Chapter Review · a rapid pre-test review (speedrun)

Functions and Their Graphs: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Relation: A set of ordered pairs $(x, y)$ , which an equation names by carving out the points satisfying it. A function pairs each input with exactly one output, so being a function belongs to a curve plus a choice of input, never to the curve alone.
Natural domain, declared domain: The natural domain is the largest set on which the formula returns a real number, assumed when none is stated. A declared domain travels with the function and overrides it.
Range, codomain: The outputs actually produced, against the set they were declared to live in. The range sits inside the codomain and can be strictly smaller.
One-to-one: Different inputs always give different outputs: $f(a) = f(b)$ forces $a = b$ . No horizontal line meets the graph twice.
Inside, outside: In $y = a\,f(b(x - h)) + k$ : the inside ( $b$ , $h$ ) acts before $f$ and moves only the first coordinate, the outside ( $a$ , $k$ ) after $f$ and only the second.

Formulas and theorems

The two line tests

$\begin{gathered} \text{vertical: reads } x \text{ as the input} \\ \text{horizontal: reads } y \text{ as the input} \end{gathered}$

Text description

The upward opening parabola y equals x squared, labelled with that equation. A vertical line meets it at one point, so this curve passes the vertical line test, and a horizontal line meets it at two points, one on each arm, so it fails the horizontal one. That is the opposite verdict from the worked example, which is about the sideways curve x equals y squared.

Use when One test asked about two different variables: the curve passes whichever reading gives each input exactly one output, and it can pass one and fail the other. Swapping the two is exactly what inverting does.

e.g. $x = y^2$ fails the vertical test ( $x = 4$ gives $y = \pm 2$ ) and passes the horizontal one.
Function equality

$\begin{gathered} f = g \iff \text{same domain,} \\ \text{same codomain, same values} \end{gathered}$

Use when All three, not the formula alone. $\dfrac{x^2}{x}$ and $x$ agree for $x \neq 0$ , but $0$ lies in only one domain, so they differ.
Additivity and scaling

$\begin{gathered} \text{in general:} \\ f(a + b) \neq f(a) + f(b) \\ f(kx) \neq k\,f(x) \end{gathered}$

Use when Only $f(x) = mx$ satisfies both. Even $f(x) = mx + c$ with $c \neq 0$ fails.

e.g. $f(x) = x^2$ : $f(2 + 3) = 25$ , while $f(2) + f(3) = 13$ .
Difference quotient

$\frac{f(x + h) - f(x)}{h}$

Use when $h \neq 0$ , with $x$ and $x + h$ in the domain. Cancelling the $h$ widens the domain, so the simplified form and the original agree only for $h \neq 0$ . It is the secant slope from $x$ to $x + h$ .

e.g. $f(x) = x^2 - 5x$ gives $2x + h - 5$ .
A finite table underdetermines a function

$\begin{gathered} g(x) = f(x) \\ +\, c\,(x - x_1)(x - x_2)\cdots(x - x_n) \end{gathered}$

Use when Any constant $c$ , any sampled inputs $x_1, \dots, x_n$ . Each factor vanishes at its own sample, so infinitely many functions fit one finite table. Join plotted points only after proving the features between them.

e.g. $x$ and $x + (x - 1)(x - 2)(x - 3)$ agree at $1, 2, 3$ , yet give $0$ and $-6$ at $x = 0$ .
Sign of a factored form

$f(x) = a\,(x - r_1)^{k_1}(x - r_2)^{k_2}\cdots$

Use when Needs a fully factored rule. The sign of $f$ on each interval is the product of its factor signs; a factor with no real root ( $x^2 + 1$ ) keeps one sign throughout, the sign of its leading coefficient. Odd multiplicity $k_i$ flips the sign at $r_i$ and the graph crosses; even $k_i$ keeps it and the graph touches.
Even and odd

$\begin{gathered} \text{even: } f(-x) = f(x) \\ \text{odd: } f(-x) = -f(x) \end{gathered}$

Use when Both need a domain holding $-x$ whenever it holds $x$ . Even: fixed by reflection across the $y$ -axis. Odd: fixed by a $180^\circ$ rotation about the origin, with $f(0) = 0$ whenever $0$ is in its domain. Exponent parity decides it only when every term shares one parity.

e.g. $x^5 - 4x$ is odd; $x^2 + x$ is neither, as $f(-x) = x^2 - x$ .
End behavior

$\begin{gathered} a_n x^n + \cdots + a_0 \\ = x^n\left(a_n + \frac{a_{n-1}}{x} + \cdots + \frac{a_0}{x^n}\right) \end{gathered}$

Use when Polynomials, $a_n \neq 0$ . Both ends track $a_n x^n$ : signs matching for even $n$ , opposite for odd $n$ , all flipped when $a_n < 0$ .

e.g. $-2x^3 + x = x^3\left(-2 + \tfrac{1}{x^2}\right)$ falls right and rises left.
Transformation master formula

$\begin{gathered} y = a\,f\big(b(x - h)\big) + k \\ (u, v) \longmapsto \left(\frac{u}{b} + h,\ av + k\right) \end{gathered}$

Use when $b \neq 0$ , with $(u, v)$ on the graph of $f$ . The halves never mix, and a negative $a$ or $b$ swaps that interval's endpoints. Outside moves run in arithmetic order (scale, then shift), inside moves in reverse.

e.g. $(6, -2)$ on $f$ lands at $(5, 10)$ on $y = -3f(x + 1) + 4$ .
Factoring a compound inside

$f(bx - c) = f\!\left(b\left(x - \frac{c}{b}\right)\right)$

Use when $b \neq 0$ . Pull $b$ out before reading the shift: it is $\tfrac{c}{b}$ , never $c$ .
Composition

$(g \circ f)(x) = g(f(x))$

Use when Reads right to left: $f$ acts first. Associative, with $\operatorname{id}(x) = x$ neutral on either side, but NOT always commutative. Every transformation is a composition.

e.g. $f(x) = x + 1$ , $g(x) = x^2$ : $(g \circ f)(x) = (x + 1)^2 \neq x^2 + 1 = (f \circ g)(x)$ .
Domain of a composition

$\begin{gathered} \operatorname{dom}(g \circ f) \\ = \{\, x \in \operatorname{dom} f : f(x) \in \operatorname{dom} g \,\} \end{gathered}$

Use when NOT $\operatorname{dom} f \cap \operatorname{dom} g$ . The second condition constrains an OUTPUT, so the range of $f$ settles it.
Inverse: the two-sided test

$g \circ f = \operatorname{id}_{\operatorname{dom} f}, \qquad f \circ g = \operatorname{id}_{\operatorname{dom} g}$

Use when Both, each on its own domain, and those domains can differ. A two-sided inverse is unique, which licenses the name $f^{-1}$ .
Inverse: existence, sets, and graph

$\begin{gathered} f \text{ invertible} \iff f \text{ one-to-one} \\ \operatorname{dom} f^{-1} = \operatorname{range} f \\ \operatorname{range} f^{-1} = \operatorname{dom} f \end{gathered}$

Text description

A curve and its inverse, mirror images across the line y = x.

Use when Needs the codomain taken to be the range. Restricting a domain to earn one-to-oneness yields a different function. The graph of $f^{-1}$ is the reflection across $y = x$ , a function exactly when $f$ is one-to-one.

e.g. $x^2$ inverts to $\sqrt{x}$ on $x \ge 0$ and to $-\sqrt{x}$ on $x \le 0$ .

Problem types, step by step

Find the natural domain of a formula

Collect every requirement: nonzero denominators, nonnegative insides for even roots.
An even root in a denominator needs its inside strictly positive, not merely nonnegative.
Intersect the requirements; a declared domain overrides all of it.

e.g. $f(x) = \dfrac{1}{\sqrt{5 - x}}$ needs $5 - x > 0$ , so the domain is $x < 5$ .

Find the range of a formula

Bound the rule from the inside out: a square, an even root, and an absolute value are all $\ge 0$ .
For each bound decide ATTAINED or only APPROACHED by solving $f(x) =$ that bound and checking whether an input in the domain produces it.
Report the set closed at every attained bound and open at every approached one.

e.g. $f(x) = \dfrac{1}{x^2 + 1}$ : $x^2 + 1 \ge 1$ , so $0 < f(x) \le 1$ , with $1$ hit at $x = 0$ and $0$ never.

Evaluate at a number, an expression, or a piecewise case

Wrap the whole argument in parentheses, copy it into every $x$ , then simplify; $f(x)$ is never $f$ times $x$ .
For a piecewise rule, the inequalities say which single case owns the input; apply that one alone. A valid rule covers the domain with no input given two outputs.
To solve $f(x) = c$ , treat it as an ordinary equation in $x$ .

e.g. $f(x) = x^2 + 1$ gives $f(2x) = 4x^2 + 1$ , not $2x^2 + 1$ .

Simplify a difference quotient

Substitute $x + h$ into the whole rule and expand; for a fractional rule, combine over a common denominator instead.
Subtract $f(x)$ . Every surviving term must carry a factor of $h$ , which checks the expansion.
Divide by $h$ and record $h \neq 0$ beside the answer.

e.g. $f(x) = \dfrac{1}{x}$ : the numerator is $\dfrac{-h}{x(x + h)}$ , so the quotient is $\dfrac{-1}{x(x + h)}$ .

Decide function of which variable, and one-to-oneness

Name the input variable; the question is incomplete without it.
Solve for the output. A surviving $\pm$ means one input has two outputs, so it is not a function of that variable.
For one-to-oneness, hunt for two inputs sharing an output, or show $f(a) = f(b)$ forces $a = b$ .

e.g. $x = y^2$ is a function of $y$ but not of $x$ , since $x = 4$ gives $y = \pm 2$ .

Read a polynomial graph off its factored form

Factor fully; the roots are the zeros and the $x$ -intercepts.
Mark the roots on a line and multiply the factor signs on each interval between consecutive roots.
Apply each root's multiplicity to decide crossing or touching.
Factor out the highest power to settle both ends, then fit a curve to those features.

e.g. $(x + 1)(x - 3)^2$ : negative on $x < -1$ , positive on both sides of $3$ , so it crosses at $-1$ and touches at $3$ .

Locate a hole

Only the inputs the denominator rejects are candidates.
Factor and cancel. A rejected input whose factor cancels leaves a hole; one whose factor survives gives a vertical asymptote.
Evaluate the simplified rule at the cancelled input for the hole's height.

e.g. $\dfrac{x^2 - 9}{x^2 - 3x} = \dfrac{x + 3}{x}$ for $x \neq 3$ : hole at $(3, 2)$ , asymptote at $x = 0$ .

Transform a point, a domain, or a range

Factor the inside into $b(x - h)$ so $b$ and $h$ are visible; read $a$ and $k$ off the outside.
Send $(u, v)$ to $\left(\tfrac{u}{b} + h,\ av + k\right)$ , and each endpoint of the domain and range the same way.
Check by putting the new input into the transformed rule and confirming the old height returns.

e.g. $y = f(3x + 6) = f(3(x + 2))$ : the point at input $9$ moves to input $\tfrac{9}{3} - 2 = 1$ .

Compose two functions, or decompose one

To compose $g \circ f$ : the right function $f$ acts first, so substitute its rule into every input slot of $g$ .
For the domain, impose $x \in \operatorname{dom} f$ , then solve $f(x) \in \operatorname{dom} g$ ; keep that domain even when the formula simplifies away every trace of it.
To decompose $h$ : peel the outermost operation off as $g$ , call what it acts on $f$ , and confirm $g(f(x))$ rebuilds $h$ . Any split that composes back is correct.

e.g. $f(x) = \sqrt{x}$ , $g(x) = \dfrac{1}{x - 3}$ : $(g \circ f)(x) = \dfrac{1}{\sqrt{x} - 3}$ on $x \ge 0$ with $x \neq 9$ .

Find an inverse and prove it

Confirm $f$ is one-to-one, restricting the domain if it is not and recording the restriction taken.
Write $y = f(x)$ and solve for $x$ : for a linear-fractional rule, clear the denominator, gather the $x$ terms, factor, divide.
Rename the input to get $f^{-1}$ , with $\operatorname{dom} f^{-1} = \operatorname{range} f$ .
Verify both compositions return $x$ , each on its own domain.

e.g. $f(x) = \dfrac{x}{x - 2}$ gives $x(y - 1) = 2y$ , so $f^{-1}(x) = \dfrac{2x}{x - 1}$ , on $x \neq 1$ .

Exam traps

Trap Letting a simplified formula overwrite the domain: calling $\dfrac{x^2 - 4}{x - 2}$ the function $x + 2$ , or calling $(g \circ f)(x) = (\sqrt{x})^2 = x$ the identity.

Fix Both wrong: the first is $x + 2$ with $x = 2$ deleted, a hole at $(2, 4)$ ; the second is $x$ on $x \ge 0$ .
Trap Reading the shift of $y = f(2x - 3)$ as right $3$ .

Fix Factor first: $2x - 3 = 2\left(x - \tfrac{3}{2}\right)$ , a compression by $2$ and a shift right $\tfrac{3}{2}$ . The point at input $4$ lands at $\tfrac{7}{2}$ , not $7$ .
Trap Taking $\operatorname{dom} f \cap \operatorname{dom} g$ as the domain of $g \circ f$ .

Fix For $f(x) = \sqrt{x}$ and $g(x) = \dfrac{1}{x - 2}$ that deletes $x = 2$ and keeps $x = 4$ ; the truth is the reverse, since $\sqrt{4} = 2$ .
Trap Declaring $g = f^{-1}$ on the strength of one composition.

Fix For $f(x) = x^2$ on all reals and $g(x) = \sqrt{x}$ , $f(g(x)) = x$ but $g(f(x)) = \lvert x \rvert$ , returning $3$ at $x = -3$ .
Trap Reading a superscript on a function name as arithmetic, so $f^{-1}$ becomes $\dfrac{1}{f}$ or $f^2$ becomes $(f(x))^2$ .

Fix Both mark composition: $f^2 = f \circ f$ . For $f(x) = 2x - 1$ : $f^2(x) = 4x - 3$ , not $4x^2 - 4x + 1$ , and $f^{-1}(x) = \tfrac{x + 1}{2}$ , not $\tfrac{1}{2x - 1}$ .
Trap Claiming the graphs of $f$ and $f^{-1}$ can meet only on the line $y = x$ .

Fix Reflection makes their shared points symmetric about that line, not confined to it: $f(x) = -x^3$ meets $f^{-1}(x) = -\sqrt[3]{x}$ at $(1, -1)$ and $(-1, 1)$ . Only $f(a) = a$ forces one onto the line.
Trap Reordering two moves on the same side of $f$ and expecting the same graph.

Fix $2f(x) + 3$ (stretch, then shift) and $2(f(x) + 3) = 2f(x) + 6$ (shift, then stretch) sit $3$ apart everywhere.
Trap Claiming a finished graph names the transformation that made it.

Fix For $f(x) = x^2$ , $y = 4x^2$ is both $f(2x)$ and $4f(x)$ ; the coincidence needs a homogeneous power and dies for $x^2 + 1$ . Describe a transformation, never "the" one.

Chapter test Questions from across the chapter

Vocabulary and notation

Formulas and theorems

The two line tests

Function equality

Additivity and scaling

Difference quotient

A finite table underdetermines a function

Sign of a factored form

Even and odd

End behavior

Transformation master formula

Factoring a compound inside

Composition

Domain of a composition

Inverse: the two-sided test

Inverse: existence, sets, and graph