Contents / Algebra / Functions & Graphs

Chapter 5

Functions & Graphs

Function notation, domain and range, linear and quadratic graphs, transformations, and inverses.

Introduction

Function notation, domain and range, linear and quadratic graphs, transformations, and inverses.

A function is a special kind of relation: it pairs every input with exactly one output. This "exactly one" rule is what makes functions predictable and powerful for modeling the real world.

Functions appear as formulas, graphs, tables, and verbal descriptions. Once you understand the definition you can decide whether any given relation qualifies as a function and can use function notation fluently.

5.1Precise Definition

Definition 5.1 (Function). A function $f$ from a set $A$ (the domain) to a set $B$ is a rule that assigns to each element $x$ in $A$ exactly one element $f (x)$ in $B$ .

No input may have two different outputs. Two different inputs may share the same output.

Intuition. A function is a machine with a strict rule: drop in one number, get exactly one number out. No arguing, no choices.

Vending machines are functions (one button → one item); a "choose any snack for $1" machine would not be.

Example 5.2. Is " $y$ is the square of $x$ " a function? Pick any input, say $x = 3$ . Squaring gives exactly one output, $y = 9$ . There is no way to get a second value, so every input has a unique output: it is a function, namely $f (x) = x^{2}$ .

Solution.

Is " $y$ is a square root of $x$ " a function? Pick $x = 9$ . Both $y = 3$ and $y = - 3$ satisfy $y^{2} = 9$ , so the single input $9$ has two outputs. That breaks the one-output rule, so it is not a function. (Sanity check: this failure happens for every $x > 0$ , since a positive number always has two real square roots.)

□

Exercises for §5.1 →

5.2The Vertical Line Test

Proposition 5.3 (Vertical line test). For graphs in the coordinate plane: if any vertical line crosses the graph more than once, the graph does not represent a function.

Proof. A vertical line at a fixed $x$ would intersect two points with the same $x$ but different $y$ 's, violating the one-output rule.∎

Intuition. Imagine a vertical stick sliding across the graph. If it ever touches the curve in two places at once, that x has two y's and is not a function.

Circles, most sideways parabolas, and figure-eights all fail the test.

Example 5.4. Does $y = x^{2}$ pass? Fix any vertical line, say $x = 3$ . The only point of the graph on that line is $(3, 9)$ , because $x = 3$ forces $y = 9$ . No matter which vertical line you pick, it meets the curve at most once, so $y = x^{2}$ is a function.

Solution.

Does $x = y^{2}$ pass? Take the vertical line $x = 4$ . Points on the graph with $x = 4$ satisfy $4 = y^{2}$ , i.e. $y = 2$ or $y = - 2$ , so the line hits the sideways parabola at both $(4, 2)$ and $(4, - 2)$ . Two hits means one $x$ maps to two $y$ 's, so $x = y^{2}$ fails the test and is not a function. (Sanity check: this matches the earlier "square root" relation, since solving $x = y^{2}$ for $y$ gives $y = \pm x$ .)

□

Exercises for §5.2 →

5.3Function Notation

Notation (Function notation). $f (x)$ names the output of the function named $f$ when the input is $x$ . It is read " $f$ of $x$ ", and it does not mean multiplication.

Evaluating: replace every occurrence of the input variable with the given number and simplify.

Intuition. f is the name of the recipe. The parentheses hold the ingredient you are feeding it.

f(3) asks "what does the recipe give when I use 3?"

Example 5.5. Let f(x) = 2x² − 5x + 1.

Solution.

f(2) = 2(4) − 5(2) + 1 = 8 − 10 + 1 = −1.
f(−1) = 2(1) − 5(−1) + 1 = 2 + 5 + 1 = 8.

□

Exercises for §5.3 →

5.4Everyday and Mathematical Examples

Area of a square: A(s) = s² (one side → one area).
Temperature conversion: C(F) = (F − 32) × 5/9.
Cost of buying n apples at $0.50 each: C(n) = 0.50n (plus any fixed fee).
The rule "the output is the input plus 3" is the function f(x) = x + 3.

Intuition. Many real quantities depend on exactly one other quantity in a rule-like way; that dependence is a function.

Once named, the function can be used, composed, graphed, and studied without rewriting the rule every time.

Example 5.6. The cost of a taxi ride that charges $2.50 per mile is the function C(m) = 2.5m. C(4) = $10 for a 4-mile ride. Different distances give different costs; each distance gives exactly one cost.

Exercises for §5.4 →

Common Mistakes

Thinking f(x) means f multiplied by x: It is function application, not multiplication.
Believing that every relation is a function: Many are not (circles, one-to-many tables).
Forgetting that two inputs can map to the same output: Functions are allowed to be many-to-one; they are forbidden only from being one-to-many.
Confusing domain with range: Domain = inputs you may feed in; range = outputs that actually come out.
Assuming the vertical line test applies only to continuous curves: It applies to any graph, including discrete points.

The domain of a function is the set of all input values for which the function rule produces a real number output. It is the "legal input list."

Knowing the domain is essential before graphing, finding range, or composing functions. Many common mistakes in algebra come from forgetting domain restrictions.

5.5Definition and Importance

Definition 5.7 (Domain). For a function $f$ , the domain is

{x ∣ f (x) is defined and real} .

Values that produce division by zero, negative numbers under even roots, or other undefined operations are excluded.

Intuition. The domain is the menu of ingredients the recipe accepts. Some ingredients (like negative numbers for square roots) make the recipe blow up.

You must know what you are allowed to plug in before you start cooking.

Example 5.8. Find the domain of $f (x) = 1/ x$ .

Solution.

Ask where the rule could fail to give a real number. The only danger is dividing by zero.
Set the denominator to zero: $x = 0$ . That input is illegal.
Every other real number is fine, so the domain is all reals except $0$ , written $(- \infty, 0) \cup (0, \infty)$ .
Sanity check: $f (2) = 1/2$ and $f (- 5) = - 1/5$ are perfectly defined, but $f (0) = 1/0$ is not, matching the exclusion.

□

Exercises for §5.5 →

5.6Finding the Domain of Rational Functions

Proposition 5.9 (Domain of a rational function). For a rational function $p (x) / q (x)$ , exclude every real root of the denominator, that is, every solution of

q (x) = 0.

If the numerator and denominator share a common factor, the simplified function has a hole rather than a vertical asymptote at that root, but the original expression is still undefined there; the value is excluded from the domain either way.

Intuition. The bottom of a fraction is a "no-fly zone" when it hits zero.

Canceling a factor removes the asymptote but the original function still had a hole at that x.

Example 5.10. Find the domain of $f (x) = \frac{x - 2}{x ^{2} - 4}$ .

Solution.

Set the denominator to zero and solve: $x^{2} - 4 = 0 \Rightarrow (x - 2) (x + 2) = 0 \Rightarrow x = 2$ or $x = - 2$ . Both are excluded from the domain.
Notice the numerator and denominator share the factor $(x - 2)$ : $f (x) = \frac{x - 2}{( x - 2 ) ( x + 2 )} = \frac{1}{x + 2}$ for $x \neq = 2$ .
The cancellation turns $x = 2$ into a hole (removable) rather than an asymptote, but $x = 2$ is still excluded because the original expression is $0/0$ there. The uncanceled factor $(x + 2)$ leaves a genuine vertical asymptote at $x = - 2$ .
Domain: all reals except $2$ and $- 2$ , i.e. $(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$ .
Sanity check: plugging $x = 2$ into the original gives $\frac{0}{0}$ (undefined) and $x = - 2$ gives $\frac{- 4}{0}$ (undefined), so both exclusions are real.

□

Exercises for §5.6 →

5.7Domain of Square Root and Even Root Functions

Proposition 5.11 (Domain of an even root). For $g (x)$ , or any even root, require

g (x) \geq 0.

Solve that inequality; its solution set is the domain.

Intuition. You cannot take the even root of a negative number in real numbers; the inside must be "non-negative or zero."

The graph of a square root function lives only where the inside expression is on or above the x-axis.

Example 5.12. Find the domain of $f (x) = 4 - x$ .

Solution.

The radicand (the expression under the root) must be non-negative: $4 - x \geq 0$ .
Solve the inequality. Subtract $4$ : $- x \geq - 4$ . Divide by $- 1$ and flip the inequality (dividing by a negative reverses it): $x \leq 4$ .
Domain: $(- \infty, 4]$ . The endpoint $4$ is included because $0 = 0$ is defined.
Sanity check: $f (4) = 0 = 0$ (valid), $f (0) = 4 = 2$ (valid), but $f (5) = - 1$ is not real, so $5$ is correctly excluded.

□

Exercises for §5.7 →

5.8Combined Restrictions

When a function contains both rational and radical parts (or other restrictions), intersect all the individual domain conditions.

Write the most restrictive set that satisfies every requirement at once.

Intuition. Each dangerous part (denominator, even root, logarithm argument, etc.) writes its own "keep out" sign. The final domain obeys every sign.

Example 5.13. Find the domain of $f (x) = \frac{x + 1}{x - 3}$ .

Solution.

Step 1 (radical condition): the radicand must be non-negative, $x + 1 \geq 0 \Rightarrow x \geq - 1$ .
Step 2 (denominator condition): the denominator must be nonzero, $x - 3 \neq = 0 \Rightarrow x \neq = 3$ .
Step 3 (intersect): keep the $x$ that satisfy both conditions at once. From $x \geq - 1$ we punch out $x = 3$ , giving $[- 1, 3) \cup (3, \infty)$ .
Sanity check: $x = - 1$ gives $\frac{0}{- 4} = 0$ (valid), $x = 5$ gives $\frac{6}{2}$ (valid), but $x = - 2$ fails the root and $x = 3$ divides by zero, so both are correctly excluded.

□

Exercises for §5.8 →

Common Mistakes

Including a value that zeros a denominator: Even if the graph has a hole, that x is not in the domain of the given expression.
Forgetting to solve the inequality for even roots: Simply writing "x ≥ 0" when the radicand is more complicated loses points.
Intersecting instead of unioning restrictions: All conditions must be satisfied simultaneously (intersection).
Assuming polynomials have restricted domains: Unless context says otherwise, polynomial domains are all reals.
Confusing domain of the function with the x-intercepts: Intercepts are where output is zero; domain is where input is allowed.

The range of a function is the set of all output values that are actually produced as the input runs through the entire domain.

While the domain is about legal inputs, the range is about the numbers that come out the other end. For many functions the range is a proper subset of the real numbers.

5.9Definition

Definition 5.14 (Range). If $f : D \to R$ , the range is

{f (x) ∣ x \in D},

the image of the domain under the function rule.

Intuition. If the domain is the list of ingredients you may use, the range is the list of actual dishes the recipe can produce.

Some recipes can make any flavor; others can only make sweet things.

Example 5.15. Find the range of $f (x) = x^{2}$ (domain: all reals).

Solution.

A square is never negative, so every output satisfies $f (x) \geq 0$ . No negative number can ever come out.
Is every non-negative number actually reached? For any $y \geq 0$ , choosing $x = y$ gives $f (y) = (y)^{2} = y$ , so yes, every $y \geq 0$ is hit.
Range: $[0, \infty)$ . Note this is a proper subset of the reals even though the domain is all of $R$ — domain and range need not match.
Sanity check: $0 = f (0)$ and $4 = f (2)$ appear, but $- 1$ never does, matching $[0, \infty)$ .

□

Exercises for §5.9 →

5.10Finding the Range from the Rule or Graph

Method 5.16 (Finding the range). Any of three methods will do:

Graph the function and project onto the y-axis: the covered $y$ -values are the range.
Solve $y = f (x)$ for $x$ in terms of $y$ , and find for which $y$ a real $x$ exists. This is the domain of the inverse.
Use calculus, or completing the square, to find minimum and maximum values for polynomials and quadratics.

Intuition. The graph is a "shadow" on the y-axis when light shines from the side; the lit part is the range.

Inverting the rule turns "outputs" back into "inputs" for the inverse function.

Example 5.17. Find the range of $f (x) = x^{2} + 4 x + 7$ by completing the square.

Solution.

Take half of the $x$ -coefficient ( $4/2 = 2$ ) and square it ( $2^{2} = 4$ ). Add and subtract it: $f (x) = (x^{2} + 4 x + 4) + 7 - 4 = (x + 2)^{2} + 3$ .
The squared part $(x + 2)^{2}$ is $\geq 0$ and equals $0$ at its smallest, when $x = - 2$ . So the smallest output is $0 + 3 = 3$ , attained at the vertex $(- 2, 3)$ .
Since the parabola opens upward ( $a = 1 > 0$ ), outputs grow without bound above the minimum, so every value $\geq 3$ is achieved.
Range: $[3, \infty)$ .
Sanity check: $f (- 2) = 4 - 8 + 7 = 3$ (the minimum) and $f (0) = 7 \geq 3$ , consistent with the range.

□

Exercises for §5.10 →

5.11Range from Common Function Families

Summary (Ranges of the common families).

Linear (non-constant): all reals.
Quadratic with $a > 0$ : $[y -value of the vertex, \infty)$ .
Absolute value: $[minimum, \infty)$ .
Rational such as $1/ x$ : all reals except $0$ , or whatever the horizontal asymptote value is.
Square root: $[0, \infty)$ , adjusted by any vertical shift.

Intuition. Each family has a characteristic "footprint" on the y-axis.

Once you know the family and its vertex or asymptotes you can read the range quickly.

Example 5.18 ((absolute value family):). $f (x) = ∣ x - 1∣ + 2$ . The absolute-value part is $\geq 0$ with minimum $0$ at $x = 1$ , so the smallest output is $0 + 2 = 2$ and outputs increase without bound above it. Range: $[2, \infty)$ . Sanity check: $f (1) = 0 + 2 = 2$ (the floor) and $f (5) = 4 + 2 = 6$ .

Example 5.19 ((rational family):). $f (x) = \frac{1}{x} + 3$ . The parent $1/ x$ produces every value except $0$ (it has horizontal asymptote $y = 0$ ). Adding $3$ shifts every output up by $3$ , so the excluded value shifts from $0$ to $3$ . Range: all reals except $3$ , i.e. $(- \infty, 3) \cup (3, \infty)$ . Sanity check: no real $x$ solves $\frac{1}{x} + 3 = 3$ (that needs $\frac{1}{x} = 0$ , impossible), confirming $3$ is missing.

Exercises for §5.11 →

5.12Restricted Domains and Their Effect on Range

When the domain itself is restricted (by context or by previous domain work), the range is the image of only that restricted set.

A function may have a smaller range when its domain is a small interval.

Intuition. If you only allow the machine to be fed numbers from 0 to 10, it may only be able to produce outputs from 5 to 30 even if with all reals it could produce more.

Example 5.20. Find the range of $f (x) = x^{2}$ on the restricted domain $[- 2, 3]$ .

Solution.

Do not just plug in the endpoints — the extremes can occur in the interior. Locate the vertex: $x = 0$ , which lies inside $[- 2, 3]$ , giving the minimum $f (0) = 0$ .
The maximum on a closed interval occurs at whichever endpoint is farther from the vertex. Here $f (- 2) = 4$ and $f (3) = 9$ , so the maximum is $9$ at $x = 3$ .
Because $f$ is continuous, it takes every value between its min and max on the interval. Range: $[0, 9]$ .
Sanity check: had we naively used only the endpoints we would have gotten $[4, 9]$ and missed the values below $4$ down to $0$ produced near $x = 0$ .

□

Exercises for §5.12 →

Common Mistakes

Confusing range with domain: Domain = inputs; range = outputs.
Assuming every function has range all reals: Quadratics, square roots, and many rationals do not.
Forgetting the effect of horizontal asymptotes: They often (but not always) indicate a value missing from the range.
Reporting the range of the simplified expression instead of the original: Holes or asymptotes from the original still affect the range of outputs.
Including y-values that are only approached but never reached: Asymptotes are not attained.

The graph of a function f is the set of all points (x, f(x)) in the plane as x runs through the domain.

The visual picture instantly reveals domain, range, intercepts, increasing/decreasing behavior, symmetry, asymptotes, and many other properties that are hard to see from the algebraic rule alone.

5.13Plotting Points and the Definition

To plot the graph, choose convenient x-values in the domain, compute f(x), and mark the point (x, f(x)).

Connect the points with a smooth curve when the function is continuous, or leave as discrete points when it is not.

Intuition. Each input x gets its own "height" f(x). Plotting all those heights gives the picture of the rule.

The graph is the "skyline" produced by the function's heights.

Example 5.21. For $f (x) = 2 x + 1$ , when $x = - 1$ , $f = - 1$ , point $(- 1, - 1)$ ; $x = 0$ , $f = 1$ , point $(0, 1)$ ; $x = 1$ , $f = 3$ , point $(1, 3)$ .

Solution.

The points lie on the line $y = 2 x + 1$ .

□

Exercises for §5.13 →

5.14Key Features Visible on the Graph

x-intercepts: where the graph crosses the x-axis (solve f(x) = 0).
y-intercept: where the graph crosses the y-axis (the point (0, f(0)), if 0 is in the domain).
Domain: the projection of the graph onto the x-axis (all x for which there is a point).
Range: the projection of the graph onto the y-axis (all y-values that are hit).

Intuition. Intercepts are where the skyline meets the ground (x-axis) or the upright pole at x = 0.

Domain and range are the "shadows" the graph casts on the two axes when light shines from the side or above.

Example 5.22. The graph of $f (x) = x^{2} - 1$ has $x$ -intercepts at $\pm 1$ , $y$ -intercept at $- 1$ , domain all reals, range $[- 1, \infty)$ .

Exercises for §5.14 →

5.15Increasing, Decreasing, and Extrema

Definition 5.23 (Increasing and decreasing). On an interval, $f$ is increasing if

x_{1} < x_{2} ⟹ f (x_{1}) < f (x_{2})

(the graph rises), and decreasing if

x_{1} < x_{2} ⟹ f (x_{1}) > f (x_{2})

(the graph falls).

Local max/min are peaks and valleys visible on the graph.

Intuition. Walking right on the x-axis, if your height is always getting bigger you are on an increasing part.

A local max is the top of a hill; after it you start going down for a while.

Example 5.24. $f (x) = - x^{2} + 4$ is increasing on $(- \infty, 0]$ and decreasing on $[0, \infty)$ . The global maximum is at the vertex $(0, 4)$ .

Exercises for §5.15 →

5.16Asymptotes and Holes

Definition 5.25 (Asymptotes and holes).

Vertical asymptote: the graph shoots to $\pm \infty$ as $x$ approaches a finite value, usually an uncancelled zero of a denominator.
Horizontal or oblique asymptote: the graph approaches a line as $x \to \pm \infty$ .
Hole: an isolated missing point, where a cancelled factor made the original expression undefined.

Intuition. Vertical asymptote: a "wall" the graph never crosses but gets infinitely tall or deep beside it.

Hole: a single missing dot; the graph jumps over that one forbidden x but is otherwise fine nearby.

Example 5.26. $f (x) = 1/ (x - 2)$ has vertical asymptote $x = 2$ and horizontal asymptote $y = 0$ . The graph never touches $y = 0$ but gets arbitrarily close as $x \to \pm \infty$ .

Exercises for §5.16 →

Common Mistakes

Plotting (f(x), x) instead of (x, f(x)): x is always the horizontal coordinate.
Assuming the graph is defined everywhere: Check domain first; gaps or asymptotes may exist.
Connecting discrete points with a curve when the function is not continuous: Step functions and rational functions with holes must not be connected.
Misidentifying asymptotes: Horizontal asymptotes are about end behavior as x → ∞; vertical are about blow-ups at finite x.
Forgetting to use open/closed circles at endpoints or holes: The circles communicate exactly which points are included.

A linear function produces a straight-line graph and has the algebraic form f(x) = mx + b, where m is the slope (constant rate of change) and b is the y-intercept.

Linear functions are the simplest non-constant functions and the foundation for understanding all other functions through transformations.

5.17Definition and Slope-Intercept Form

Definition 5.27 (Linear function). A linear function is a degree-one polynomial

f (x) = m x + b,

whose slope is the constant rate of change

m = \frac{Δ y}{Δ x} = \frac{rise}{run} .

When m > 0 the function is increasing; m < 0 decreasing; m = 0 constant.

Intuition. Every time x increases by 1, y increases by exactly m (the slope).

b is where the line crosses the y-axis when you start at x = 0.

Example 5.28. f(x) = 2x − 1. When x increases by 1, f increases by 2. f(0) = −1, so y-intercept −1. Graph is a line rising 2 for every 1 right.

Exercises for §5.17 →

5.18Finding the Equation from Two Points or Slope and Intercept

Method 5.29 (Finding the equation of a line).

Given the slope $m$ and a point $(x_{0}, y_{0})$ : write $y - y_{0} = m (x - x_{0})$ , then convert to slope-intercept form if you want it.
Given two points: first compute

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}},

then use either point to find $b$ .

Intuition. Two points determine a unique straight line (unless vertical).

Slope tells the tilt; one point anchors the line in the plane.

Example 5.30. Points (1, 4) and (3, 10).

Solution.

m = (10 − 4)/(3 − 1) = 6/2 = 3.
Using (1, 4): 4 = 3(1) + b → b = 1.
f(x) = 3x + 1.

□

Exercises for §5.18 →

5.19Special Cases: Horizontal and Vertical Lines

Horizontal: f(x) = b (slope 0, constant function, passes vertical line test).

Vertical: x = k (undefined slope, fails vertical line test, not a function).

Intuition. A horizontal line is perfectly flat; y never changes no matter how far you walk in x.

A vertical line is a wall; for one x you would need every y, which a function cannot do.

Example 5.31. f(x) = 5 is the horizontal line y = 5. Every input gives the same output 5. It is a (very boring) function.

Exercises for §5.19 →

5.20Real-World Linear Functions

Constant speed, constant cost per unit, temperature scales, etc. are linear.

The slope has units: miles per hour, dollars per item, degrees per minute, etc.

Intuition. Linear functions describe steady, no-surprise change.

"For every extra mile I drive, the cost goes up exactly the same amount."

Example 5.32. A phone plan costs $20 per month plus $0.10 per text. f(t) = 0.10t + 20. After 150 texts the bill is 0.10*150 + 20 = $35.

Exercises for §5.20 →

Common Mistakes

Confusing slope with y-intercept: m is the tilt; b is the starting height.
Using (y2 − y1)/(x1 − x2): The denominator must be in the same order as the numerator (consistent points).
Thinking every straight line is the graph of a function: Vertical lines are not.
Forgetting units when interpreting slope: "3" by itself is meaningless; "3 meters per second" tells the story.
Plotting only one point and the slope without a second point: Always plot the intercept and at least one more point obtained from the slope.

A quadratic function is a polynomial of degree exactly two: f(x) = ax² + bx + c, a ≠ 0.

Its graph is a parabola, one of the most important curves in mathematics and physics. Parabolas describe projectile paths, satellite dishes, suspension bridges, and the shape of many data relationships.

5.21Standard Form and Direction

Definition 5.33 (Quadratic function). A quadratic function is a polynomial of degree exactly two,

f (x) = a x^{2} + b x + c, a \neq = 0,

and its graph is a parabola.

The sign of a tells whether the parabola opens upward (a > 0, has minimum) or downward (a < 0, has maximum).

The magnitude |a| controls width: |a| > 1 narrows, 0 < |a| < 1 widens relative to y = x².

Intuition. The x² term is the "bowl maker." Positive a makes a right-side-up bowl; negative makes an upside-down bowl.

Steeper bowls (large |a|) change fast; wide bowls change slowly.

Example 5.34. f(x) = −2x² + 8 has a = −2 < 0 so opens down, narrower than y = −x².

Exercises for §5.21 →

5.22Vertex and Axis of Symmetry

The vertex is the turning point of the parabola, its minimum or its maximum.

Theorem 5.35 (Vertex and axis of symmetry). The vertex of $f (x) = a x^{2} + b x + c$ sits at

x = - \frac{b}{2 a}, y = f (- \frac{b}{2 a}),

and the axis of symmetry is the vertical line $x = - \frac{b}{2 a}$ .

Intuition. The vertex is the bottom or top of the bowl; everything is symmetric left and right of it.

The axis is the mirror line down the middle of the parabola.

Example 5.36. f(x) = x² − 4x + 3.

Solution.

x = 4/2 = 2.
f(2) = 4 − 8 + 3 = −1.
Vertex (2, −1), axis x = 2.

□

Exercises for §5.22 →

5.23Intercepts

Proposition 5.37 (Intercepts of a parabola). The y-intercept is always $(0, c)$ , found by plugging in $x = 0$ .

The x-intercepts are the solutions of $a x^{2} + b x + c = 0$ , found by the quadratic formula or by factoring; the number of real x-intercepts equals the number of real roots, so $0$ , $1$ or $2$ .

Intuition. The y-intercept is where the bowl crosses the upright pole at x = 0.

The x-intercepts are where the bowl touches or crosses the ground (y = 0).

Example 5.38. f(x) = x² − 5x + 6 has y-intercept 6.

Solution.

Roots of x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0, x = 2, 3. Two x-intercepts.

□

Exercises for §5.23 →

5.24Transformations of y = x²

Definition 5.39 (Vertex form). The vertex form of a quadratic is

f (x) = a (x - h)^{2} + k,

whose vertex is at $(h, k)$ . The $(x - h)$ shifts the parabola left or right — right when $h$ is positive; $a$ stretches, compresses and reflects it; and $+ k$ shifts it up or down.

Intuition. Start with the basic bowl y = x² sitting at (0,0).

Move the bottom to (h, k), then stretch or flip it.

Example 5.40. f(x) = −(x − 4)² + 1 has vertex (4, 1), opens down, same width as y = x².

Exercises for §5.24 →

Common Mistakes

Using −b/a instead of −b/(2a) for the vertex x: Forgets the 2 in the denominator.
Thinking every parabola crosses the x-axis twice: Many have zero or one (tangent) intercept.
Confusing vertex form shift direction: (x − 3) moves the graph right 3, not left.
Forgetting that $a$ affects both direction and width: the sign of $a$ sets the direction and $∣ a ∣$ sets the width. For example $a = 2$ opens upward (positive) and is narrow ( $∣ a ∣ > 1$ ); $a = - \frac{1}{2}$ opens downward and is wide.
Plotting only the vertex and intercepts without checking another point: Always verify the shape with one more point.

A piecewise function uses different rules on different parts of its domain. It is written with a large left brace and lists each piece together with the interval on which it applies.

Piecewise definitions let us express functions that behave differently in different regions, such as absolute value, step functions, and many real tax or shipping cost schedules.

5.25Definition and Notation

Definition 5.41 (Piecewise function). A piecewise function is defined by cases,

f (x) = ⎩ ⎨ ⎧ expression_{1}, expression_{2}, ⋮ condition_{1}, condition_{2},

each expression paired with the interval on which it applies.

The conditions partition the domain (or at least cover it without overlap except possibly at endpoints).

Intuition. The function is like a choose-your-own-adventure book: depending on which interval your x lands in, you follow a different rule.

The brace is the "menu" that tells you which page to turn to.

Example 5.42. The absolute value function is the piecewise function

Solution.

f(x) = x if x ≥ 0
```
   −x  if x < 0
```

□

Exercises for §5.25 →

5.26Evaluating Piecewise Functions

Method 5.43 (Evaluating a piecewise function). To evaluate $f (a)$ :

Determine which condition $a$ satisfies.
Use only the corresponding expression.
Ignore all the other pieces.

Intuition. You only follow the rule for the "room" your input number is in.

Crossing from one room to another may change which formula you use.

Example 5.44. f(x) = 3x if x < 1, x + 1 if x ≥ 1.

Solution.

f(−2) uses first piece: 3(−2) = −6.
f(4) uses second piece: 4 + 1 = 5.

□

Exercises for §5.26 →

5.27Graphing Piecewise Functions

Method 5.45 (Graphing a piecewise function).

Graph each piece only on the interval where its rule applies.
Use open circles at endpoints that are not included, and closed circles at included endpoints.
Do not connect two pieces unless the definition and the limits agree at the junction.

Intuition. Each piece is a "guest" that is only allowed to sit in its assigned chairs (the interval).

The circles are like "reserved" or "open" signs at the boundary seats.

Example 5.46. f(x) = x if x ≤ 1, 3 if x > 1.

Solution.

Graph the line y = x up to and including (1,1), then a horizontal ray y = 3 starting just after x = 1 (open circle at (1,3) because f(1) = 1, not 3).

□

Exercises for §5.27 →

5.28Continuity and Differentiability at Junctions

A piecewise function may have corners or jumps even when each piece is smooth.

Definition 5.47 (Continuity at a junction). A piecewise function is continuous at a junction $c$ if

x \to c^{-} lim f (x) = x \to c^{+} lim f (x) = f (c) .

It is differentiable at $c$ only if the left and right derivatives also match.

Intuition. Continuity means the pieces meet without a jump; you can draw the whole graph without lifting the pencil at the join.

A corner (like |x|) is continuous but has no single tangent line at the corner.

Example 5.48 ((checking continuity at a junction):). Let $f (x) = 2 x$ if $x < 1$ and $f (x) = x + 1$ if $x \geq 1$ . Is $f$ continuous at the junction $x = 1$ ?

Solution.

Step 1 (left limit): as $x \to 1^{-}$ use the first piece, $2 x \to 2 (1) = 2$ .
Step 2 (right limit): as $x \to 1^{+}$ use the second piece, $x + 1 \to 1 + 1 = 2$ .
Step 3 (value): $f (1) = 1 + 1 = 2$ (the $x \geq 1$ piece owns the endpoint).
All three equal $2$ , so $f$ is continuous at $x = 1$ . (Had the second piece been $x + 5$ , the right limit would be $6 \neq = 2$ and the graph would jump.)

□

Example 5.49 ((continuity vs. differentiability):). $∣ x ∣$ is continuous everywhere — the two pieces $- x$ and $x$ both approach $0$ at the junction, matching $f (0) = 0$ , so there is no jump. But it is not differentiable at $0$ : the left piece has slope $- 1$ and the right piece has slope $+ 1$ , and $- 1 \neq = + 1$ , so there is no single tangent line at the corner.

Exercises for §5.28 →

Common Mistakes

Evaluating with the wrong piece: Always check the condition for the specific input value; do not average or blend pieces.
Graphing a piece outside its interval: The rule is invalid elsewhere.
Using closed circles when the piece definition uses < or >: Match the inequality in the definition.
Assuming piecewise functions are always discontinuous: Many are continuous (e.g. |x|).
Forgetting that the pieces must cover the whole intended domain: Gaps mean the function is not defined there.

Transformations let us take the graph of a known parent function and quickly produce the graph of a new related function by shifting, stretching, compressing, or flipping it. Instead of plotting point by point every time, we apply a small number of rules to move or resize the entire graph.

Mastering transformations is essential for understanding families of functions (quadratics, exponentials, absolute value, etc.) and for graphing without a calculator. The general form is $g (x) = a f (b (x - h)) + k$ .

5.29Horizontal and Vertical Shifts

Proposition 5.50 (Shifts). Replacing the input $x$ with $(x - h)$ shifts the graph $h$ units to the right if $h > 0$ , and to the left if $h < 0$ . Adding a constant $k$ outside the function, as in $f (x) + k$ , moves the graph up by $k$ , or down if $k$ is negative. Together,

g (x) = f (x - h) + k

sends every point $(x, y)$ of the original graph to $(x + h, y + k)$ .

Shifts do not change the shape, only the position.

Intuition. Imagine the parent graph is drawn on a transparency. Sliding the sheet right by 3 moves every point right 3; that is a horizontal shift.

Adding to the output value lifts the whole picture straight up.

Example 5.51. Start with $f (x) = x^{2}$ . Then $g (x) = f (x - 2) + 3 = (x - 2)^{2} + 3$ is the parabola shifted right 2 units and up 3. The vertex moves from $(0, 0)$ to $(2, 3)$ . All other points move the same amount.

Example 5.52. For $f (x) = ∣ x ∣$ , $h (x) = ∣ x + 4∣ - 1$ shifts left 4 and down 1. The corner is now at $(- 4, - 1)$ .

Exercises for §5.29 →

5.30Stretches, Compressions, and Reflections

Proposition 5.53 (Stretches, compressions and reflections). Multiplying the output by a constant $a$ produces a vertical stretch when $∣ a ∣ > 1$ , and a compression when $0 < ∣ a ∣ < 1$ ; if $a < 0$ the graph is also reflected over the x-axis.

Replacing $x$ with $b x$ , as in $f (b x)$ , produces a horizontal stretch or compression by the factor $1/∣ b ∣$ ; if $b < 0$ we also get a reflection over the y-axis.

Reflections flip the graph like a mirror: $- f (x)$ over x-axis, $f (- x)$ over y-axis. These preserve distances but reverse orientation.

Intuition. Vertical stretch by 2 makes the graph twice as tall (skinny or tall V or U). Compression by 1/2 squashes it toward the x-axis.

Horizontal scaling is the counterintuitive one: a larger $∣ b ∣$ squeezes the graph sideways toward the $y$ -axis (compression by $1/∣ b ∣$ ), while a small $∣ b ∣$ spreads it out.

Example 5.54. From $f (x) = x^{2}$ , $g (x) = - 2 f (x) = - 2 x^{2}$ stretches vertically by 2 and reflects over x-axis. The parabola opens downward and is narrower.

Example 5.55. $g (x) = f (2 x)$ for $f (x) = ∣ x ∣$ gives $g (x) = ∣2 x ∣ = 2∣ x ∣$ , which compresses the V horizontally by factor 1/2 (steeper sides).

Exercises for §5.30 →

5.31Combining Multiple Transformations

When several transformations are combined, the order matters.

Method 5.56 (Applying several transformations). Write the function as $a f (b (x - h)) + k$ , identify $a$ , $b$ , $h$ , $k$ , and then work from the inside out:

Horizontal scaling and reflection (the $b$ inside).
Horizontal shift (the $h$ ).
Vertical scaling and reflection (the $a$ ).
Vertical shift (the $k$ ).

Write the function in the form $a f (b (x - h)) + k$ and apply changes from inside out. Always identify $a, b, h, k$ first.

Intuition. Think of the function as a recipe. Do the inside changes (what you feed in) before the outside changes (what comes out).

Horizontal moves happen before you stretch vertically; otherwise the amounts get distorted.

Example 5.57. Transform $f (x) = x$ into $g (x) = - 3 2 (x - 1) + 4$ .

Solution. $b = 2$ : horizontal compression by 1/2. $h = 1$ : shift right 1. $a = - 3$ : vertical stretch by 3 and reflect over x-axis. $k = 4$ : shift up 4.

The domain becomes $x \geq 1$ , range $y \leq 4$ .□

Example 5.58. Start with $f (x) = x^{3}$ . Apply: reflect over y-axis, shift left 1, stretch vertically by 1/2, shift down 2.

Solution.

Result: $g (x) = \frac{1}{2} f (- (x + 1)) - 2 = \frac{1}{2} (- (x + 1))^{3} - 2 = - \frac{1}{2} (x + 1)^{3} - 2$ .

□

Exercises for §5.31 →

Common Mistakes

Reversing horizontal shift direction: $f (x - 3)$ is right, not left. The sign inside is opposite the direction you expect.
Confusing vertical and horizontal scales: $a$ is always vertical (multiplies y). The inside $b$ is horizontal.
Forgetting that order inside vs outside matters: Doing the vertical stretch before the horizontal shift changes the visual result.
Ignoring the sign of $a$ or $b$ : Negative $a$ always flips vertically; negative $b$ flips horizontally.
Assuming stretches always make things bigger: A factor between 0 and 1 compresses.
Losing track of the parent function: Always start from the basic shape before layering transformations.

Even and odd functions are classified by their symmetry.

Even functions are symmetric about the y-axis: the left and right sides are mirror images.

Odd functions are symmetric about the origin: rotating the graph 180 degrees around (0,0) leaves it unchanged.

These symmetries simplify integrals, Fourier analysis, and many algebraic manipulations.

5.32Even Functions

Definition 5.59 (Even function). A function $f$ is even when

f (- x) = f (x)

for every $x$ in the domain.

Examples: $f (x) = x^{2}$ , $f (x) = cos x$ , $f (x) = ∣ x ∣$ , and the constant functions.

Intuition. Even: the function does not care about the sign of the input; only how far you are from zero matters.

If you walk the same distance left or right of center you get the same height.

Example 5.60. Test whether $f (x) = x^{4} + 3$ is even.

Solution.

Replace $x$ with $- x$ : $f (- x) = (- x)^{4} + 3$ .
Simplify. Since $(- x)^{4} = x^{4}$ (an even power kills the sign), $f (- x) = x^{4} + 3 = f (x)$ .
Because $f (- x) = f (x)$ for every $x$ , the function is even. Its graph is a single U-shaped bowl (sitting 3 units up) that is a mirror image across the $y$ -axis.
Sanity check: $f (2) = 16 + 3 = 19$ and $f (- 2) = 16 + 3 = 19$ agree, as an even function requires.

□

Exercises for §5.32 →

5.33Odd Functions

Definition 5.61 (Odd function). A function $f$ is odd when

f (- x) = - f (x)

for every $x$ in the domain.

Examples: $f (x) = x^{3}$ , $f (x) = sin x$ , and $f (x) = x$ , the identity.

Note that the zero function is both even and odd.

Intuition. Odd: going the same distance in the opposite direction flips the sign of the output.

The graph looks the same after a 180-degree spin around the origin.

Example 5.62. Test whether $f (x) = x^{3} - x$ is odd.

Solution.

Compute $f (- x) = (- x)^{3} - (- x) = - x^{3} + x$ .
Factor out $- 1$ : $- x^{3} + x = - (x^{3} - x) = - f (x)$ .
Since $f (- x) = - f (x)$ for every $x$ , the function is odd.
Sanity check: $f (2) = 8 - 2 = 6$ and $f (- 2) = - 8 + 2 = - 6 = - f (2)$ , exactly the sign-flip an odd function demands.

□

Exercises for §5.33 →

5.34Symmetry on the Graph

Proposition 5.63 (Symmetry of even and odd graphs). An even function has reflection symmetry across the y-axis. An odd function has point symmetry — rotational symmetry of order two — about the origin.

These visual tests are quick ways to classify a graph without algebra.

Intuition. Y-axis mirror: left half is the flip of the right half (even).

Origin spin: every point has a matching point directly opposite through the center (odd).

Example 5.64. The graph of y = x^2 is a mirror image across the y-axis. The graph of y = x^3 looks the same after rotating the paper 180 degrees around (0,0).

Exercises for §5.34 →

5.35Neither or Both

Most functions are neither even nor odd.

A function can be both only if f(x) = 0 for all x (the zero function).

To test algebraically, compute f(−x) and compare with f(x) and −f(x).

Intuition. "Neither" is the default; the function has no special left-right or origin symmetry.

Only the flat zero line at height 0 satisfies both mirror and spin symmetries at once.

Example 5.65. f(x) = x + 1.

Solution.

f(−x) = −x + 1, which equals neither f(x) = x + 1 nor −f(x) = −x − 1.
Hence neither.

□

Exercises for §5.35 →

Common Mistakes

Forgetting the domain must be symmetric about 0: Even/odd definitions require that if x is in the domain then −x is too.
Checking only at one point: Must hold for all x (or a general x).
Confusing even with "even degree polynomial": A polynomial of even degree can be odd (e.g. x^4 is even function; x^3 is odd function of odd degree).
Assuming |x| is odd: |−x| = |x|, so even.
Thinking constants other than zero are odd: A nonzero constant c satisfies f(−x) = c = f(x) so even, not odd.

An inverse function undoes what the original function does. If $f$ takes an input $x$ and produces $y$ , then the inverse $f^{- 1}$ takes that $y$ and returns the original $x$ . In symbols, $f (f^{- 1} (y)) = y$ and $f^{- 1} (f (x)) = x$ .

Not every function has an inverse. Only functions that are one-to-one (different inputs always produce different outputs) are invertible. Inverses are powerful for solving equations and for understanding the symmetry between a function and its "reverse."

5.36Definition and Notation

Definition 5.66 (Inverse function). An inverse function $f^{- 1}$ satisfies both composition identities on the appropriate domains:

f (f^{- 1} (y)) = y, f^{- 1} (f (x)) = x .

The superscript $- 1$ means "inverse," not "reciprocal." The reciprocal of $f$ would be written $1/ f (x)$ .

The inverse is itself a function only when the original is one-to-one.

Intuition. Think of $f$ as a machine that converts input to output. The inverse machine runs the conversion backward.

If $f$ turns 4 into 12, the inverse must turn 12 back into 4.

Example 5.67. Let $f (x) = 4 x + 1$ . Then $f (2) = 9$ . The inverse must satisfy $f^{- 1} (9) = 2$ .

Solution.

Solving $y = 4 x + 1$ by swapping and solving gives $f^{- 1} (x) = (x - 1) /4$ .
Check: $f (f^{- 1} (9)) = 4 * ((9 - 1) /4) + 1 = 9$ .

□

Exercises for §5.36 →

5.37One-to-One Functions and the Horizontal Line Test

Definition 5.68 (One-to-one function). A function is one-to-one (injective) when

f (a) = f (b) ⟹ a = b;

equivalently, no two distinct inputs share the same output.

Proposition 5.69 (Horizontal line test). If you can draw any horizontal line that intersects the graph of $f$ more than once, then $f$ is not one-to-one and has no inverse.

Many common functions fail this (e.g., $x^{2}$ , $sin x$ , $∣ x ∣$ on all reals) unless we restrict their domains.

Intuition. One-to-one is like each output has exactly one "owner" input. If two people have the same phone number in the directory, you cannot look up who called from the number alone.

The horizontal line test is a quick visual: crossing the same height more than once means duplicate outputs.

Example 5.70. $f (x) = x^{2}$ on all reals fails the test (horizontal line $y = 4$ crosses at $x = 2$ and $x = - 2$ ).

Solution.

Restrict domain to $x \geq 0$ and it becomes one-to-one; the inverse is the square-root function (principal branch).

□

Exercises for §5.37 →

5.38Finding the Inverse Algebraically

Method 5.71 (Finding an inverse algebraically).

Replace $f (x)$ with $y$ .
Swap $x$ and $y$ .
Solve the equation for $y$ .
Replace $y$ with $f^{- 1} (x)$ .

Always state the domain of the inverse (it is the range of the original). Check by composing both ways.

For more complicated functions the algebra may produce a quadratic or require care with signs and restrictions.

Intuition. Swapping x and y literally exchanges the roles of input and output.

Solving for the new y is "undoing" the original recipe step by step.

Example 5.72. Find the inverse of $f (x) = \frac{2 x + 3}{x - 1}$ .

Solution.

Set $y = (2 x + 3) / (x - 1)$ .
Swap: $x = (2 y + 3) / (y - 1)$ .
Multiply both sides by $(y - 1)$ : $x (y - 1) = 2 y + 3$ .
$x y - x = 2 y + 3$ .
Collect y terms: $x y - 2 y = x + 3$ .
$y (x - 2) = x + 3$ .
$y = (x + 3) / (x - 2)$ .
Thus $f^{- 1} (x) = (x + 3) / (x - 2)$ , with domain $x \neq = 2$ (the value that would have been the original range restriction point).

□

Example 5.73. For $f (x) = x^{3} - 5$ , swap and solve: $x = y^{3} - 5$ , $y^{3} = x + 5$ , $y = 3 x + 5$ . So $f^{- 1} (x) = 3 x + 5$ .

Exercises for §5.38 →

5.39Graphs and Properties of Inverses

Theorem 5.74 (Graph of an inverse). The graph of $f^{- 1}$ is the reflection of the graph of $f$ across the line

y = x,

so every point $(a, b)$ on $f$ becomes $(b, a)$ on the inverse.

Consequently, the domain of $f^{- 1}$ equals the range of $f$ , and the range of $f^{- 1}$ equals the domain of $f$ . If $f$ is increasing, so is $f^{- 1}$ ; if decreasing, so is its inverse.

Intuition. Reflection over y = x is like turning the paper over the diagonal so that x and y labels swap places.

The "input becomes output" picture is literally mirrored.

Example 5.75. The graph of $f (x) = 2 x - 4$ is a line through $(0, - 4)$ with slope 2. Its inverse $f^{- 1} (x) = (x + 4) /2$ is a line through $(0, 2)$ with slope $1/2$ . Reflecting the first line over $y = x$ swaps the intercepts and the slope becomes its reciprocal.

Exercises for §5.39 →

Common Mistakes

Writing $f^{- 1} (x) = 1/ f (x)$ : This is the reciprocal, not the inverse (except for the special case f(x) = 1/x).
Forgetting to swap x and y before solving: You end up with the original function again.
Ignoring domain restrictions: The inverse formula may be defined more widely than the actual inverse; restrict to the range of the original.
Assuming every function has an inverse: Quadratics, cubics without domain restriction, periodic functions, and constant functions do not.
Not checking the composition: Always verify $f (f^{- 1} (x)) = x$ and $f^{- 1} (f (x)) = x$ with a number or symbolically.
Confusing the graph reflection: Reflection is over y = x, not over an axis.
Losing the order when solving complex expressions: Multiply by denominators before collecting like terms with the new y.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter