Ax² + Cy² + Dx + Ey + F = 0
• The key is the product AC (the coefficients of x² and y²).

Condition	Conic section
Only one variable squared	Parabola
A = C (both squared, same coefficient)	Circle
A ≠ C but A and C have the same sign (AC > 0)	Ellipse
A and C have different signs (AC < 0)	Hyperbola

• Example: 9x² − 4y² − 36x − 24y − 36 = 0 has A = 9, C = −4, so AC = −36 < 0 → hyperbola.
• Don't confuse: if both A and C are positive but unequal, it's an ellipse (e.g., 9x² + 4y² = 36); if one is positive and one is negative, it's a hyperbola.

🔍 Which standard form to use

• If the x² term has the positive coefficient (or equivalently, the y² term is subtracted), use the horizontal hyperbola form.
• If the y² term has the positive coefficient (or equivalently, the x² term is subtracted), use the vertical hyperbola form.

🌍 Applications: positioning problems

🌍 Trilateration and LORAN

• Principle: if an observer at one focus hears a signal earlier than an observer at another focus, the source is closer to the first observer.
• The time difference multiplied by the speed of the signal gives the difference in distances.
• This difference defines a hyperbola with the two observers at the foci.

🌍 Locating Sasquatch (Example 7.5.4)

• Setup: Jeff at (−5, 0), Carl at (5, 0); Jeff hears a call 9 seconds earlier than Carl.
• Speed of sound: 760 miles/hour = 760/3600 miles/second.
• Distance difference: (760/3600) × 9 = 1.9 miles.
• This means the source is 1.9 miles closer to Jeff, so it lies on a hyperbola with foci at Jeff and Carl and fixed distance d = 1.9.
• Since d = 2a, we have a = 0.95, so a² = 0.9025.
• Distance from center to focus: c = 5.
• Using c² = a² + b²: b² = 25 − 0.9025 = 24.0975.
• Equation: x²/0.9025 − y²/24.0975 = 1.
• Since Sasquatch is closer to Jeff (on the left), it lies on the western branch (left branch) of the hyperbola.

🌍 Adding a third observer (Example 7.5.5)

• Kai at (−5, 6) hears the call 18 seconds after Jeff.
• Distance difference: (760/3600) × 18 = 3.8 miles.
• New hyperbola with foci at Jeff (−5, 0) and Kai (−5, 6); center at (−5, 3).
• This is a vertical hyperbola: d = 2b = 3.8, so b = 1.9, b² = 3.61.
• Distance from center to focus: c = 3.
• Using c² = a² + b²: a² = 9 − 3.61 = 5.39.
• Equation: (y − 3)²/3.61 − (x + 5)²/5.39 = 1.
• Kai heard the call later, so Sasquatch is on the southern branch (below the center).
• The intersection of the western branch of the first hyperbola and the southern branch of the second hyperbola gives the exact location: approximately (−0.9629, −0.8113).
• Don't confuse: you need at least three observers (two hyperbolas) to pinpoint a location; one hyperbola alone only narrows the source to a curve.

🌍 Earthquake epicenter location

• Similar principle: seismic P-waves travel at a known speed; recording stations measure arrival times.
• Time differences between stations define hyperbolas with the stations at the foci.
• The epicenter is the intersection of multiple hyperbolas.
• Note: this technique assumes the Earth is flat over small distances; for larger distances, the curvature of the Earth must be considered.

🏗️ Other applications

🏗️ Hyperboloids of revolution

• Natural draft cooling towers are often shaped as hyperboloids (3D surfaces formed by rotating a hyperbola around an axis).
• Each vertical cross-section is a hyperbola.
• Example problem (Exercise 32): given the tower's width at the base, top, and narrowest point, determine the height by finding the hyperbola equation and solving for the appropriate y-coordinate.

🏗️ Cassegrain telescope

• Uses the reflective property of the hyperbola (light rays directed toward one focus reflect toward the other focus) combined with a parabolic mirror.
• This design creates a compact, powerful telescope.

🔢 Eccentricity of hyperbolas

🔢 Definition

Eccentricity e: e = (distance from center to focus) / (distance from center to vertex) = c/a

• For hyperbolas, since c > a (because c² = a² + b² and b > 0), we have e > 1 always.
• Compare with ellipses: for ellipses, e < 1; for circles, e = 0; for parabolas, e = 1.

🔢 What eccentricity tells you

• Larger e means the branches open more widely (the hyperbola is "flatter").
• Smaller e (closer to 1) means the branches are more tightly curved around the vertices.
• Example: if vertices are at (±3, 0) and e = 2, then c/a = 2, so c = 2a = 6. Using c² = a² + b²: 36 = 9 + b², so b² = 27. Equation: x²/9 − y²/27 = 1.

🧭 Overview

🧠 One-sentence thesis

Gaussian Elimination systematically transforms any system of linear equations into triangular form through row operations, enabling us to determine whether the system has one solution, infinitely many solutions, or no solution at all.

📌 Key points (3–5)

• What a linear equation is: an equation of the form a₁x₁ + a₂x₂ + ... + aₙxₙ = c where variables appear only to the first power with number coefficients.
• Three possible outcomes: consistent independent (exactly one solution), consistent dependent (infinitely many solutions with free variables), or inconsistent (no solution, contradiction).
• The algorithm: use three moves—swap equations, multiply an equation by a nonzero number, or replace an equation with itself plus a multiple of another—to reach triangular form.
• Common confusion: when elimination produces 0 = 0 vs. 0 = 6—the first means the equations are equivalent (dependent system), the second is a contradiction (inconsistent system).
• Why triangular form works: once in triangular form, back-substitution becomes straightforward because each equation has a leading variable with coefficient 1 and increasing subscripts.

📐 Linear equations and systems

📐 What counts as linear

Linear equation in n variables: an equation of the form a₁x₁ + a₂x₂ + ... + aₙxₙ = c where a₁, a₂, ..., aₙ and c are real numbers and at least one of the a-coefficients is nonzero.

• The key requirement: variables appear only to the first power.
• Coefficients must be fixed numbers, not expressions involving variables.
• Examples of non-linear equations: x² + y = 1 (variable squared), xy = 5 (variables multiplied together), e^(2x) + ln(y) = 1 (transcendental functions).
• Special cases: x = 5 can be considered linear in two variables by setting a₁ = 1, a₂ = 0, c = 5.
• Don't confuse: identities (0 = 0, always true) and contradictions (0 ≠ 0, never true) are not considered linear equations even though they can arise during solving.

🔗 Systems of equations

System of equations: a collection of two or more equations where we seek all pairs (or tuples) of values that satisfy all equations simultaneously, written with a curly bracket notation.

• The goal is to find intersection points where all equations are satisfied at once.
• For two variables: geometrically finding where lines intersect.
• For three variables: geometrically finding where planes intersect in three-dimensional space.
• Example: The system {y = f(x), y = g(x)} asks for all (x, y) satisfying both equations.

🎯 The three types of solutions

✅ Consistent independent

• What it means: exactly one solution exists; no free variables.
• The system has a unique intersection point.
• Example from the excerpt: solving {2x - y = 1, y = 3} gives the single solution (2, 3).
• Algebraically: after elimination, every variable has a determined value.
• Geometrically (two variables): two non-parallel lines intersecting at one point.

♾️ Consistent dependent

• What it means: infinitely many solutions exist; at least one free variable (parameter).
• During elimination, you get an identity like 0 = 0.
• Example from the excerpt: {2x - 4y = 6, 3x - 6y = 9} reduces to 0 = 0 because the equations are equivalent.
• The solution is expressed parametrically: let a free variable equal t, then express other variables in terms of t.
• Example solution format: {(t, ½t - 3/2) | -∞ < t < ∞} means for any value of t, you get a solution.
• Geometrically (two variables): the two equations represent the same line.
• Don't confuse: 0 = 0 (dependent) with 0 = 6 (inconsistent)—the first means equations are equivalent, the second means they contradict.

❌ Inconsistent

• What it means: no solution exists; the system contains a contradiction.
• During elimination, you get a false statement like 0 = 6 or 0 = -18.
• Example from the excerpt: {6x + 3y = 9, 4x + 2y = 12} leads to 0 = -18 after elimination.
• Geometrically (two variables): parallel distinct lines that never intersect.
• The system is impossible to satisfy.

🔧 The three allowed moves (Theorem 8.1)

🔄 Interchange equations

• What: swap the position of any two equations.
• Why it's safe: the order of equations doesn't change which points satisfy all of them.
• When to use: to get a convenient leading variable (especially one with coefficient 1) into the top position.
• Example from the excerpt: swapping E₁ and E₃ to get x (with coefficient 1) in the first equation.

✖️ Multiply an equation by a nonzero constant

• What: replace an equation with a nonzero multiple of itself.
• Why it's safe: multiplying both sides by the same nonzero number preserves equality.
• When to use: to make the leading variable's coefficient equal to 1.
• Example from the excerpt: replacing E₂ with (-1/2)E₂ to change -2y into y.
• Critical restriction: the multiplier must be nonzero (multiplying by zero would lose information).

➕ Add a multiple of one equation to another

• What: replace an equation with itself plus a nonzero multiple of another equation.
• Why it's safe: adding equals to equals preserves equality.
• When to use: to eliminate a variable from an equation.
• Example from the excerpt: replacing E₂ with -2E₁ + E₂ to eliminate x from the second equation.
• This is the core elimination step in Gaussian Elimination.

📋 Triangular form (Definition 8.3)

📋 The four requirements

A system is in triangular form when:

1. Variable subscripts increase left to right in each equation.
2. Leading variable has coefficient 1 in each equation.
3. Leading variable subscripts increase going down (each equation's leading variable has a larger subscript than the one above).
4. Equations without variables go at the bottom (identities like 0 = 0 or contradictions like 0 = 6).

🎯 Why triangular form matters

• Once in triangular form, back-substitution becomes mechanical.
• The last equation gives you one variable directly.
• Substitute upward into earlier equations to find remaining variables.
• Example from the excerpt: {x - y + z = 5, y - ½z = -3, z = 6} immediately gives z = 6, then y = 0, then x = -1.

🔍 What triangular form reveals

• If you get 0 = 0: the system is dependent; the variable that would have been the leading variable in that equation becomes free.
• If you get 0 = (nonzero): the system is inconsistent; stop—there's no solution.
• If every variable becomes a leading variable: the system is independent; exactly one solution exists.

🔄 The Gaussian Elimination algorithm

🔄 The systematic process

1. Start with the first variable (usually x or x₁).
2. Get coefficient 1 on that variable in the first equation (swap or multiply).
3. Eliminate that variable from all equations below (add multiples).
4. Move to the next variable and repeat steps 2–3 for the second equation.
5. Continue until you reach triangular form or discover a contradiction/identity.

📝 Tracking your work

• Label equations: E₁, E₂, E₃, etc., to track which equation is which.
• Document each move: "Replace E₂ with -2E₁ + E₂" makes your work verifiable.
• Example from the excerpt shows each step explicitly with arrows indicating the transformation.

⚠️ Common pitfalls

• Forgetting to check the original equations: after solving, substitute your answer into the original system, not the transformed one.
• Losing track of free variables: if an equation vanishes (becomes 0 = 0), remember that variable is free and must be parameterized.
• Inefficiency: the algorithm always works but may take extra steps; sometimes you can spot contradictions or dependencies earlier.

🧪 Parametric solutions

🧪 When parameters appear

• When a system is consistent dependent, at least one variable is free.
• A free variable is one that never becomes a leading variable in any equation.
• We assign the free variable a parameter (usually t, s, etc.) and express other variables in terms of it.

🧪 How to write parametric solutions

• Identify free variables: any variable without an equation where it's the leading variable.
• Assign parameters: let x₃ = s, x₄ = t, etc.
• Solve for other variables in terms of parameters using back-substitution.
• Write the solution set: {(expression in t, expression in t, t) | -∞ < t < ∞}.
• Example from the excerpt: {(t, ½t - 3/2) | -∞ < t < ∞} means x = t and y = ½t - 3/2 for any real t.

🧪 Multiple parameterizations

• Not unique: you can solve for different variables and get different-looking but equivalent solution sets.
• Example from the excerpt: solving 2x - 4y = 6 for y gives {(t, ½t - 3/2)}, solving for x gives {(2t + 3, t)}.
• Both describe the same line, just parameterized differently.
• Always check: substitute your parametric expressions back into the original equations to verify.

🧮 Special system types

🧮 Overdetermined systems

Overdetermined: more equations than variables.

• Example from the excerpt: three equations in two variables (x and y).
• Can be consistent or inconsistent: having extra equations doesn't guarantee a contradiction.
• Example: {x - y = 0, x + y = 2, -2x + y = -2} is overdetermined and inconsistent because the first two intersect at (1, 1) but the third doesn't pass through that point.

🧮 Underdetermined systems

Underdetermined: more variables than equations.

• Example from the excerpt: two equations in four variables (x₁, x₂, x₃, x₄).
• If consistent, necessarily dependent: not enough equations to determine all variables, so free variables must exist.
• Example solution: {(2 - s - t, 3s + 2t, s, t) | -∞ < s, t < ∞} with two free variables.

🧪 Application: mixture problems

🧪 Setting up mixture equations

• Identify unknowns: amounts of each component (e.g., x mL of 30% solution, y mL of 90% solution, w mL of water).
• Write volume equation: sum of all components equals total volume.
• Write concentration equation: sum of active ingredient from each component equals required amount in final mixture.
• Example from the excerpt: x + y + w = 500 (total volume) and 0.30x + 0.90y = 200 (total acid).

🧪 Interpreting parametric solutions

• Mixture problems often yield dependent systems with one free variable.
• The parameter represents a degree of freedom in the recipe.
• Physical constraints: amounts must be nonnegative (x ≥ 0, y ≥ 0, w ≥ 0).
• These constraints restrict the parameter range.
• Example from the excerpt: the algebraic solution allows -∞ < t < ∞, but physical constraints give 0 ≤ t ≤ 2500/9.

🧪 Practical use

• Flexibility: if you have limited stock of one ingredient, set that variable to the available amount and solve for the parameter.
• Example from the excerpt: if only 100 mL of 90% solution is available, set y = 100 and solve for t, then find required amounts of other components.

🧭 Overview

🧠 One-sentence thesis

Augmented matrices streamline solving systems of linear equations by encoding coefficients and constants into a rectangular array, allowing row operations to replace equation manipulations and ultimately producing solutions through row echelon or reduced row echelon form.

📌 Key points (3–5)

• What an augmented matrix is: a rectangular array that encodes a system of linear equations, separating variable coefficients from constants with a vertical bar.
• Why matrices help: row operations on matrices mirror equation operations but focus only on coefficients, eliminating the need to rewrite variables repeatedly.
• Row echelon vs. reduced row echelon form: row echelon form produces triangular systems requiring back-substitution; reduced row echelon form yields solutions directly without back-substitution.
• Common confusion: row echelon form is not unique (a matrix has infinitely many), but reduced row echelon form is unique for any given matrix.
• Gauss-Jordan Elimination: the process of putting a matrix into reduced row echelon form, which automates solution-finding for linear systems.

📐 What matrices are and how they encode systems

📐 Matrix definition and structure

Matrix: a rectangular array of real numbers enclosed in square brackets.

• Size (dimension): described as rows × columns (e.g., a 2 × 3 matrix has 2 rows and 3 columns).
• Entries: individual numbers in the matrix, labeled with double subscripts (first = row number, second = column number).
  • Rows are numbered top to bottom; columns are numbered left to right.
  • Example: if A = [3 0 -1; 2 -5 10], then a₁₁ = 3, a₁₂ = 0, a₁₃ = -1, a₂₁ = 2, a₂₂ = -5, a₂₃ = 10.
• Matrices are denoted by uppercase letters (A, B, C); entries by the corresponding lowercase letter with subscripts.

🔗 Encoding a system into an augmented matrix

• Each equation becomes a row; each variable and the constant gets its own column.
• A vertical bar | separates the left-hand side (variable coefficients) from the right-hand side (constants).
• Important: if a variable is missing from an equation, record its coefficient as 0.

Example: The system
2x + 3y - z = 1
10x - z = 2
4x - 9y + 2z = 5
encodes as:

  2   3  -1 |  1
 10   0  -1 |  2
  4  -9   2 |  5

(Note the 0 for y in the second equation.)

Augmented matrix: the matrix formed by appending the constant column to the coefficient matrix.

🔄 Row operations and their purpose

🔄 The three allowed row operations

The excerpt presents Theorem 8.2, which lists operations that produce an equivalent system:

Operation	Description	Effect
Interchange two rows	Swap any two rows	Reorders equations
Multiply a row by a nonzero constant	Replace a row with a nonzero multiple of itself	Scales an equation
Add a multiple of one row to another	Replace a row with itself plus a nonzero multiple of another row	Eliminates variables

• These operations mirror the equation-level moves from Gaussian Elimination (Theorem 8.1).
• The goal is to transform the matrix into a form where the solution is easier to read.

🎯 Why row operations work

• All moves are determined entirely by the coefficients of the variables, not the variables themselves.
• This is analogous to synthetic division (Section 3.2): a bookkeeping shortcut that avoids rewriting symbols.
• The excerpt emphasizes that row operations on the augmented matrix correspond exactly to operations on the equations.

Example (switching rows):
System:
3x - y + z = 3
2x - 4y + 3z = 16
x - y + z = 5
Switch E₁ and E₃ ↔ Switch R₁ and R₃ in the matrix.

Example (replacing a row with a multiple of another plus itself):
System:
x + (3/2)y - (1/2)z = 1/2
10x - z = 2
4x - 9y + 2z = 5
Replace E₂ with -10E₁ + E₂ ↔ Replace R₂ with -10R₁ + R₂ in the matrix.

🪜 Row echelon form and triangular systems

🪜 What row echelon form is

Leading entry: the first nonzero entry (if it exists) in a row.

Row echelon form (Definition 8.4): a matrix satisfying:

1. The first nonzero entry in each row is 1 (called a "leading 1").
2. The leading 1 of a given row must be to the right of the leading 1 of the row above it.
3. Any row of all zeros cannot be placed above a row with nonzero entries.

• This is the matrix analog of "triangular form" for systems of equations.
• The leading 1 in each row corresponds to the "leading variable" in each equation.

🔍 Solving via row echelon form

• Once the matrix is in row echelon form, decode it back into equations.
• The system is now triangular, so you can solve by back-substitution (starting from the last equation and working upward).

Example (from Example 8.2.1):
The system
3x - y + z = 8
x + 2y - z = 4
2x + 3y - 4z = 10
is encoded, then row operations produce:

 1  2  -1 |  4
 0  1 -4/7| 4/7
 0  0   1 | -1

Decoding gives:
x + 2y - z = 4
y - (4/7)z = 4/7
z = -1
Back-substitute: z = -1, then y = 0, then x = 3. Solution: (3, 0, -1).

⚠️ Don't confuse: row echelon form is not unique

• A matrix has infinitely many row echelon forms (different sequences of row operations can produce different-looking triangular forms).
• However, all row echelon forms of the same matrix will decode to equivalent systems with the same solution.

✨ Reduced row echelon form and direct solutions

✨ What reduced row echelon form is

Reduced row echelon form (Definition 8.5): a matrix satisfying:

1. The matrix is in row echelon form.
2. The leading 1s are the only nonzero entry in their respective columns (i.e., all entries above and below each leading 1 are 0).

• This is the "cleanest" form: each leading 1 stands alone in its column.
• Key fact: while a matrix has many row echelon forms, it has only one reduced row echelon form.

🎁 Why reduced row echelon form is powerful

• When you decode a matrix in reduced row echelon form, you get the solution instantly without any back-substitution.
• Each row directly states the value of one variable (or a simple relationship if there are free variables).

Example (continuing Example 8.2.1):
Starting from row echelon form:

 1  2  -1 |  4
 0  1 -4/7| 4/7
 0  0   1 | -1

Further row operations (using R₃ to zero out entries above it, then R₂ to zero out entries above it) produce:

 1  0  0 | 3
 0  1  0 | 0
 0  0  1 | -1

Decoding gives:
x = 3
y = 0
z = -1
No back-substitution needed!

🔧 Gauss-Jordan Elimination

Gauss-Jordan Elimination: the process of putting a matrix into reduced row echelon form.

• Start by getting the matrix into row echelon form (Gaussian Elimination).
• Then work upward: use each leading 1 to zero out all entries above it.
• Tip from the excerpt: start with the last leading 1 and work upward; this produces more zeros faster, speeding up calculations.

Example (from Example 8.2.2):
The system
x₂ - 3x₁ + x₄ = 2
2x₁ + 4x₃ = 5
4x₂ - x₄ = 3
encodes as:

 -3  1  0  1 |  2
  2  0  4  0 |  5
  0  4  0 -1 |  3

After Gauss-Jordan Elimination:

 1  0  0 -5/12| -5/12
 0  1  0 -1/4 |  3/4
 0  0  1  5/24| 35/24

Decoding gives:
x₁ - (5/12)x₄ = -5/12
x₂ - (1/4)x₄ = 3/4
x₃ + (5/24)x₄ = 35/24
x₄ is free (assign it parameter t), so:
x₁ = (5/12)t - 5/12
x₂ = (1/4)t + 3/4
x₃ = -(5/24)t + 35/24
Solution: {((5/12)t - 5/12, (1/4)t + 3/4, -(5/24)t + 35/24, t) : -∞ < t < ∞}.

🖩 Practical applications and calculator use

🖩 Automating row operations

• The excerpt notes that putting a matrix into row echelon or reduced row echelon form is easily programmed into a calculator.
• Graphing calculators typically have built-in functions for this.
• This is especially useful for systems with messy fractions or many variables.

📈 Finding functions through data points

The excerpt illustrates using augmented matrices to find a quadratic function passing through given points.

Example (from Example 8.2.3):
Find the quadratic function f(x) = ax² + bx + c passing through (-1, 3), (2, 4), (5, -2).

• Substitute each point into f(x):
  • f(-1) = 3 → a(-1)² + b(-1) + c = 3 → a - b + c = 3
  • f(2) = 4 → 4a + 2b + c = 4
  • f(5) = -2 → 25a + 5b + c = -2
• Encode into an augmented matrix:

  1 -1  1 |  3
  4  2  1 |  4
 25  5  1 | -2

• Use a calculator to find reduced row echelon form (the excerpt says "we've tortured you enough already with fractions!").
• Solution: a = -7/18, b = 13/18, c = 37/9.
• The quadratic is f(x) = -(7/18)x² + (13/18)x + 37/9.
• Verify: f(-1) = 3, f(2) = 4, f(5) = -2. ✓

🔍 Why this works

• Each point (x, y) on the graph gives one linear equation in the unknowns a, b, c.
• Three points → three equations → a system solvable by augmented matrix methods.
• The excerpt notes you can verify the solution both analytically (by substitution) and graphically (by plotting the function and the data points).

🧭 Overview

🧠 One-sentence thesis

Matrix arithmetic extends familiar algebraic operations to rectangular arrays of numbers, enabling systematic manipulation of linear systems through addition, scalar multiplication, and a specially-defined matrix multiplication that serves the needs of solving equations.

📌 Key points (3–5)

• Matrix equality: Two matrices are equal only if they have the same dimensions and identical entries in corresponding positions.
• Addition and scalar multiplication: These operations work entry-by-entry and inherit commutative, associative, and distributive properties from real numbers.
• Matrix multiplication is different: Unlike addition, matrix multiplication requires the number of columns in the first matrix to match the number of rows in the second, and the operation is generally not commutative (AB may not equal BA).
• Common confusion: Matrix multiplication does not follow all real-number rules—the zero-product property fails (two non-zero matrices can multiply to give zero), and AB ≠ BA in general.
• Why it matters: Matrix notation converts systems of linear equations into the compact form AX = B, setting up the solution methods explored in later sections.

📐 Equality and basic operations

📏 What matrix equality means

Matrix Equality: Two matrices are equal if they are the same size and their corresponding entries are equal.

• Formally: A = [aᵢⱼ]ₘₓₙ and B = [bᵢⱼ]ₚₓᵣ are equal when m = p, n = r, and aᵢⱼ = bᵢⱼ for all positions i,j.
• It's not enough for matrices to "look similar"—they must have identical dimensions and identical numbers in every matching spot.
• Example: A 2×3 matrix can never equal a 3×2 matrix, even if they contain the same numbers.

➕ Matrix addition

Matrix Addition: The sum of two same-sized matrices is obtained by adding corresponding entries.

• Only defined when both matrices have exactly the same dimensions.
• Formula: A + B = [aᵢⱼ]ₘₓₙ + [bᵢⱼ]ₘₓₙ = [aᵢⱼ + bᵢⱼ]ₘₓₙ
• Example: Adding a 2×3 matrix to another 2×3 matrix produces a 2×3 matrix where each entry is the sum of the corresponding entries.
• Don't confuse: You cannot add matrices of different sizes—there's no way to match up the entries.

✖️ Scalar multiplication

Scalar Multiplication: Multiplying a matrix by a real number (scalar) means multiplying every entry by that number.

• Formula: kA = k[aᵢⱼ]ₘₓₙ = [kaᵢⱼ]ₘₓₙ
• The word "scalar" refers to real numbers; it comes from the idea of "scaling" (magnifying or shrinking).
• Geometric interpretation: Multiplying a position matrix by scalar k scales all coordinates by factor k.
• Example: If P represents point (−2, 1), then 3P represents (−6, 3), which is the original point scaled by factor 3.

🔢 Properties inherited from real numbers

🔄 Commutative and associative properties

For addition:

• Commutative: A + B = B + A (order doesn't matter)
• Associative: (A + B) + C = A + (B + C) (grouping doesn't matter)
• These hold because real number addition is commutative and associative, and matrix addition is defined entry-by-entry.

For scalar multiplication:

• Associative with scalars: (kr)A = k(rA)
• Identity: 1A = A
• Additive inverse: −A = (−1)A

📦 Distributive properties

Two distributive laws connect scalar multiplication with addition:

• Scalar over scalar addition: (k + r)A = kA + rA
• Scalar over matrix addition: k(A + B) = kA + kB

These allow algebraic manipulation similar to ordinary algebra with variables.

🆔 Identity and inverse elements

Additive identity:

• The zero matrix 0ₘₓₙ (all entries are 0) satisfies A + 0ₘₓₙ = A for any m×n matrix A.

Additive inverse:

• For every matrix A = [aᵢⱼ]ₘₓₙ, there exists −A = [−aᵢⱼ]ₘₓₙ such that A + (−A) = 0ₘₓₙ.
• To get the additive inverse, simply negate every entry.

Subtraction:

• Defined as A − B = A + (−B), which amounts to subtracting corresponding entries.

⚙️ Matrix multiplication mechanics

🎯 Row times column product

Product of a row and column: Multiply corresponding entries and sum the results.

• For row Rᵢ from matrix A and column Cⱼ from matrix B: Rᵢ · Cⱼ = aᵢ₁b₁ⱼ + aᵢ₂b₂ⱼ + ... + aᵢₙbₙⱼ
• Critical requirement: The row and column must have the same number of entries.
• Example: Row [2, 0, −1] times column [3, 4, 5] gives (2)(3) + (0)(4) + (−1)(5) = 6 + 0 − 5 = 1.

🔗 Full matrix multiplication

Matrix Multiplication: The product AB has entry (i,j) equal to row i of A times column j of B.

• Dimension requirement: To compute AB, the number of columns in A must equal the number of rows in B.
• If A is m×n and B is n×r, then AB is m×r.
• Formula: AB = [Rᵢ · Cⱼ]ₘₓᵣ where Rᵢ is the i-th row of A and Cⱼ is the j-th column of B.
• Don't confuse: Even when both AB and BA are defined, they are usually not equal and may not even have the same dimensions.

⚠️ Non-commutative nature

Key differences from real numbers:

• AB ≠ BA in general (no commutative property)
• AB = 0 does not imply A = 0 or B = 0 (zero-product property fails)
• Example from the excerpt: Two non-zero matrices can multiply to give the zero matrix.

🏛️ Properties of matrix multiplication

🔗 What properties do hold

Associative property:

• (AB)C = A(BC) when all products are defined
• This means we can write ABC without ambiguity about grouping.

Associative with scalars:

• k(AB) = (kA)B = A(kB)

Distributive properties:

• A(B ± C) = AB ± AC (left distributive)
• (A ± B)C = AC ± BC (right distributive)

🆔 Identity matrix

Identity matrix Iₖ: A k×k matrix with 1's on the main diagonal and 0's elsewhere.

• Main diagonal: positions where row number equals column number (i = j)
• Formula: Iₖ = [dᵢⱼ]ₖₓₖ where dᵢⱼ = 1 if i = j, and 0 otherwise
• Property: For any m×n matrix A, we have IₘA = AIₙ = A
• Note: The size of the identity matrix must match appropriately—use Iₘ on the left and Iₙ on the right.

🧮 Polynomial expressions with matrices

• Matrix "polynomials" like C² − 5C + 10I₂ are evaluated by:
  • Computing powers (C² means C times itself)
  • Performing scalar multiplication
  • Adding/subtracting matrices
  • Using the identity matrix for constant terms
• Example: Expanding (M − 2I₄)(M + 3I₄) gives M² + M − 6I₄, similar to expanding (x − 2)(x + 3) = x² + x − 6.

🔄 Applications and connections

🌀 Geometric transformations

Rotation example:

• A 2×1 position matrix [x, y] represents point (x, y)
• Multiplying by a rotation matrix R transforms the point
• The excerpt shows R with entries involving √2/2 rotates points counterclockwise by 45°
• This can transform entire curves: the hyperbola x² − y² = 4 becomes y = 2/x after rotation.

📊 Systems of linear equations

Matrix equation form:

• A system like {3x − y + z = 8, x + 2y − z = 4, 2x + 3y − 4z = 10} can be written as AX = B
• A is the coefficient matrix (entries from variable coefficients)
• X is the unknowns matrix [x, y, z]
• B is the constant matrix [8, 4, 10]
• Solving the system becomes equivalent to solving the matrix equation AX = B
• This sets up the question: Can we "divide" by matrix A? (addressed in the next section)

🎲 Stochastic processes

The exercises mention using matrix multiplication for modeling:

• State matrices track quantities over time
• Transition matrices describe how states change
• Repeated multiplication Qⁿ X models evolution over n time steps
• Steady states occur when QXₛ = Xₛ (the state doesn't change)

🔺 Special matrix types

📐 Square matrices

• A matrix with the same number of rows and columns
• Only square matrices can have powers (A², A³, etc.) that are always defined
• The identity matrix is always square

🔺 Triangular matrices

Upper triangular: All entries below the main diagonal are zero. Lower triangular: All entries above the main diagonal are zero.

• Example: A matrix with entries only on and above the diagonal is upper triangular
• Products of upper triangular matrices remain upper triangular
• Products of lower triangular matrices remain lower triangular
• Some matrices are both upper and lower triangular (diagonal matrices with zeros off the main diagonal)

🧭 Overview

🧠 One-sentence thesis

Matrix inverses provide a powerful method to solve systems of linear equations by transforming the equation AX = B into X = A⁻¹B, enabling us to solve entire families of systems with different constants using a single inverse calculation.

📌 Key points (3–5)

• Core method: Rewrite a system as AX = B, then solve by finding X = A⁻¹B, where A⁻¹ is the inverse of the coefficient matrix.
• How to find A⁻¹: Use the super-sized augmented matrix [A | Iₙ] and apply Gauss-Jordan elimination until it becomes [Iₙ | A⁻¹].
• When it works: A matrix A is invertible if and only if AX = B has a unique solution for every matrix B—meaning the system is consistent and independent.
• Common confusion: Not all matrices are invertible; only square matrices can be invertible, and even then, not all square matrices have inverses.
• Why it matters: Finding A⁻¹ once lets you solve infinitely many systems with the same coefficient matrix but different constants, simply by computing A⁻¹B.

🔄 From systems to matrix equations

🔄 Rewriting systems as AX = B

The excerpt shows how to encode a system of linear equations into matrix form:

• Coefficient matrix A: contains all the coefficients of the variables.
• Unknowns matrix X: a column matrix of the variables (x, y, z, etc.).
• Constants matrix B: a column matrix of the constants from the right-hand side.

Example: The system {2x − 3y = 16; 3x + 4y = 7} becomes AX = B where A = [2 −3; 3 4], X = [x; y], B = [16; 7].

🔄 Motivation from real numbers

To understand matrix inverses, the excerpt revisits solving 3x = 5:

• Divide by 3, which is the same as multiplying by 3⁻¹ (the multiplicative inverse of 3).
• Steps: 3⁻¹(3x) = 3⁻¹(5) → (3⁻¹ · 3)x = 3⁻¹(5) → 1 · x = 3⁻¹(5) → x = 3⁻¹(5).
• The process uses associativity, the inverse property (3⁻¹ · 3 = 1), and the multiplicative identity (1 · x = x).

For matrices, we need the same structure: associativity (Theorem 8.5), a multiplicative identity (Iₙ), and a multiplicative inverse (A⁻¹).

🧩 What is an invertible matrix

🧩 Definition of invertible

Invertible matrix: An n × n matrix A is invertible if there exists a matrix A⁻¹ (read "A-inverse") such that A⁻¹A = AA⁻¹ = Iₙ.

• Only square matrices can be invertible (because A and A⁻¹ must have the same dimensions).
• The inverse A⁻¹ must satisfy both equations: A⁻¹A = Iₙ and AA⁻¹ = Iₙ (matrix multiplication is not necessarily commutative, so these are two separate conditions).
• Not all square matrices are invertible (see exercises).

🧩 Key properties (Theorem 8.6)

Property	What it means
Uniqueness	If A is invertible, then A⁻¹ is unique—there is only one inverse.
Unique solutions	A is invertible if and only if AX = B has a unique solution for every n × r matrix B.

Implication for systems: If the coefficient matrix A is invertible, the system is consistent and independent (exactly one solution).

Don't confuse: "A is square" does not guarantee "A is invertible." The excerpt notes that being square is necessary but not sufficient.

🛠️ How to find the inverse

🛠️ The super-sized augmented matrix method

To find A⁻¹ for an n × n matrix A:

1. Form the augmented matrix [A | Iₙ] by placing A on the left and the identity matrix Iₙ on the right.
2. Apply Gauss-Jordan elimination (row operations) to transform the left side into Iₙ.
3. When the left side becomes Iₙ, the right side becomes A⁻¹: [A | Iₙ] → [Iₙ | A⁻¹].

Why this works: The identity matrix Iₙ in [A | Iₙ] keeps a "running memory" of all the row operations needed to "undo" multiplication by A. These operations produce exactly A⁻¹.

🛠️ Detailed example walkthrough

The excerpt works through finding A⁻¹ for A = [2 −3; 3 4]:

• Start with [A | I₂] = [2 −3 | 1 0; 3 4 | 0 1].
• The process involves solving two systems simultaneously: AA⁻¹ = I₂ means finding columns of A⁻¹ such that multiplying by A gives the columns of I₂.
• This produces two systems with the same coefficient matrix A but different constants (the columns of I₂).
• By combining both systems into one super-sized augmented matrix, we do all the row operations once.
• Result: A⁻¹ = [4/17 3/17; −3/17 2/17].

Verification: Check both AA⁻¹ = I₂ and A⁻¹A = I₂ to confirm the inverse is correct.

🛠️ Solving systems with A⁻¹

Once you have A⁻¹, solving AX = B is simple:

• Multiply both sides by A⁻¹: A⁻¹(AX) = A⁻¹B.
• Simplify: (A⁻¹A)X = A⁻¹B → I₂X = A⁻¹B → X = A⁻¹B.

Example: For the system {2x − 3y = 16; 3x + 4y = 7}, compute X = A⁻¹B = [4/17 3/17; −3/17 2/17][16; 7] = [5; −2], so (x, y) = (5, −2).

🔁 Solving families of systems

🔁 One inverse, many systems (Example 8.4.1)

The excerpt demonstrates solving three different systems with the same coefficient matrix A = [3 1 2; 0 −1 5; 2 1 4]:

• Find A⁻¹ once using [A | I₃] → [I₃ | A⁻¹].
• Result: A⁻¹ = [9/13 2/13 −7/13; −10/13 −8/13 15/13; −2/13 1/13 3/13].
• Solve three systems by computing X = A⁻¹B for three different B matrices:
  • System (a): B = [26; 39; 117] → X = [−39; 91; 26].
  • System (b): B = [4; 2; 5] → X = [5/13; 19/13; 9/13].
  • System (c): B = [1; 0; 0] → X = [9/13; −10/13; −2/13] (which is the first column of A⁻¹).

Key advantage: The row operations to find A⁻¹ are determined entirely by A, not by the constants in B. So one inverse calculation solves infinitely many systems with different constants.

🔁 Real-world application: circuit analysis (Example 8.4.2)

The excerpt presents an electronics problem with mesh currents in a circuit:

• The system relates battery voltages (VB₁, VB₂) and resistances (R₁ through R₆) to four mesh currents (i₁, i₂, i₃, i₄).
• With all resistances equal to 1 kΩ, the coefficient matrix A is 4 × 4.
• Find A⁻¹ using a calculator: A⁻¹ = [1.625 1.25 1.125 1; 1.25 1.5 1.25 1; 1.125 1.25 1.625 1; 1 1 1 1].
• Solve four different scenarios by changing VB₁ and VB₂:
  • (a) VB₁ = 10 V, VB₂ = 5 V → currents are 10.625, 6.25, 3.125, 5 mA.
  • (b) VB₁ = 10 V, VB₂ = 0 V → currents are 16.25, 12.5, 11.25, 10 mA.
  • (c) VB₁ = 0 V, VB₂ = 10 V → currents are −11.25, −12.5, −16.25, −10 mA (negative means opposite direction).
  • (d) VB₁ = 10 V, VB₂ = 10 V → currents are 5, 0, −5, 0 mA (batteries push in opposite directions, canceling some flows).

Inverse problem: Given currents and voltages, find resistances. This produces a 4 × 6 system (4 equations, 6 unknowns), which is not invertible but is consistent and dependent. The solution involves free variables and constraints (resistances must be positive).

🧮 Practical computation

🧮 Using calculators

The excerpt shows that for larger matrices (3 × 3 and above), calculators can compute A⁻¹ directly:

• Input matrix A into the calculator.
• Use the inverse function to get A⁻¹.
• Multiply A⁻¹ by various B matrices to solve multiple systems quickly.

Example: In Example 8.4.2, the calculator gives A⁻¹ with decimal entries, which are then used to solve four systems by simple matrix multiplication.

🧮 Verification

Always check your inverse by computing both AA⁻¹ and A⁻¹A to confirm they equal the identity matrix Iₙ.

Example: For A = [3 1 2; 0 −1 5; 2 1 4] and A⁻¹ = [9/13 2/13 −7/13; −10/13 −8/13 15/13; −2/13 1/13 3/13], the excerpt verifies A⁻¹A = I₃ and AA⁻¹ = I₃.

🔍 Important distinctions

🔍 Invertible vs. non-invertible

• Invertible: A is square, and [A | Iₙ] can be row-reduced to [Iₙ | A⁻¹]. The system AX = B has a unique solution for any B.
• Non-invertible: A is not square, or A is square but cannot be row-reduced to Iₙ. The system AX = B may have no solution or infinitely many solutions.

Example: In the circuit problem, when unknowns are resistances (4 equations, 6 unknowns), the coefficient matrix is 4 × 6 and therefore not invertible. The system is consistent but dependent (infinitely many solutions with free variables).

🔍 Short-run vs. long-run use

• Short-run: If you only need to solve one system, using Gauss-Jordan elimination on the augmented matrix [A | B] is faster than finding A⁻¹.
• Long-run: If you need to solve many systems with the same A but different B, finding A⁻¹ once and then computing X = A⁻¹B for each B is much more efficient.

Don't confuse: The effort to find A⁻¹ is comparable to solving one system, but it pays off when you solve multiple systems with the same coefficient matrix.

🧭 Overview

🧠 One-sentence thesis

The determinant of a square matrix is a single real number that determines whether the matrix is invertible and enables solving systems of linear equations through Cramer's Rule and finding matrix inverses through adjoints.

📌 Key points (3–5)

• What the determinant is: a real number assigned to each square matrix, computed recursively by reducing larger matrices to smaller ones.
• Connection to invertibility: a matrix is invertible if and only if its determinant is nonzero; determinant zero means the matrix cannot be "divided by."
• Cramer's Rule application: when the coefficient matrix has nonzero determinant, each unknown can be solved individually as a ratio of determinants.
• Common confusion: the determinant is not just a formula to memorize—it fundamentally determines whether a system has a unique solution; row operations change the determinant in predictable ways (sign flip for row swap, scalar multiple for row scaling, unchanged for row replacement).
• Alternative inverse formula: the inverse of a matrix can be expressed as one over the determinant times the adjoint (transpose of the cofactor matrix).

🔢 Definition and computation

🔢 Recursive definition

Determinant of A, denoted det(A): For a 1×1 matrix [a₁₁], det(A) = a₁₁. For larger n×n matrices, det(A) = a₁₁·det(A₁₁) − a₁₂·det(A₁₂) + ... + (−1)^(1+n)·a₁ₙ·det(A₁ₙ), where Aᵢⱼ is the (n−1)×(n−1) matrix formed by deleting row i and column j of A.

• The definition is recursive: you compute the determinant of an n×n matrix by breaking it down into determinants of (n−1)×(n−1) matrices.
• For 2×2 matrices, this simplifies to a memorable formula: det([a b; c d]) = ad − bc.
• The initial definition expands along the first row, using entries a₁₁, a₁₂, ..., a₁ₙ.
• Example: For a 2×2 matrix [4 −3; 2 1], det = 4·det([1]) − (−3)·det([2]) = 4(1) + 3(2) = 10.

🧮 Minors and cofactors

ij minor of A, denoted Mᵢⱼ: Mᵢⱼ = det(Aᵢⱼ).

ij cofactor of A, denoted Cᵢⱼ: Cᵢⱼ = (−1)^(i+j)·Mᵢⱼ = (−1)^(i+j)·det(Aᵢⱼ).

• The minor is just the determinant of the smaller matrix after deleting a row and column.
• The cofactor adds a sign based on position: (−1)^(i+j) creates a checkerboard pattern of + and − signs.
• The sign pattern for 3×3 matrices is: [+ − +; − + −; + − +].
• Don't confuse: the sign in the cofactor formula is only part of the story—the actual sign of a term in the determinant expansion also depends on the sign of the matrix entry and the minor itself.

🔄 Sign pattern

The alternating signs (−1)^(i+j) follow a checkerboard:

• 2×2: [+ −; − +]
• 3×3: [+ − +; − + −; + − +]
• 4×4: [+ − + −; − + − +; + − + −; − + − +]

🛠️ Properties and computational techniques

🛠️ Expanding along any row

• Theorem 8.7 (first property): You can compute det(A) by expanding along any row k, not just the first: det(A) = aₖ₁·Cₖ₁ + aₖ₂·Cₖ₂ + ... + aₖₙ·Cₖₙ.
• Strategy: choose a row with zeros to simplify computation.
• Example: If row 2 of a 3×3 matrix is [0 −1 5], expanding along row 2 means you only compute two cofactors (the zero term contributes nothing).

🔀 Row operations and determinants

Theorem 8.7 (row operation properties):

Row operation	Effect on determinant
Interchange two rows	det(A′) = −det(A)
Multiply a row by nonzero scalar c	det(A′) = c·det(A)
Replace a row with itself plus a multiple of another row	det(A′) = det(A)

• These properties make determinants easier to compute: use row operations to create an upper triangular matrix, then multiply the diagonal entries.
• Example: Transform a 3×3 matrix to upper triangular using row replacements (determinant unchanged), then det = product of diagonal entries.
• Don't confuse: only the replacement operation leaves the determinant unchanged; swapping rows flips the sign, and scaling a row scales the determinant.

🔺 Special cases

Theorem 8.7 (special determinants):

• If A has two identical rows or a row of all zeros, then det(A) = 0.
• If A is upper or lower triangular, det(A) = product of the main diagonal entries.
• Example: The identity matrix Iₙ has det(Iₙ) = 1 (all diagonal entries are 1, product is 1).

🔗 Product and inverse properties

Theorem 8.7 (multiplicative properties):

• det(AB) = det(A)·det(B) for any n×n matrices A and B.
• det(Aⁿ) = [det(A)]ⁿ for any natural number n.
• If A is invertible, det(A⁻¹) = 1/det(A).

🎯 Invertibility connection

🎯 The fundamental criterion

Theorem 8.7 (invertibility): A is invertible if and only if det(A) ≠ 0.

• This is why the determinant determines invertibility.
• If det(A) ≠ 0, the system AX = B has a unique solution (consistent and independent).
• If det(A) = 0, the matrix is not invertible—you cannot "divide" by it, just as you cannot divide a real number by zero.

🔍 Why this works

• When you row-reduce A to find A⁻¹, the determinant of intermediate matrices differs from det(A) by at most a nonzero multiple.
• If det(A) ≠ 0, then the reduced row echelon form of A must also have nonzero determinant, which means all diagonal entries are 1, so the reduced form is Iₙ (A is invertible).
• Conversely, if A is invertible, it can be transformed to Iₙ, and since det(Iₙ) = 1 ≠ 0, the same logic implies det(A) ≠ 0.

📐 Cramer's Rule

📐 The theorem

Theorem 8.8 (Cramer's Rule): Suppose AX = B is the matrix form of a system of n linear equations in n unknowns. If det(A) ≠ 0, then the system is consistent and independent, and the solution for unknown xⱼ is:

xⱼ = det(Aⱼ) / det(A)

where Aⱼ is the matrix A with its jth column replaced by the constants in B.

• In words: solve for each unknown one at a time by computing a ratio of determinants.
• The numerator det(Aⱼ) uses a modified coefficient matrix where the column of coefficients for xⱼ is replaced by the constant column B.
• The denominator is always det(A), the determinant of the original coefficient matrix.

📐 How to apply Cramer's Rule

Step-by-step:

1. Write the system in matrix form AX = B and identify the coefficient matrix A, unknowns X, and constants B.
2. Compute det(A). If det(A) = 0, Cramer's Rule does not apply (system is not consistent independent).
3. For each unknown xⱼ, form Aⱼ by replacing the jth column of A with B.
4. Compute det(Aⱼ) and then xⱼ = det(Aⱼ) / det(A).

Example: For the system {2x₁ − 3x₂ = 4; 5x₁ + x₂ = −2}, A = [2 −3; 5 1], B = [4; −2]. Then A₁ = [4 −3; −2 1] (replace first column with B) and A₂ = [2 4; 5 −2] (replace second column with B). Compute det(A) = 17, det(A₁) = −2, det(A₂) = −24, so x₁ = −2/17 and x₂ = −24/17.

📐 When to use Cramer's Rule

• Cramer's Rule is useful when you need to solve for one or a few unknowns without finding all of them.
• Example: To find only z in a 3×3 system, identify z as x₃, form A₃ by replacing the third column of A with B, and compute z = det(A₃)/det(A).
• Don't confuse: Cramer's Rule requires det(A) ≠ 0; if det(A) = 0, the system either has no solution or infinitely many solutions, and Cramer's Rule does not apply.

🔄 Matrix inverse via adjoint

🔄 The adjoint matrix

Adjoint of A, denoted adj(A): the matrix whose ij-entry is the ji cofactor of A, Cⱼᵢ.

• In formula: adj(A) = [C₁₁ C₂₁ ... Cₙ₁; C₁₂ C₂₂ ... Cₙ₂; ...; C₁ₙ C₂ₙ ... Cₙₙ].
• Notice the reversal of subscripts: the ij-entry of adj(A) is Cⱼᵢ, not Cᵢⱼ (this is a transpose of the cofactor matrix).
• The adjoint collects all the cofactors of A in a specific arrangement.

🔄 Inverse formula

Theorem 8.9: If A is an invertible n×n matrix, then:

A⁻¹ = (1/det(A)) · adj(A)

• This provides an alternative method for finding the inverse: compute the determinant and all the cofactors, arrange them into the adjoint, then divide by det(A).
• For 2×2 matrices, this simplifies to: [a b; c d]⁻¹ = (1/(ad−bc)) · [d −b; −c a].
• Example: For A = [3 1 2; 0 −1 5; 2 1 4] with det(A) = −13, compute all nine cofactors (C₁₁ = −9, C₂₁ = −2, C₃₁ = 7, C₁₂ = 10, C₂₂ = 8, C₃₂ = −15, C₁₃ = 2, C₂₃ = −1, C₃₃ = −3), then adj(A) = [−9 −2 7; 10 8 −15; 2 −1 −3], so A⁻¹ = (−1/13)·adj(A).

🔄 Connection to Cramer's Rule

• The adjoint formula for A⁻¹ is deeply connected to Cramer's Rule.
• When solving AX = Iₙ (to find A⁻¹), each column of A⁻¹ can be found using Cramer's Rule, and the determinants that appear are exactly the cofactors of A.
• This explains why the denominators in A⁻¹ are all det(A) and the numerators are cofactors.

🧮 Practical computation tips

🧮 Choosing the best row or column

• Always expand along a row (or column) with the most zeros to minimize computation.
• Example: If a 4×4 matrix has a row [0 0 3 0], expanding along that row means computing only one 3×3 determinant instead of four.

🧮 Using row operations strategically

• Transform the matrix to upper triangular form using row replacements (which preserve the determinant), then multiply the diagonal entries.
• Keep track of row swaps (flip sign) and row scalings (multiply determinant by the scalar).
• Example: If you swap two rows once and then reach upper triangular form with diagonal [3, −1, 13/3], the determinant is −1 · 3 · (−1) · (13/3) = 13 (the negative from the swap).

🧮 Don't confuse

• Determinant vs absolute value: det(A) is not the same as |A| (absolute value notation); det(A) can be negative.
• Expanding along rows vs columns: Theorem 8.7 allows expansion along any row; in a full Linear Algebra course, you learn you can also expand along columns, but this text focuses on rows.
• Cofactor sign: the sign (−1)^(i+j) is part of the cofactor definition, but the final sign of a term in the determinant also depends on the matrix entry and the minor's value.

🧭 Overview

🧠 One-sentence thesis

Partial fraction decomposition reverses the common-denominator process to rewrite a single rational function as a sum of simpler fractions, which is essential for certain calculus applications.

📌 Key points (3–5)

• What the technique does: transforms a rational function from a single fraction (like (x² - x - 6)/(x⁴ + x²)) into a sum of simpler fractions (like terms with denominators x, x², and x² + 1).
• Two-step process: first determine the form of the decomposition by factoring the denominator; then solve for the unknown coefficients by equating like powers of x.
• Form depends on factorization: linear factors (like x - 1) contribute terms with constant numerators; irreducible quadratics (like x² + 1) contribute terms with linear numerators (Ax + B).
• Common confusion: repeated factors require multiple terms—if (x - 1)² appears, you need both A/(x - 1) and B/(x - 1)²; don't assume one term is enough.
• Why it matters: the decomposed form is "more palatable to Calculus students" for integration and other operations, even though the original form is better for algebra tasks like finding zeros and asymptotes.

🔍 The core idea and motivation

🔍 Reversing the LCD process

• In intermediate algebra, you combine fractions by finding a common denominator:
  • Example: A/x + B/x² + (Cx + D)/(x² + 1) → (single fraction with denominator x²(x² + 1))
• Partial fraction decomposition does the reverse:
  • Start with (x² - x - 6)/(x⁴ + x²) and break it back into A/x + B/x² + (Cx + D)/(x² + 1).
• The excerpt emphasizes this is needed for calculus, not for algebra tasks like sign diagrams or finding asymptotes.

📐 Proper rational functions

A rational function R(x) = N(x)/D(x) is proper when the degree of the numerator N(x) is less than the degree of the denominator D(x).

• Only proper rational functions can be decomposed directly.
• If the function is not proper (degree of numerator ≥ degree of denominator), use polynomial long division first to separate the polynomial part from the proper remainder.
• Example: 4x³/(x² - 2) is not proper; long division gives 4x + 8x/(x² - 2), and only the remainder 8x/(x² - 2) is decomposed.

🧩 Determining the form of the decomposition

🧩 Factor the denominator completely

• Factor D(x) into linear factors (ax + b) and irreducible quadratics (ax² + bx + c).
• Linear factors correspond to real zeros; irreducible quadratics correspond to pairs of non-real complex conjugate zeros.
• Check irreducibility using the discriminant: if b² - 4ac < 0, the quadratic has no real zeros and is irreducible.

🔢 Multiplicity matters

• Multiplicity = how many times a factor appears.
• Example: x²(x² + 1) = x · x · (x² + 1) means x = 0 is a zero of multiplicity 2.
• Watch for hidden repeated factors: (x - 1)(1 - x) both correspond to x = 1, so rewrite (1 - x) = -(x - 1) to see x = 1 has multiplicity 2.

📋 Theorem 8.10: The form rules

Factor type	Multiplicity m	Terms in decomposition
Linear (ax + b)	m	A₁/(ax + b) + A₂/(ax + b)² + ... + Aₘ/(ax + b)ᵐ
Irreducible quadratic (ax² + bx + c)	m	(B₁x + C₁)/(ax² + bx + c) + (B₂x + C₂)/(ax² + bx + c)² + ... + (Bₘx + Cₘ)/(ax² + bx + c)ᵐ

• Key distinction: linear factors get constant numerators (A, B, ...); irreducible quadratics get linear numerators (Bx + C).
• Each power of the factor up to m must appear in the decomposition.
• Example: if the denominator is x²(x² + 1), the form is A/x + B/x² + (Cx + D)/(x² + 1).

⚠️ Don't duplicate terms

• The excerpt shows a common mistake: writing (Bx + C)/x² can be split as Bx/x² + C/x² = B/x + C/x², so the B/x term duplicates the A/x term already present.
• Solution: drop the redundant term and use only A/x + B/x² (not A/x + B/x + C/x²).

🔧 Solving for the unknown coefficients

🔧 Clear denominators

• Multiply both sides by the common denominator to eliminate fractions.
• Example: starting from (x² - x - 6)/(x²(x² + 1)) = A/x + B/x² + (Cx + D)/(x² + 1), multiply through by x²(x² + 1):
  • x² - x - 6 = Ax(x² + 1) + B(x² + 1) + (Cx + D)x²

📊 Expand and collect like terms

• Expand the right side and group by powers of x.
• Example: x² - x - 6 = (A + C)x³ + (B + D)x² + Ax + B

⚖️ Equate coefficients (Theorem 8.11)

Theorem 8.11: If two polynomials are equal for all x in an interval, then the coefficients of corresponding powers of x are equal.

• Match coefficients of x³, x², x¹, and x⁰ (the constant term) on both sides.
• This gives a system of linear equations.
• Example:
  • Coefficient of x³: A + C = 0
  • Coefficient of x²: B + D = 1
  • Coefficient of x: A = -1
  • Constant: B = -6

🧮 Solve the system

• The excerpt recommends substitution (from intermediate algebra) as the most efficient method for these systems, rather than the matrix methods from earlier sections.
• Example: from A = -1 and A + C = 0, substitute to get C = 1; from B = -6 and B + D = 1, get D = 7.

🎯 Worked examples and special cases

🎯 Example with repeated linear factor

• Problem: 3/(x³ - 2x² + x)
• Factor: x(x - 1)²
• Form: A/x + B/(x - 1) + C/(x - 1)²
• After clearing and equating: A = 3, B = -3, C = 3.
• Don't confuse: you need both B/(x - 1) and C/(x - 1)² because the multiplicity is 2.

🎯 Example with irreducible quadratic

• Problem: 3/(x³ - x² + x)
• Factor: x(x² - x + 1); check discriminant of x² - x + 1: (-1)² - 4(1)(1) = -3 < 0, so irreducible.
• Form: A/x + (Bx + C)/(x² - x + 1)
• Solution: A = 3, B = -3, C = 3.
• Don't confuse: the irreducible quadratic gets a linear numerator (Bx + C), not just a constant.

🎯 Example with repeated irreducible quadratic

• Problem: (x³ + 5x - 1)/(x⁴ + 6x² + 9)
• Factor: (x² + 3)²
• Form: (Ax + B)/(x² + 3) + (Cx + D)/(x² + 3)²
• Solution: A = 1, B = 0, C = 2, D = -1.

🎯 Example requiring clever factoring

• Problem: 8x²/(x⁴ + 16)
• x⁴ + 16 does not factor as a sum of squares, but rewrite:
  • x⁴ + 16 = (x²)² + 4² = (x² + 4)² - 8x² (add and subtract 8x²)
  • = (x² + 4 - 2x√2)(x² + 4 + 2x√2)
  • = (x² - 2x√2 + 4)(x² + 2x√2 + 4)
• Both quadratics have discriminant -8 < 0, so both are irreducible.
• Form: (Ax + B)/(x² - 2x√2 + 4) + (Cx + D)/(x² + 2x√2 + 4)
• Solution involves a 4×4 system; final answer has A = √2, B = 0, C = -√2, D = 0.

🎯 Example with improper rational function

• Problem: 4x³/(x² - 2)
• Not proper (degree 3 ≥ degree 2), so use long division first:
  • 4x³/(x² - 2) = 4x + 8x/(x² - 2)
• Now decompose only the proper remainder 8x/(x² - 2).
• Factor x² - 2 = (x - √2)(x + √2) (reducible, despite not factoring "nicely").
• Form: A/(x - √2) + B/(x + √2)
• Final answer: 4x + 4/(x + √2) + 4/(x - √2).

⚠️ Common pitfalls and cautions

⚠️ Reducible vs irreducible quadratics

• Irreducible means no real zeros (discriminant < 0).
• Reducible means it factors over the reals, even if the factorization involves irrational numbers like √2.
• Example: x² - 2 is reducible (factors as (x - √2)(x + √2)), but x² + 1 is irreducible (no real zeros).
• Don't assume: just because a quadratic "doesn't factor nicely" doesn't mean it's irreducible—check the discriminant.

⚠️ Mathematical vs real-world solutions

• The excerpt includes a cautionary example (Exercise 23): Cramer's Rule gave a solution with a negative number of servings, which is impossible in the real-world context.
• Lesson: partial fraction decomposition (and linear algebra methods) always produce a mathematical answer, but you must check whether the answer makes sense in the applied context.

⚠️ Why this section matters

• The excerpt warns against a common algebraic error: thinking that 8/(x² - 9) = 8/x² - 8/9.
• If that were true, "this section would have no need to exist"—you could just split the denominator arbitrarily.
• The whole point: you cannot split a denominator that way; you must use the systematic partial fraction decomposition process.

🧭 Overview

🧠 One-sentence thesis

Unlike linear systems, non-linear systems have no universal algorithm and require a flexible combination of substitution, elimination, graphing, and algebraic insight to solve, often yielding multiple discrete solutions or no solutions at all.

📌 Key points (3–5)

• No general algorithm exists: Non-linear systems cannot be solved by a single routine method; each problem demands its own approach.
• Multiple discrete solutions are common: A consistent non-linear system can have more than one solution without having infinitely many (unlike linear systems).
• All hazards return: Extraneous solutions, domain restrictions, and unusual function behavior reappear.
• Common confusion: Don't assume "dependent/independent" labels apply—these terms are generally not used for non-linear systems.
• Graphing helps avoid errors: Sketching curves can reveal the number of solutions and prevent algebraic dead ends.

🔧 Solution techniques for non-linear equations

🔧 Elimination when variables appear in similar forms

• If both equations contain the same powers (e.g., x² and y² only), elimination can remove one variable.
• Example: Given x² + y² = 4 and 4x² + 9y² = 36, multiply the first equation by –4 and add to the second to eliminate x².
• Result: 5y² = 20, so y = ±2; substitute back to find x = 0.
• Don't confuse: Elimination doesn't always remove a variable completely; sometimes it only removes troublesome cross-terms like xy.

🔄 Substitution when one equation is simple

• Solve the simpler equation for one variable, then substitute into the other.
• Example: From y – 2x = 0, get y = 2x; substitute into x² + y² = 4 to obtain x² + (2x)² = 4, yielding 5x² = 4.
• Solve for x, then back-substitute to find y.
• Caution: Substitution can lead to complicated expressions; sometimes elimination is cleaner even if not obvious at first.

🎯 Avoiding division by a variable

• Never divide both sides by a variable (e.g., dividing yz = y by y) because it assumes the variable is non-zero.
• Instead, factor: yz – y = 0 becomes y(z – 1) = 0, giving two cases: y = 0 or z = 1.
• Work through each case separately to avoid losing solutions.

🖼️ Using graphs to guide algebra

Making a quick sketch of the problem situation can save a lot of time and effort.

• Before heavy algebra, sketch both curves to see how many intersections to expect and where they lie.
• Example: A circle and a parabola intersecting twice, both in positive-y regions, tells you to discard negative-y solutions from the quadratic formula.
• Graphs also confirm whether a system is inconsistent (no intersection).

🧮 Worked examples and strategies

🧮 Eliminating cross-terms

• When equations contain xy terms that prevent direct elimination, eliminate the xy and constant terms together to find a simpler relationship.
• Example: x² + 2xy – 16 = 0 and y² + 2xy – 16 = 0. Subtract the first from the second: y² – x² = 0, so y = ±x.
• Substitute y = x into the first equation: x² + 2x² – 16 = 0, yielding x² = 16/3.
• Substitute y = –x: x² – 2x² – 16 = 0 gives x² = –16 (no real solution).

🔢 Substitution with exponential/logarithmic equations

• For systems mixing exponentials and polynomials, solve one equation for a simple variable and substitute.
• Example: y + 4e^(2x) = 1 and y² + 2e^x = 1. Solve the first for y: y = 1 – 4e^(2x). Substitute into the second: (1 – 4e^(2x))² + 2e^x = 1.
• Expand and simplify: 16e^(4x) – 8e^(2x) + 2e^x = 0. Factor: 2e^x(8e^(3x) – 4e^x + 1) = 0.
• Since e^x is never zero, solve 8u³ – 4u + 1 = 0 (where u = e^x) using factoring or the quadratic formula after synthetic division.

🧩 Three-variable systems

• Label equations E1, E2, E3 and look for the simplest equation to start.
• Example: yz = y rewrites as y(z – 1) = 0, giving two cases: y = 0 or z = 1.
• Case 1 (y = 0): Substitute into the other two equations and solve the resulting two-variable system.
• Case 2 (z = 1): Substitute and check for contradictions; if one equation becomes false (e.g., –2 = 0), discard this case.
• Combine solutions from all valid cases.

📉 Systems of non-linear inequalities

📉 Regions instead of intervals

Instead of obtaining a few numbers which divide the real number line into intervals, we get an equation of a curve which divides the plane into two regions.

• To solve x² + y² – 4 < 0, first find the boundary curve: x² + y² = 4 (a circle).
• The curve divides the plane into "inside" and "outside" regions.
• Use test points in each region (e.g., (0, 0) inside, (0, 3) outside) to determine which region satisfies the inequality.
• If the test point makes the inequality true, shade that region; if false, do not shade.

🖊️ Boundary included or not

• If the inequality is ≤ or ≥, the boundary curve is part of the solution (draw it solid).
• If the inequality is < or >, the boundary is not part of the solution (draw it dashed).
• Example: x² + y² – 4 < 0 means the circle itself does not satisfy the inequality (points on the circle give 0 < 0, which is false).

🔗 Compound inequalities and intersections

• A system of inequalities requires finding the set-theoretic intersection of the solution regions.
• Solve each inequality separately, then shade only the region where both (or all) conditions hold.
• Example: y² – 4 ≤ x and x < y + 2. The first gives the region to the right of (inside) a parabola; the second gives the region above a line. The solution is where these overlap.
• Find intersection points of the boundary curves algebraically to make the graph precise.

🧪 Example workflow

1. Rewrite each inequality with zero on one side (e.g., x² + y² – 4 ≥ 0).
2. Identify the boundary curve (set the left side equal to zero).
3. Sketch the curve (circle, parabola, ellipse, hyperbola, line, etc.).
4. Choose test points in each region created by the curve.
5. Substitute test points into the inequality to determine which region(s) to shade.
6. For systems, find the intersection of all shaded regions.
7. Solve the system of boundary equations to mark exact intersection points on the graph.

⚠️ Common pitfalls and reminders

⚠️ Extraneous solutions

• Non-linear algebra (squaring, substitution, factoring) can introduce solutions that don't satisfy the original system.
• Always check solutions in both original equations.
• Graphing beforehand helps: if the graph shows two intersections in positive-y regions, discard any negative-y algebraic solutions.

⚠️ Domain restrictions

• Functions like square roots, logarithms, and rational expressions impose domain constraints.
• Example: e^x = (–1 – √5)/4 has no real solution because the right side is negative and e^x is always positive.
• Check domain validity before proceeding with algebraic solutions.

⚠️ Labeling and terminology

• Non-linear systems are called "consistent" (has solutions) or "inconsistent" (no solutions).
• The terms "dependent" and "independent" are generally not used for non-linear systems.
• A consistent non-linear system can have 1, 2, 3, or more solutions—any finite number—without being "dependent."

⚠️ When to use which method

Situation	Recommended method	Why
Both equations have x² and y² only	Elimination	Can remove one squared variable cleanly
One equation is linear or simple	Substitution	Easy to isolate one variable
Equations have xy cross-terms	Elimination of cross-terms	Simplifies to y = ±x or similar
Complicated expressions	Graph first	Avoid algebraic dead ends; see solution count
Three or more variables	Case-by-case from simplest equation	Organize work; avoid losing solutions

⚠️ Reflection symmetry trick

• If equation 2 can be obtained from equation 1 by swapping x and y, the graph of equation 2 is the reflection of equation 1 across the line y = x.
• Example: x² + 2xy – 16 = 0 and y² + 2xy – 16 = 0. The second graph is the first reflected across y = x.
• This saves graphing effort and confirms intersection points lie on y = x or symmetrically placed.

🧭 Overview

🧠 One-sentence thesis

Sequences are functions whose domains are the natural numbers, and they can be described explicitly by formulas, recursively by initial values and recursion equations, or classified as arithmetic (constant addition) or geometric (constant multiplication) patterns.

📌 Key points (3–5)

• What a sequence is: a function mapping natural numbers to terms, written as {aₙ} or aₙ, n ≥ 1.
• Two main explicit patterns: arithmetic sequences add a fixed number d each step; geometric sequences multiply by a fixed number r each step.
• Recursive vs explicit definitions: some sequences are defined by starting values and recursion equations (e.g., factorials, Fibonacci); others have direct formulas for the nth term.
• Common confusion: not every sequence is arithmetic or geometric—many are neither, and some can be broken into separate patterns for numerators and denominators.
• Why sequences matter: they model real-world processes like compound interest and population growth, and provide the foundation for series and financial mathematics.

📐 What sequences are

📐 Definition and notation

Sequence: a function a whose domain is the natural numbers {1, 2, 3, ...}.

• The value a(n) is written as aₙ and called the nth term.
• Notation: {aₙ}∞ₙ₌₁ or aₙ, n ≥ 1.
• Example: the sequence 1/2, −3/4, 9/8, −27/16, ... assigns a₁ = 1/2, a₂ = −3/4, a₃ = 9/8, etc.

🔢 Terms and indexing

• Each number in the list is a term: first term, second term, third term, etc.
• The index (usually n, k, i, or j) counts which term you're looking at.
• Some sequences start at n = 0 or another value; the excerpt notes this is technically a relaxed interpretation of the definition, but widely accepted.

📊 Graphing sequences

• Plot points (n, aₙ) for each natural number n in the domain.
• Do not connect the dots—the domain is discrete (only whole numbers), not continuous intervals.
• Example: the sequence bₖ = (−1)^k / (2k + 1), k ≥ 0, produces isolated points, not a curve.

🔁 Recursive sequences

🔁 How recursion works

• A recursive definition gives:
  • One or more initial values (e.g., a₁ = 7).
  • A recursion equation that builds new terms from previous ones (e.g., aₙ₊₁ = 2 − aₙ).
• You compute terms step-by-step: use the initial value to find the second term, then use the second to find the third, etc.

🔁 Factorial example

n-factorial: 0! = 1 and n! = n · (n − 1)! for n ≥ 1.

• Written out: 1! = 1, 2! = 2·1 = 2, 3! = 3·2·1 = 6, 4! = 4·3·2·1 = 24.
• Informal shorthand: n! = n · (n − 1) · (n − 2) · ... · 2 · 1, with 0! = 1 as a special case.
• This is a recursive sequence where each term depends on the previous term.

🔁 Fibonacci sequence

• Defined by F₀ = 1, F₁ = 1, Fₙ = Fₙ₋₁ + Fₙ₋₂ for n ≥ 2.
• Each term is the sum of the two preceding terms.
• The excerpt mentions this is "world famous" and explored further in exercises.

➕ Arithmetic sequences

➕ Definition and pattern

Arithmetic sequence: a sequence where aₙ₊₁ = aₙ + d for all n ≥ k, for some fixed number d called the common difference.

• You move from term to term by adding the same number d every time.
• Rewritten: aₙ₊₁ − aₙ = d (hence "common difference").
• Example: 1, 3, 5, 7, ... is arithmetic with d = 2 (the odd numbers).

➕ Explicit formula

Formula: aₙ = a + (n − 1)d, n ≥ 1, where a is the first term.

• Why it works: to reach the nth term, you add d to a exactly (n − 1) times.
• Example: for the sequence 3, 5, 7, 9, ..., a = 3 and d = 2, so aₙ = 3 + (n − 1)·2 = 2n + 1 (after simplifying).

➕ How to recognize arithmetic sequences

• Compute successive differences: a₂ − a₁, a₃ − a₂, a₄ − a₃, ...
• If all differences equal the same number d, the sequence is arithmetic.
• Example: for 1, 3, 5, 7, differences are 2, 2, 2 → arithmetic with d = 2.

✖️ Geometric sequences

✖️ Definition and pattern

Geometric sequence: a sequence where aₙ₊₁ = r·aₙ for all n ≥ k, for some fixed number r ≠ 0 called the common ratio.

• You move from term to term by multiplying by the same number r every time.
• Rewritten (if r ≠ 0): aₙ₊₁ / aₙ = r (hence "common ratio").
• Example: 1/2, −3/4, 9/8, −27/16, ... has r = −3/2.

✖️ Explicit formula

Formula: aₙ = a·r^(n−1), n ≥ 1, where a is the first term and r ≠ 0.

• Why it works: to reach the nth term, you multiply a by r exactly (n − 1) times.
• Example: for 1/2, −3/4, 9/8, −27/16, ..., a = 1/2 and r = −3/2, so aₙ = (1/2)·(−3/2)^(n−1) = (−3)^(n−1) / 2^n.
• Why r ≠ 0: to avoid the indeterminate form 0⁰.

✖️ How to recognize geometric sequences

• Compute successive ratios: a₂/a₁, a₃/a₂, a₄/a₃, ...
• If all ratios equal the same number r, the sequence is geometric.
• Example: for 1, −1/3, 1/5, −1/7, ratios are −1/3 and −3/5 → not geometric (different ratios).

✖️ Alternating sequences

• A sequence like 1, −1/3, 1/5, −1/7, ... alternates signs.
• This is often produced by a factor like (−1)^k in the formula.
• Example: bₖ = (−1)^k / (2k + 1) alternates because (−1)^k switches between +1 and −1.

🔍 Finding explicit formulas

🔍 When in doubt, write it out

• Rule of thumb: list the first several terms to spot patterns.
• Look for arithmetic (constant differences), geometric (constant ratios), or other regularities.
• Example: 2, 5, 10, 17, ... can be rewritten as 1+1, 4+1, 9+1, 16+1, ... → aₙ = n² + 1.

🔍 Breaking into parts

• If a sequence is neither arithmetic nor geometric overall, try separating numerators and denominators (or other components).
• Example: 2/5, 2/1, −2/3, −2/7, ...
  • Numerators: all 2 (constant).
  • Denominators: 5, 1, −3, −7 form an arithmetic sequence with d = −4 → dₙ = 9 − 4n.
  • Combined: aₙ = 2 / (9 − 4n).

🔍 Combining arithmetic and geometric

• Example: 1/1, −2/7, 4/13, −8/19, ...
  • Numerators: 1, −2, 4, −8 → geometric with r = −2 → cₙ = (−2)^(n−1).
  • Denominators: 1, 7, 13, 19 → arithmetic with d = 6 → dₙ = 6n − 5.
  • Combined: aₙ = (−2)^(n−1) / (6n − 5).

🔍 Non-uniqueness warning

• Important: given only a finite sample of terms, infinitely many formulas can generate the same initial terms.
• Example: 2, 5, 10, 17, ... could be aₙ = n² + 1 or bₙ = (−1/4)n⁴ + (5/2)n³ − (31/4)n² + (25/2)n − 5 (both produce the same first four terms).
• Your answer is correct if it produces the required terms in the correct order, even if it differs from a textbook answer.

💰 Application: compound interest

💰 Sequences in finance

• Recall the compound interest formula: Aₖ = P·(1 + r/n)^k, where:
  • P = principal (initial deposit).
  • r = annual percentage rate.
  • n = number of compounding periods per year.
  • k = number of compounding periods elapsed.
• This is a geometric sequence with first term P·(1 + r/n) and common ratio (1 + r/n).

💰 Annuities (preview)

• An annuity involves regular additional deposits, not just one initial amount.
• The value grows according to a more complex pattern than a simple geometric sequence.
• The excerpt notes this will be covered in the next section after developing more tools.

⚠️ Common confusions

⚠️ Arithmetic vs geometric

Feature	Arithmetic	Geometric
Operation	Add d each step	Multiply by r each step
Test	Check if aₙ₊₁ − aₙ is constant	Check if aₙ₊₁ / aₙ is constant
Formula	aₙ = a + (n − 1)d	aₙ = a·r^(n−1)
Example	1, 3, 5, 7 (d = 2)	2, 6, 18, 54 (r = 3)

• Don't confuse: many sequences are neither arithmetic nor geometric (e.g., 1, −2/7, 4/13, −8/19).

⚠️ Indexing starting points

• Some sequences start at n = 1, others at n = 0 or another value.
• Technically, Definition 9.1 requires natural numbers {1, 2, 3, ...}, but the excerpt adopts a relaxed view.
• To shift a sequence starting at k = 0 to start at k = 1, replace k with k − 1 in the formula.

⚠️ Recursive vs explicit

• Recursive: gives starting value(s) and a rule to build terms from previous ones (e.g., Fibonacci).
• Explicit: gives a direct formula for the nth term without needing previous terms (e.g., aₙ = 2n − 1).
• Some sequences are easier to define recursively; others are easier to define explicitly.

🧭 Overview

🧠 One-sentence thesis

Summation notation provides a compact way to express the sum of many terms following a pattern, enabling efficient calculation of arithmetic and geometric sums and applications like annuities and infinite series.

📌 Key points (3–5)

• What summation notation does: compresses long sums into a single expression using an index variable, limits, and a formula for each term.
• Index variable flexibility: the index letter (n, k, j, etc.) is a "dummy variable" that can be changed without affecting the sum's value.
• Formulas for special sums: arithmetic and geometric sequences have closed-form sum formulas that avoid adding term-by-term.
• Common confusion: distinguishing when geometric series converge (absolute value of r less than 1) versus when they diverge (absolute value of r greater than or equal to 1).
• Real-world application: annuities use geometric sum formulas to calculate future value when regular payments earn compound interest.

📐 Understanding summation notation

📐 Basic structure and meaning

Summation notation: a compact way to write the sum of terms by specifying an index variable, starting value (lower limit), ending value (upper limit), and a formula for each term.

• Example from the excerpt: the sum 5 + 7 + 9 + 11 can be written as the summation from n = 3 to 6 of (2n − 1).
• To evaluate: substitute each integer from the lower limit to the upper limit into the formula and add all resulting terms.
• The notation replaces writing out many terms with ellipses ("...").

🔄 The dummy variable property

• The index variable is called a "dummy variable" because changing its letter does not change the sum.
• Example: summation from n = 3 to 6 of (2n − 1) equals summation from k = 3 to 6 of (2k − 1) equals summation from j = 3 to 6 of (2j − 1).
• Don't confuse: the index variable with other variables in the formula—only the index gets replaced during summation; other letters (like x in a polynomial) remain as variables.

📝 Uses in definitions

• Summation notation appears in mathematical definitions to express general forms concisely.
• Example: polynomials can be defined as f(x) = summation from k = 0 to n of (a_k times x to the power k).
• Matrix operations also use summation: the product of row i and column j is the summation from k = 1 to n of (a_ik times b_kj).

🔢 Working with summations

🔢 Evaluating summations step-by-step

The excerpt provides three evaluation examples:

1. Summation from k = 1 to 4 of (13 divided by 100 to the power k): substitute k = 1, 2, 3, 4 into the formula and add to get 0.13131313.
2. Summation from n = 0 to 4 of (n factorial divided by 2): recall that n factorial is n times (n−1) times ... times 1; evaluating gives 17.
3. Summation from n = 1 to 5 of ((−1) to the power (n+1) times n times (x−1) to the power n): produces an alternating sum involving the variable x.

Key technique: replace the index with each integer in the range and compute each term before adding.

✍️ Writing sums in summation notation

To convert a written-out sum into summation notation:

1. Identify the pattern: determine the formula for the nth term (often using arithmetic or geometric sequence formulas).
2. Find the lower limit: the value of the index that produces the first term.
3. Find the upper limit: the value of the index that produces the last term.

Example from the excerpt: the sum 1 + 3 + 5 + ... + 117 is an arithmetic sequence with first term 1 and common difference 2, so the nth term is 2n − 1. Setting 2n − 1 = 117 gives n = 59, so the sum is the summation from n = 1 to 59 of (2n − 1).

Common pitfall: when the upper limit itself is a variable (like n), choose a different letter for the index (like k) to avoid confusion.

🧮 Formulas for arithmetic and geometric sums

🧮 Arithmetic sequence sum formula

For an arithmetic sequence with first term a and common difference d, the sum S of the first n terms is:

• Formula 1: S = n times (first term plus nth term) divided by 2.
• Formula 2: S = (n divided by 2) times (2a plus (n−1) times d).

Why it works: write the sum forwards and backwards, add the two equations vertically; every pair sums to the same value (2a plus (n−1) times d), and there are n such pairs, giving 2S = n times that constant.

Helpful memory aid: the sum equals the number of terms times the average of the first and last terms.

Special case: the sum of the first n natural numbers (1 + 2 + 3 + ... + n) is n times (n+1) divided by 2. Example: the sum of the first 100 natural numbers is 5050.

🧮 Geometric sequence sum formula

For a geometric sequence with first term a and common ratio r, the sum S of the first n terms depends on r:

Condition	Formula	Derivation insight
r ≠ 1	S = a times (1 − r to the power n) divided by (1 − r), or equivalently (first term minus (n+1)th term) divided by (1 − r)	Multiply the sum by r, subtract from the original sum; most terms cancel
r = 1	S = n times a	All terms equal a, so the sum is just a added n times

Key step in derivation: comparing S and r times S causes all middle terms to cancel, leaving only the first and last terms.

📊 Properties of summation notation

The excerpt lists four properties (proofs left as exercises):

1. Linearity: summation of (a_n ± b_n) = summation of a_n ± summation of b_n.
2. Constant factor: summation of (c times a_n) = c times summation of a_n.
3. Splitting the range: summation from m to p can be split into summation from m to j plus summation from (j+1) to p.
4. Index shifting: summation from m to p of a_n equals summation from (m+r) to (p+r) of a_(n−r).

Advice from the excerpt: "When in doubt, write it out!"—expand the summation to see why these properties hold.

💰 Application: ordinary annuities

💰 What an annuity is

Ordinary annuity: an investment plan where equal payments P are deposited at the end of each compounding period into an account earning interest at rate r compounded n times per year.

• Differs from simple compound interest because payments are added regularly, not just once.
• Let i = r divided by n (interest rate per period).

💰 Deriving the annuity formula

The excerpt traces the account balance A_k after k periods:

• After period 1: A_1 = P (first payment, no interest yet).
• After period 2: A_2 = P times (1 + i) + P (first payment grew by interest, plus second payment).
• After period 3: A_3 = P times (1 + i) squared times (1 + 1/(1+i) + 1/(1+i) squared).
• Pattern: A_k = P times (1 + i) to the power (k−1) times a geometric sum.

The sum in parentheses is a geometric series with first term 1 and ratio 1/(1+i). Using the geometric sum formula and simplifying gives:

Future value formula: A = P times ((1 + i) to the power (nt) − 1) divided by i, where t is the number of years and k = nt.

Annuity-due variation: if payments are made at the beginning of each period instead of the end, the formula changes (derivation left as an exercise).

💰 Example calculation

Given: 6% annual interest compounded monthly, $50 monthly payments, 30 years.

• i = 0.06 divided by 12 = 0.005.
• nt = 12 times 30 = 360.
• A = 50 times ((1.005) to the power 360 − 1) divided by 0.005 ≈ $50,225.75.

To find how long to reach $100,000: set A = 100,000, isolate the exponential term, take natural logarithms, and solve for t ≈ 40.06 years.

Observation: in just 10 additional years (from 30 to 40), the annuity nearly doubles—a lesson about the power of compounding over time.

∞ Infinite series and geometric series

∞ From finite sums to infinite series

Series: an infinite sum, written as summation from k = 1 to infinity.

• Example: the repeating decimal 0.9999... equals 0.9 + 0.09 + 0.009 + ..., which is summation from k = 1 to n of (9 divided by 10 to the power k).
• Using the geometric sum formula: summation from k = 1 to n of (9 divided by 10 to the power k) = 1 − 1/(10 to the power (n+1)).
• As n approaches infinity, 1/(10 to the power (n+1)) approaches 0, so the infinite sum equals 1.
• Conclusion: 0.9999... = 1 (which may be surprising but follows from the logic of infinite sums).

∞ Geometric series convergence theorem

For a geometric sequence with first term a and ratio r:

• If absolute value of r < 1: the infinite sum converges to a divided by (1 − r).
• If absolute value of r ≥ 1: the sum is not defined (does not converge).

Why the condition matters: when absolute value of r < 1, r to the power n approaches 0 as n approaches infinity, so the formula from the finite case simplifies. When absolute value of r ≥ 1, the terms do not shrink to zero, so the sum grows without bound or oscillates.

Application: any non-terminating decimal can be expressed as an infinite geometric series with denominators that are powers of 10, allowing conversion of repeating decimals to fractions.

∞ Don't confuse: convergent vs divergent series

• Convergent: the partial sums approach a finite limit (happens when absolute value of r < 1 for geometric series).
• Divergent: the partial sums do not approach a finite limit (happens when absolute value of r ≥ 1).
• The excerpt notes that "what goes wrong" when absolute value of r ≥ 1 is left to Calculus, but exercises explore specific cases.

🧭 Overview

🧠 One-sentence thesis

The Principle of Mathematical Induction (PMI) allows us to prove formulas for all natural numbers by establishing a base case and showing that truth for any number k implies truth for k+1, much like climbing stairs one step at a time.

📌 Key points (3–5)

• What PMI proves: formulas or properties that hold for all natural numbers n (or all n beyond some starting point).
• Two required steps: (1) prove the base case P(1) is true, and (2) prove that whenever P(k) is true, P(k+1) must also be true (the inductive step).
• The induction hypothesis: assuming P(k) is true is not circular reasoning—it's showing that truth "propagates" from one number to the next.
• Common confusion: the induction hypothesis feels like assuming what you want to prove, but you're only assuming it for k to deduce it for k+1; the base case + propagation together cover all natural numbers.
• Why it matters: PMI rigorously proves formulas that were previously only motivated by patterns (e.g., arithmetic sequence formulas, summation formulas).

🏗️ The structure of an induction proof

🏗️ What PMI says

The Principle of Mathematical Induction (PMI): Suppose P(n) is a sentence involving the natural number n. IF (1) P(1) is true and (2) whenever P(k) is true, it follows that P(k+1) is also true, THEN the sentence P(n) is true for all natural numbers n.

• PMI is a principle (or axiom)—a property of natural numbers we choose to accept.
• P(n) acts like function notation: if P(n) is "n squared plus 1 equals 3," then P(1) is "1 squared plus 1 equals 3" (which is false).

🪜 The staircase metaphor

• Base step: show you can get on the stairs (prove P(1) is true).
• Inductive step: show you can climb from any step k to the next step k+1 (prove P(k) true implies P(k+1) true).
• Conclusion: if both hold, you can climb the entire staircase—P(n) is true for all natural numbers n.

🔄 How the inductive step works

• You assume P(k) is true (the induction hypothesis).
• You deduce P(k+1) is true using that assumption.
• This is not circular: you're showing that truth at k forces truth at k+1.
• Example chain: P(1) true → P(2) true → P(3) true → … and so on forever.

🧮 Proving the arithmetic sequence formula

🧮 The formula to prove

• Arithmetic sequences are defined recursively: start with a₁ = a, then aₙ₊₁ = aₙ + d for n ≥ 1.
• The pattern suggests aₙ = a + (n − 1)d, but how do we prove this?

✅ Base case (n = 1)

• P(1) says a₁ = a + (1 − 1)d.
• This simplifies to a₁ = a, which is true by the definition of the sequence.

➡️ Inductive step (k → k+1)

• Assume P(k) is true: aₖ = a + (k − 1)d.
• Show P(k+1) is true: aₖ₊₁ = a + ((k+1) − 1)d = a + kd.
• By the recursive definition, aₖ₊₁ = aₖ + d.
• Substitute the induction hypothesis: aₖ₊₁ = (a + (k−1)d) + d = a + kd. ✓
• Hence P(k+1) is true.

🎯 Conclusion

• By PMI, aₙ = a + (n−1)d is true for all natural numbers n.

📚 Example proofs using PMI

📚 Sum formula for arithmetic sequences

• Goal: prove the sum from j=1 to n of (a + (j−1)d) equals (n/2)(2a + (n−1)d).
• Base case (n=1): left side = a, right side = (1/2)(2a) = a. ✓
• Inductive step: assume the formula holds for k terms; show it holds for k+1 terms.
  • Sum of first k+1 terms = (sum of first k terms) + (k+1)st term.
  • Use the induction hypothesis for the first k terms, then add the (k+1)st term.
  • Algebraic manipulation shows both sides equal ((k+1)/2)(2a + kd). ✓

🔢 Conjugate of a power

• Goal: prove (z̄)ⁿ = z̄ⁿ for complex numbers z and n ≥ 1.
• Base case (n=1): (z̄)¹ = z̄¹ is trivially true.
• Inductive step: assume (z̄)ᵏ = z̄ᵏ; show (z̄)ᵏ⁺¹ = z̄ᵏ⁺¹.
  • (z̄)ᵏ⁺¹ = (z̄)ᵏ · z̄ = z̄ᵏ · z̄ (by induction hypothesis).
  • Use the product rule for conjugates: z̄ᵏ · z̄ = z̄ᵏ⁺¹. ✓

📈 Exponential inequality (base case ≠ 1)

• Goal: prove 3ⁿ > 100n for n > 5.
• Wrinkle: the base case is n = 6, not n = 1 (PMI still works; conclusion holds for all n ≥ 6).
• Base case (n=6): 3⁶ = 729 > 600 = 100(6). ✓
• Inductive step: assume 3ᵏ > 100k; show 3ᵏ⁺¹ > 100(k+1).
  • 3ᵏ⁺¹ = 3 · 3ᵏ > 3(100k) = 300k (by induction hypothesis).
  • Need to show 300k > 100(k+1), i.e., k > 1/2, which is true for k ≥ 6. ✓

🧩 Determinant property (case analysis)

• Goal: if A′ is obtained from an n×n matrix A by replacing row R with cR, then det(A′) = c·det(A).
• Base case (n=1): A = [a], A′ = [ca], so det(A′) = ca = c·det(A). ✓
• Inductive step: assume true for k×k matrices; show true for (k+1)×(k+1) matrices.
  • Case 1: R is the first row. The first row of A′ is c times the first row of A; cofactors are identical. Factoring out c gives det(A′) = c·det(A). ✓
  • Case 2: R is not the first row. The first rows of A and A′ are the same, but the cofactor matrices satisfy the induction hypothesis. Again, det(A′) = c·det(A). ✓
• (Note: Case 1 doesn't use the induction hypothesis; the proof could be restructured, but this approach is more intuitive.)

🔑 Key takeaways

🔑 When to use PMI

• Whenever you see a property stated "for all natural numbers n," PMI is likely needed for a rigorous proof.
• PMI is often hidden in the background of proofs in advanced courses.

🔑 Don't confuse: assuming vs. proving

• Common mistake: "Aren't we assuming what we want to prove?"
• Clarification: You assume P(k) only to deduce P(k+1). The base case + the propagation rule together establish P(n) for all n.
• Think of it as a chain reaction: P(1) is the spark, and the inductive step is the rule that spreads the fire to every subsequent number.

🔑 Variations on PMI

• The base case can start at any natural number (e.g., n = 4 or n = 6), not just n = 1.
• The conclusion then holds for all n greater than or equal to that base case.

🧭 Overview

🧠 One-sentence thesis

The Binomial Theorem provides a formula for expanding powers of binomials (a + b) to the nth power using binomial coefficients, which can be computed via factorials or Pascal's Triangle, and the theorem is proven by mathematical induction.

📌 Key points (3–5)

• What the theorem does: gives a formula to expand (a + b) raised to any natural number n without repeated multiplication.
• Factorials and binomial coefficients: binomial coefficients are computed as n! / (j! (n − j)!) and count the number of ways to select j items from n items.
• Pascal's Triangle: a quick visual method to generate binomial coefficients row by row, where each interior number is the sum of the two numbers above it.
• Common confusion: binomial coefficients versus factorial notation—binomial coefficient (n choose j) is a ratio of factorials, not a single factorial.
• Why it matters: the theorem generalizes polynomial expansion, appears in counting problems, and connects algebra to combinatorics.

🔢 Factorials and their properties

🔢 Definition of factorial

For a whole number n, n factorial (denoted n!) is the term f_n of the sequence f_0 = 1, f_n = n · f_(n−1) for n ≥ 1.

• Base case: 0! = 1.
• Recursive: n! = n · (n − 1)! for n ≥ 1.
• Informally, n! = n · (n − 1) · (n − 2) · ... · 2 · 1.
• Example: 3! = 3 · 2 · 1 = 6; 4! = 4 · 3! = 24.

🧮 Simplifying factorial expressions

• Key technique: cancel common factorial factors in numerator and denominator.
• Example: 7! / 5! = (7 · 6 · 5!) / 5! = 7 · 6 = 42.
• Example: 1000! / (998! · 2!) = (1000 · 999 · 998!) / (998! · 2!) = (1000 · 999) / 2 = 499500.
• Example: (k + 2)! / (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! / (k − 1)! = k(k + 1)(k + 2).
• Don't confuse: the stipulation k ≥ 1 ensures all factorials are defined (no negative factorials).

📈 Growth rate of factorials

• Factorials grow extremely quickly.
• The excerpt proves by induction that n! > 3^n for all n ≥ 7.
• Base case: 7! = 5040 > 3^7 = 2187.
• Inductive step: assume k! > 3^k; then (k + 1)! = (k + 1) · k! > (k + 1) · 3^k ≥ 8 · 3^k > 3 · 3^k = 3^(k+1).
• For any real x > 0, n! eventually overtakes x^n and the ratio x^n / n! approaches 0 as n approaches infinity.

🎵 Counting applications

• Factorials count arrangements (order matters).
• Example: 50 songs arranged in a playlist → 50! different playlists.
• Example: select 10 songs from 50 in order → 50! / 40! playlists.
• Example: select 10 songs from 50 where order does not matter → 50! / (40! · 10!) ways.

🎲 Binomial coefficients

🎲 Definition and interpretation

Given two whole numbers n and j with n ≥ j, the binomial coefficient (n choose j) is the whole number given by (n choose j) = n! / (j! (n − j)!).

• Read as "n choose j."
• Physical interpretation: the number of ways to select j items from n items where order is unimportant.
• Example: (5 choose 2) = 5! / (2! · 3!) = (5 · 4) / 2 = 10.
• This counts the number of ways to choose 2 friends from 5 friends (or equivalently, which 3 friends stay home).

🔗 Key theorem for binomial coefficients

Theorem 9.3: For natural numbers n and j with n ≥ j, (n choose (j − 1)) + (n choose j) = ((n + 1) choose j).

• This is the additive relationship used to build Pascal's Triangle.
• Proof is computational: expand both sides using the definition of binomial coefficients, use the recursive property of factorials, find a common denominator, and simplify.
• Example: (1 choose 0) + (1 choose 1) = (2 choose 1).

🏛️ The Binomial Theorem

🏛️ Statement of the theorem

Theorem 9.4 (Binomial Theorem): For nonzero real numbers a and b, (a + b)^n = sum from j = 0 to n of (n choose j) · a^(n−j) · b^j for all natural numbers n.

• The binomial coefficients are the coefficients in the expansion.
• Example for n = 4: (a + b)^4 = (4 choose 0) a^4 + (4 choose 1) a^3 b + (4 choose 2) a^2 b^2 + (4 choose 3) a b^3 + (4 choose 4) b^4.
• Pattern: in each term, the total of the exponents is n; the exponent on a starts at n and decreases by 1; the exponent on b starts at 0 and increases by 1.
• The lower number in (n choose j) matches the exponent of b in that term.

🔍 Why the theorem works (intuition)

• Think of (a + b)^4 as multiplying (a + b)(a + b)(a + b)(a + b).
• The answer is the sum of all products with exactly 4 factors (some a, some b).
• To count the number of ways to obtain 1 factor of b out of 4 possible factors (forcing the remaining 3 to be a), the answer is (4 choose 1).
• Hence the term (4 choose 1) a^3 b appears in the expansion.
• The other terms cover the remaining cases (0 b's, 2 b's, 3 b's, 4 b's).

🧩 Proof by induction

• Let P(n) be the expansion formula in the theorem.
• Base case (n = 1): (a + b)^1 = (1 choose 0) a + (1 choose 1) b = a + b. True.
• Inductive step: assume P(k) is true; show P(k + 1) is true.
  • (a + b)^(k+1) = (a + b) · (a + b)^k.
  • Apply the induction hypothesis to (a + b)^k.
  • Distribute a and b through the sum.
  • Separate the first and last terms (which cannot be combined).
  • Shift the index on one summation so both start at j = 1.
  • Combine the two sums using Theorem 9.3.
  • Recognize that ((k+1) choose 0) = 1 and ((k+1) choose (k+1)) = 1.
  • Result: (a + b)^(k+1) = sum from j = 0 to k+1 of ((k+1) choose j) · a^((k+1)−j) · b^j.
• By induction, the theorem holds for all natural numbers n.

🧪 Examples of using the theorem

Example 1: Expand (x − 2)^4.

• Identify a = x, b = −2, n = 4.
• (x − 2)^4 = sum from j = 0 to 4 of (4 choose j) · x^(4−j) · (−2)^j.
• Result: x^4 − 8x^3 + 24x^2 − 32x + 16.

Example 2: Compute 2.1^3.

• Write 2.1^3 = (2 + 0.1)^3.
• Identify a = 2, b = 0.1 = 1/10, n = 3.
• (2 + 1/10)^3 = sum from j = 0 to 3 of (3 choose j) · 2^(3−j) · (1/10)^j.
• Result: 8 + 1.2 + 0.06 + 0.001 = 9.261.

Example 3: Find the term containing x^3 in (2x + y)^5.

• Identify a = 2x, b = y, n = 5.
• The exponents must add to 5; if the exponent on x is 3, the exponent on y must be 2.
• This corresponds to j = 2.
• Term: (5 choose 2) · (2x)^(5−2) · y^2 = 10 · (2x)^3 · y^2 = 80 x^3 y^2.

🔺 Pascal's Triangle

🔺 Structure and construction

• Pascal's Triangle arranges binomial coefficients in a triangular pattern.
• Each row corresponds to a value of n; the entries in row n are (n choose 0), (n choose 1), ..., (n choose n).
• Each row begins and ends with 1 (since (n choose 0) = 1 and (n choose n) = 1).
• Interior numbers are generated using Theorem 9.3: each number is the sum of the two numbers directly above it.
• Example: row 0 is 1; row 1 is 1 1; row 2 is 1 2 1; row 3 is 1 3 3 1; row 4 is 1 4 6 4 1.

🚀 Using Pascal's Triangle for expansion

• To expand (a + b)^n, use the nth row of Pascal's Triangle for the coefficients.
• The exponent of a starts at n and decreases; the exponent of b starts at 0 and increases.
• Example: expand (3x − y)^4 using row 4 (1 4 6 4 1).
  • (3x − y)^4 = 1·(3x)^4 + 4·(3x)^3·(−y) + 6·(3x)^2·(−y)^2 + 4·(3x)·(−y)^3 + 1·(−y)^4.
  • Result: 81x^4 − 108x^3 y + 54x^2 y^2 − 12xy^3 + y^4.

⚖️ When to use Pascal's Triangle vs the Binomial Theorem

Method	Best for
Pascal's Triangle	Expanding an entire binomial quickly (small n)
Binomial Theorem	Finding one or a few specific terms (any n)

• Don't confuse: Pascal's Triangle is efficient for full expansions; the Binomial Theorem is better for extracting individual terms without computing the entire expansion.