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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).

Chapter 5 · pp. 11–110

Chapter 1: Functions

This chapter develops the idea of a function and the tools for working with one. It starts with what a function is and how it is written, then covers domain and range, rates of change, the shape and behavior of graphs, composition of functions, transformations of graphs, and inverse functions. This first part covers the basic definitions, function notation, and the four ways to show a function: words, tables, graphs, and formulas.

What a function is

A function is a rule linking an input to an output so that each input gives exactly one output. We say the output is a function of the input. The input is the independent variable. The output is the dependent variable, because it is found from the input (p. 11).

Example: height is a function of age. On a given birthday, a person has one height, so each age gives one height (p. 11).

A relationship can work as a function in one direction but not in reverse. Age is not a function of height, because a person can be 70 inches tall at both age 20 and age 21. One input (70 inches) would then have two outputs (pp. 11–12). A coffee menu works the same way. Price is a function of the item, since each item has one price. Item is not a function of price if two items cost the same (p. 12).

A one-to-one function is stricter. Each input gives one output, and each output also comes from only one input. Suppose each percentage in a course gave its own grade point, and each grade point came from only one percentage. That would be one-to-one. If many percentages earn the same letter grade, it is still a function but not one-to-one (pp. 12, 15).

Function notation

We write a function as output = f(input), read "f of input." The letter f is the function's name. To say that height h depends on age a, we can write h = f(a), or name the function h and write h(a). Descriptive letters help us remember what the quantities mean (p. 13).

A common mistake is to read the parentheses in f(a) as multiplication, "f times a." They show that a is being put into f. For example, if d(m) gives the days in month m, then d(March) = 31. March is the input, and 31 is the output. Nothing is multiplied (p. 13).

A statement like f(2005) = 300 pairs one input with one output. If N = f(y) is the number of police officers in a town in year y, then f(2005) = 300 says the town had 300 officers in 2005. The number inside the parentheses is the input, and the number after the equals sign is the output (p. 13).

Tables as functions

To test a table, look down the input column. If any input appears with two different outputs, the table is not a function (pp. 14–15).

Inputs 2, 5, 8 with outputs 1, 3, 6: a function, and one-to-one.
Inputs −3, 0, 4 with outputs 5, 1, 5: a function, but not one-to-one, because the output 5 comes from two inputs.
Inputs 1, 5, 5 with outputs 0, 2, 4: not a function, because input 5 has outputs 2 and 4.

Only a table that is a function can be checked for one-to-one (p. 15).

Evaluating and solving

Two tasks come up again and again. To evaluate, you know the input and find the output. This always gives exactly one result. To solve, you know the output and find the input or inputs that produce it. This can give several answers, because different inputs may share an output (p. 15).

Example: let g have the table n = 1, 2, 3, 4, 5 and Q = 8, 6, 7, 6, 8. Evaluating g(3) gives 7. Solving g(n) = 6 gives n = 2 and n = 4 (p. 15).

Graphs as functions

By convention, the input goes on the horizontal axis and the output on the vertical axis. So a graph with y = f(x) shows y as a function of x (p. 16).

The vertical line test checks whether a graph is a function. Imagine vertical lines sliding across the picture. If any one of them crosses the graph more than once, some input has more than one output, so it is not a function. A parabola lying on its side fails this test (p. 17).

The horizontal line test is used only on graphs that already pass the vertical test. If any horizontal line crosses the graph more than once, one output comes from several inputs, so the function is not one-to-one. A regular parabola fails, since the line y = 4 meets it at two x-values (p. 17).

To evaluate from a graph, start at the input on the horizontal axis and go up or down to the curve, then read the height. To solve, start at the output on the vertical axis and go across, then read the x-value of every point at that height. Suppose a graph contains (2, 1), (−1, 4) and (3, 4). Then f(2) = 1, and solving f(x) = 4 gives x = −1 or x = 3. Two inputs share the output 4, so this function is not one-to-one (p. 17).

Formulas as functions

To write a relationship as a function, solve for the output variable so that it equals an expression in the input. For 2n + 6p = 12, subtract 2n to get 6p = 12 − 2n, then divide by 6:

$p = 2 - \frac{n}{3}$

Now p = f(n) can be found by direct evaluation, with no further solving (p. 18).

Not every relationship can be written this way. Solving $x^2 + y^2 = 1$ for y gives $y = \pm\sqrt{1 - x^2}$ . The input x = 0 would give both y = 1 and y = −1. So the equation is not a single function y = f(x) (p. 18).

To evaluate a formula function, replace the input variable with the given value and simplify. To solve, set the formula equal to the given output and solve for the input. For $k(t) = t^3 + 2$ :

Evaluate: $k(2) = 2^3 + 2 = 10$ .
Solve $k(t) = 1$ : $t^3 + 2 = 1$ , so $t^3 = -1$ and $t = -1$ .
Check: $k(-1) = -1 + 2 = 1$ . ✓ (p. 19)

If the formula has a quadratic, move everything to one side so it equals zero, factor, and use the zero factor theorem. This says that if a product is zero, at least one factor is zero. For $h(p) = p^2 + 2p$ , solving h(p) = 3 gives $p^2 + 2p - 3 = 0$ . This factors as $(p+3)(p-1) = 0$ , so p = −3 or p = 1. Two inputs give the same output, so h is not one-to-one (pp. 19–20).

The toolkit functions and the circle

The toolkit functions are basic functions whose equations, graphs, and key values you should know well. Later topics build on them (pp. 20–22).

Constant: $f(x) = c$
Identity: $f(x) = x$
Absolute value: $f(x) = |x|$
Quadratic: $f(x) = x^2$
Cubic: $f(x) = x^3$
Reciprocal: $f(x) = \frac{1}{x}$
Reciprocal squared: $f(x) = \frac{1}{x^2}$
Square root: $f(x) = \sqrt{x}$
Cube root: $f(x) = \sqrt[3]{x}$

The circle is an important equation that is not a function. Every point (x, y) on a circle is the same distance r from the center (h, k). The distance formula turns this into

$(x-h)^2 + (y-k)^2 = r^2$

For center (3, −9) and radius 6, the equation is $(x-3)^2 + (y+9)^2 = 36$ . A circle fails the vertical line test, so it is not a function y = f(x) (p. 21).

What Domain and Range Mean

A function is only a model of something real, so it usually works for some inputs and not for others. The domain is the set of possible inputs. The range is the set of possible outputs. (p. 32)

Take a table that links a tree's circumference to its height. A negative circumference makes no sense. Neither does a 3000-foot tree. So a sensible domain starts at 0 and stops at a realistic largest circumference. A sensible range also starts at 0 and stops at the tallest tree known, about 379 feet. (p. 32)

Sometimes the outputs come from a short list. Letter postage depends on weight $w$ in ounces. Weights up to 3.5 ounces are allowed, so the domain is $0 < w \le 3.5$ . We leave out 0 because mailing nothing is not a real case. The only prices are $0.58, $0.78, $0.98 and $1.18. So the range is the list {0.58, 0.78, 0.98, 1.18}, not an interval. (p. 33)

Ways to Write Sets of Numbers

There are three common ways to describe domains and ranges.

Inequalities. For example, $10 \le x < 30$ means all numbers from 10 up to, but not including, 30. (p. 34)

Set-builder notation. The same set is written $\{x \mid 10 \le x < 30\}$ . The braces mean "the set of" and the bar means "such that." The range can use its own variable, such as $\{h \mid 0 \le h \le 379\}$ for tree height. A domain with no limits is $\{x \mid x \text{ is a real number}\}$ . (p. 34)

Interval notation. This lists only the endpoints. A square bracket includes an endpoint. A parenthesis excludes it. (p. 34)

$5 < h \le 10$ is $(5, 10]$
$5 \le h < 10$ is $[5, 10)$
$5 < h < 10$ is $(5, 10)$
$h > 10$ is $(10, \infty)$
$h \le 10$ is $(-\infty, 10]$
all real numbers is $(-\infty, \infty)$

Infinity is not a number, so it can never be included. It always gets a parenthesis. (p. 34)

To join two separate pieces, inequalities and set-builder notation use the word "or." Interval notation uses the union symbol $\cup$ . For example, " $1 \le x \le 3$ or $x > 5$ " becomes $[1, 3] \cup (5, \infty)$ . (pp. 34–35)

Reading Domain and Range from a Graph

The domain is every input shown along the horizontal axis. The range is every output shown along the vertical axis. (p. 35)

A graph of oil production runs from 1975 to 2008 and has outputs from about 180 to 2010. Its domain is $[1975, 2008]$ and its range is about $[180, 2010]$ . The range endpoints are estimates because they do not land exactly on grid lines. (p. 36)

Be careful: the graph may continue past the edge of the window. If so, the true domain and range could be larger than what you see. Also, the variables are not always $x$ and $y$ . Descriptive letters, such as $y$ for year and $b$ for barrels, help you remember the context. (pp. 35–36)

Domain and Range of the Toolkit Functions

Each toolkit function has a domain and range that come from how it works.

Constant $f(x) = c$ : domain $(-\infty, \infty)$ . The only output is $c$ , so the range is $[c]$ . (p. 37)
Identity $f(x) = x$ and cubic $f(x) = x^3$ : domain and range are both $(-\infty, \infty)$ . (p. 37)
Quadratic $f(x) = x^2$ : domain $(-\infty, \infty)$ , range $[0, \infty)$ . A number times itself is never negative. (p. 37)
Absolute value $f(x) = |x|$ : the same domain and range as the quadratic. It measures distance from 0, which cannot be negative. (p. 38)
Reciprocal $f(x) = \frac{1}{x}$ : we cannot divide by 0, so $x = 0$ is out of the domain. One divided by a number is never 0, so 0 is out of the range too. Both are $(-\infty, 0) \cup (0, \infty)$ . (p. 38)
Reciprocal squared $f(x) = \frac{1}{x^2}$ : the domain is the same, $(-\infty, 0) \cup (0, \infty)$ . Squaring makes every output positive, so the range is $(0, \infty)$ . (p. 38)
Square root $f(x) = \sqrt{x}$ : real square roots of negative numbers do not exist. Domain and range are both $[0, \infty)$ . (p. 38)
Cube root $f(x) = \sqrt[3]{x}$ : the cube root of a negative number is real, so domain and range are both $(-\infty, \infty)$ . (p. 38)

Finding the Domain from a Formula

Two problems limit the domain of a formula. One is a negative number under a square root. The other is a zero denominator. Exclude the inputs that cause them. (p. 38)

Example 1: $f(x) = \sqrt{2x + 4}$ . We need $2x + 4 \ge 0$ , so $x \ge -2$ . The domain is $[-2, \infty)$ .

Example 2: $g(x) = \frac{3}{6 - 3x}$ . The denominator is 0 when $x = 2$ . The domain is $(-\infty, 2) \cup (2, \infty)$ . (p. 38)

Piecewise Functions

A piecewise function uses a different formula on different parts of its domain. Absolute value is the basic example:

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

The second piece changes a negative input into a positive output. (p. 39)

Real prices can work this way too. A museum charges $5 per person for groups of 1 to 9 and a flat $50 for 10 or more:

$C(n) = \begin{cases} 5n & \text{if } 1 \le n < 10 \\ 50 & \text{if } n \ge 10 \end{cases}$

(p. 39)

Evaluating a Piecewise Function

First decide which piece's domain contains the input. Then use only that formula. A data plan is defined by $C(g) = 25$ for $0 < g \le 2$ and $C(g) = 25 + 10(g - 2)$ for $g > 2$ .

For $g = 1.5$ : this is in the first piece, so $C(1.5) = \$ 25$.
For $g = 4$ : this is in the second piece, so $C(4) = 25 + 10(4 - 2) = \$ 45$. (p. 40)

Graphing a Piecewise Function

Graph each piece only on its own domain, then put them all on one set of axes. At an endpoint, use an open circle if the inequality is strict ( $<$ or $>$ ). Use a closed circle if it includes equality ( $\le$ or $\ge$ ). (p. 40)

Suppose one piece is $6 - x$ for $x > 2$ . This is a line. At $x = 2$ , the value is $6 - 2 = 4$ , but 2 is not allowed, so draw an open circle at $(2, 4)$ . At $x = 3$ , the value is $3$ , so plot $(3, 3)$ . Draw the line through them. Pieces from toolkit functions keep their usual shapes, cut off at the edges of their domains. (p. 40)

Rates of Change

A rate of change tells how an output quantity changes compared with the input quantity. Its units are always "output units per input units." Examples are dollars per hour, miles per gallon, or whales lost per year. (p. 46)

Most real quantities do not change by the same amount each step, so we usually find an average rate of change. Gas prices rose from $1.47 in 2002 to$ 2.14 in 2009. That is a rise of $0.67 over 7 years, or about$ 0.096 per year (roughly 9.6 cents). (p. 46)

In function notation, the average rate of change of $f$ on the interval $[a, b]$ is

$\frac{f(b)-f(a)}{b-a}=\frac{\text{change in output}}{\text{change in input}}$

On a graph, this is the slope of the line through the points at $x=a$ and $x=b$ . For distance versus time, it is the average speed. On a road trip, the distance went from 10 miles at $t=0$ to 292 miles at $t=6$ . The average speed is $\frac{292-10}{6}=47$ miles per hour. (pp. 47–48)

The sign matters. If outputs fall as inputs rise, the change in output is negative, so the average rate of change is negative. Gas cost $2.64 in 2007 and$ 2.14 in 2009. The change was $-0.50 over 2 years, so the rate is$ -0.25$ dollars per year. (p. 47)

Finding Average Rates of Change from Formulas

To find the average rate of change of a formula:

Evaluate $f$ at both endpoints.
Subtract the outputs, using the same order as the endpoints.
Divide by the change in input.

Example: let $F(d)=\frac{2}{d^2}$ be the force between magnets. On $[2, 6]$ , $F(6)=\frac{1}{18}$ and $F(2)=\frac12$ . Then

$\frac{\frac1{18}-\frac12}{6-2}=\frac{-\frac49}{4}=-\frac19$

The force drops by $\frac19$ newton per centimeter on average. (p. 49)

If an endpoint is a symbol, simplify and cancel. For $g(t)=t^2+3t+1$ on $[0, a]$ :

$\frac{g(a)-g(0)}{a-0}=\frac{(a^2+3a+1)-1}{a}=\frac{a(a+3)}{a}=a+3$

This one expression covers every interval starting at 0. For $[0, 5]$ it gives 8. (pp. 49–50)

Increasing and Decreasing

A function is increasing on an interval if $f(b)>f(a)$ whenever $b>a$ in that interval. Its average rate of change is positive. A function is decreasing if $f(b)<f(a)$ whenever $b>a$ , and its average rate of change is negative. (p. 50)

We write these intervals with open intervals, such as $(1, 3)$ . At a turning point the function is neither increasing nor decreasing, so the endpoint is left out. (p. 50)

Local Maxima and Minima

A local maximum is a point where the function switches from increasing to decreasing. A local minimum is a point where it switches from decreasing to increasing. Together they are the local extrema. (p. 51)

In the gas table, prices rose up to $t=8$ and then fell at $t=9$ . So the function is decreasing on $(8, 9)$ and has a local maximum at $t=8$ . (p. 51)

For harder formulas, use graphing technology to estimate extrema. In one example, a curve had a local minimum near $x\approx 2.449$ and a matching local maximum near $x\approx -2.449$ . The function was therefore increasing on $(-\infty, -2.449)$ and on $(2.449, \infty)$ . Different tools may agree only to a few decimal places because they use different approximation methods. (pp. 51–52)

Concavity

Two companies can both have rising sales and still grow very differently. To see the difference, compute the average rate of change over successive intervals.

Company A's weekly rates were 5, 2.1, 1.6, 1.3. They are shrinking.
Company B's weekly rates were 0.5, 1.5, 2.5, 3.5. They are growing.

A function is concave down if its rate of change is decreasing. It bends downward like a frown. A function is concave up if its rate of change is increasing. It bends upward like a smile. So A is concave down and B is concave up. A point where the concavity changes is an inflection point. (pp. 52–53, 55)

Concavity from a table. Compute the rate of change on each consecutive interval, then see whether the rates grow or shrink.

For $h(t)=144-16t^2$ with outputs 144, 128, 80, 0 at $t=0,1,2,3$ , the rates are $-16, -48, -80$ . These get smaller (more negative), so $h$ is concave down. (p. 54)
For a car's value, the rates were $-1829, -1590, -1382.5, -1201.5$ . These get less negative, so they are increasing. The function is concave up. (p. 54)

Increasing/Decreasing Is Separate from Concavity

The two ideas are independent, and all four combinations can happen. The car's value is decreasing but concave up. The falling object's height is decreasing and concave down. A graph can even switch several times. In one example, the graph was concave up on $(-\infty, -1)$ and $(2, \infty)$ , and concave down on $(-1, 2)$ . It had inflection points at $x=-1$ and $x=2$ . (pp. 53, 55)

Behavior of Some Toolkit Functions

Quadratic, $x^2$ : decreasing on $(-\infty, 0)$ and increasing on $(0, \infty)$ . It has a minimum at $x=0$ and is concave up everywhere. (p. 56)
Cubic, $x^3$ : increasing everywhere. It is concave down on $(-\infty, 0)$ and concave up on $(0, \infty)$ , so $(0, 0)$ is an inflection point. (p. 56)
Square root, $\sqrt{x}$ : increasing and concave down on $(0, \infty)$ . (p. 56)

Composition: Using One Function's Output as Another's Input

Suppose T(d) gives the average temperature on day d, and C(T) gives the cost of heating a house at temperature T. To find the heating cost on day 5, first compute T(5), then feed that temperature into C. Chaining functions like this is called composition. It is written $C(T(5))$ and read "C of T of 5" (p. 61).

The general form is $f(g(x))$ , read "f of g of x." An equivalent notation is $(f \circ g)(x)$ . In both, g is the inside function and f is the outside function (p. 61).

A composition only makes sense when the units match. The output units of the inside function must be the input units of the outside function. Say f(x) gives miles driven in x hours, and g(y) gives gallons used for y miles. Then g(f(x)) works: hours go in, miles come out, and miles go into g to give gallons. But f(g(y)) would push gallons into a function that expects hours, so it means nothing (pp. 61–62).

Evaluating Compositions Numerically

Always work from the inside out. Find the inner value first, then use it as the input of the outer function (pp. 62–63).

With tables: if g(3) = 2 and f(2) = 8, then f(g(3)) = f(2) = 8.
With graphs: find the input on the horizontal axis, read the output off the graph, then use that output as the next input.
Order matters. If f(4) = 1 and g(1) = 3, then g(f(4)) = 3. Usually f(g(x)) and g(f(x)) are different.

With formulas, the method is the same (p. 64). Let $f(t) = t^2 - t$ and $h(x) = 3x + 2$ . Then h(1) = 5, so $f(h(1)) = f(5) = 25 - 5 = 20$ . Put parentheses around every substituted input to avoid sign and exponent mistakes.

Evaluating a Function at an Expression

To build a single formula for a composition, we first need to evaluate a function at something other than a number. The rule: replace every occurrence of the input variable with the whole input, in parentheses (p. 65).

For $f(t) = t^2 - t$ :

$f(a) = a^2 - a$
$f(x+2) = (x+2)^2 - (x+2) = x^2 + 4x + 4 - x - 2 = x^2 + 3x + 2$
$f(t^3) = (t^3)^2 - t^3 = t^6 - t^3$

It does not matter that the new input uses the same letter as the old variable. The substitution works the same way.

Formulas for Compositions

To find $f(g(x))$ , write out the formula for g(x), then substitute it for the input variable of f (pp. 66–67). Let $f(x) = x^2$ and $g(x) = \frac{1}{x} - 2x$ . Then

$f(g(x)) = \left(\frac{1}{x} - 2x\right)^2, \qquad g(f(x)) = \frac{1}{x^2} - 2x^2.$

The two results are different, which confirms that order matters.

A single formula saves work in applications. If tax revenue is R(p) for a population of p thousand, and the population is p(t) after t years, then $R(p(t))$ gives revenue directly from the year. You never have to compute the middle value (the population) separately (pp. 66–67).

Domain of a Composition

The domain of $f(g(x))$ must satisfy two conditions. The input must be allowed in the inside function g, and the composite expression itself must be defined (pp. 67–68).

Take $f(x) = \frac{1}{\sqrt{x} - 1}$ and $g(x) = x - 2$ . Then $f(g(x)) = \frac{1}{\sqrt{x-2} - 1}$ .

The square root needs $x - 2 \ge 0$ , so $x \ge 2$ .
The denominator is zero when $\sqrt{x-2} = 1$ , which happens at $x = 3$ . So $x \ne 3$ .

The domain is $[2, 3) \cup (3, \infty)$ . A common mistake is to check only the final formula and forget the restriction that comes from the inside function.

Decomposing a Function

Decomposing means writing a function as a composition of two simpler ones. Look for an inner expression that the rest of the formula acts on (p. 68). For $f(x) = \sqrt{5 - x^2} + 3$ , the inside of the root is $h(x) = 5 - x^2$ . The outer step is $g(x) = \sqrt{x} + 3$ . Then $f(x) = g(h(x))$ . Check by recomposing: $g(5 - x^2) = \sqrt{5 - x^2} + 3$ . The answer is not unique. For example, $h(x) = x^2$ with $g(x) = \sqrt{5 - x} + 3$ also works.

Vertical and Horizontal Shifts

Shifts are compositions with a simple function that adds a number to the input or the output.

Vertical shift. $g(x) = f(x) + k$ moves the graph up by k if k > 0 and down if k < 0. Only outputs change, and the inputs stay the same. If V(t) is the vent area in square feet, then V(t) + 20 is 20 square feet more at every time. In a table, $g(x) = f(x) - 3$ subtracts 3 from every output (pp. 64–65).

Horizontal shift. $g(x) = f(x + k)$ moves the graph left by k if k > 0 and right if k < 0. The change is inside the function, so it acts on inputs. For $g(x) = f(x - 3)$ , the input must be 3 larger to produce the same output. So if f(2) = 1, then g(5) = 1, and the graph moves right 3. Starting a venting plan 2 hours sooner is $F(t) = V(t + 2)$ (pp. 66–67).

A Common Misconception About f(x + 2)

Many students see the plus sign and think f(x + 2) moves right. It moves left. Inside changes act opposite to what you might expect. With x + 2, the function receives an input 2 larger, so each output appears at an input 2 smaller. To check, take a parabola with vertex at (2, 0). A new graph with its vertex at x = 2 would need g(2) = 0, which requires f(2 − 2). So $g(x) = f(x - 2)$ is what shifts right (pp. 67–68).

Inside Versus Outside Changes

Changes outside the function affect outputs and the vertical position of the graph. Changes inside affect inputs and the horizontal position. Let G(m) be the gallons needed for m miles. Then G(m) + 10 means 10 more gallons than needed for m miles, which is an output change. G(m + 10) means the gallons for 10 more miles, which is an input change. Inside changes alter the domain, and outside changes alter the range (pp. 68–69).

Reflections

A negative sign gives a reflection, and where it sits decides which kind.

$-f(x)$ flips the graph over the x-axis. Every output changes sign.
$f(-x)$ flips the graph over the y-axis. Every input changes sign.

For $s(t) = \sqrt{t}$ , the function $-\sqrt{t}$ has range $(-\infty, 0]$ , while $\sqrt{-t}$ has domain $(-\infty, 0]$ . In a table, f(−x) negates each input and keeps the outputs unchanged (pp. 70–71).

Even and odd functions

Some functions have a built-in symmetry. A function is even if $f(-x) = f(x)$ for every input. Its graph is symmetric about the y-axis, so the left side mirrors the right side. A function is odd if $f(-x) = -f(x)$ . Its graph is symmetric about the origin, which means it looks the same after a half-turn. The functions $x^2$ and $|x|$ are even. The functions $x^3$ and $1/x$ are odd. A function such as $2^x$ has neither symmetry, so it is neither even nor odd. (p. 73)

To test a function, compute $f(-x)$ . Then compare it with $f(x)$ and with $-f(x)$ . Take $f(x) = x^3 + 2x$ . Then $f(-x) = -x^3 - 2x$ . This is not $f(x)$ , so $f$ is not even. It does equal $-f(x)$ , so $f$ is odd. (pp. 73–74)

Vertical stretches and compressions

Earlier shifts added to inputs or outputs. Now we multiply them. Multiplying the outputs is an "outside" change. If $g(x) = k\,f(x)$ , every output is multiplied by $k$ and the inputs stay the same.

If $k > 1$ , the graph is stretched away from the x-axis.
If $0 < k < 1$ , the graph is compressed toward the x-axis.
If $k < 0$ , there is also a vertical reflection.

For example, a fruit fly population $Q$ that starts twice as large as $P$ is $Q(t) = 2P(t)$ . Every point doubles its distance from the horizontal axis. For $g(x) = \tfrac{1}{2}f(x)$ , an output of 3 at $x = 4$ becomes $\tfrac{3}{2}$ . (pp. 84–85)

You can use this to find a formula from a graph. Pick a clear point. Suppose a graph has the point $(2, 2)$ , but $x^3$ gives 8 at $x = 2$ . The outputs are $\tfrac{1}{4}$ as large, so $g(x) = \tfrac{1}{4}x^3$ . (pp. 85–86)

Horizontal stretches and compressions

Multiplying the inside has the opposite effect. If $g(x) = f(kx)$ , the graph is compressed horizontally by a factor of $1/k$ when $k > 1$ . It is stretched by $1/k$ when $0 < k < 1$ . A negative $k$ also reflects the graph horizontally. For example, $R(t) = P(2t)$ is a population that moves through its life twice as fast. Its graph is squeezed to half the width. (pp. 86–87)

The stretch factor and the coefficient are reciprocals. To stretch a graph horizontally by 4, use $f\left(\tfrac{x}{4}\right)$ . The input must be four times larger to get the same output. For $g(x) = f\left(\tfrac{1}{2}x\right)$ , the input 4 gives $f(2)$ . So a table's x-values are doubled. (pp. 87–88)

Combining transformations

Order matters within a group of changes.

Vertical changes. For $a f(x) + k$ , follow the order of operations. First stretch by $a$ , then shift by $k$ . For $2f(x) + 3$ , double each output, then add 3. Shifting first and stretching second gives a different graph. (p. 89)

Horizontal changes. For $f(bx + p)$ , we would have to shift first and then stretch, which is awkward. So factor the inside into $f(b(x + h))$ . For example, $f(2x + 3) = f\left(2\left(x + \tfrac{3}{2}\right)\right)$ . Compress by $\tfrac{1}{2}$ , then shift left $\tfrac{3}{2}$ . (p. 89)

Independence. Horizontal and vertical changes do not affect each other. You may do the horizontal group or the vertical group first. Only the order inside each group matters. (p. 90)

Tables. Work from the inside out. Take $g(x) = 2f(3x) + 1$ . Multiply each x-value by $\tfrac{1}{3}$ , then multiply each output by 2, then add 1. A table entry $f(6) = 10$ becomes the point $(2, 21)$ . (p. 90)

Graphs. Take $k(x) = f\left(\tfrac{1}{2}x + 1\right) - 3$ . Factor the inside to get $f\left(\tfrac{1}{2}(x + 2)\right) - 3$ . Then follow these steps:

Stretch horizontally by 2. The point $(0, 2)$ stays put because twice 0 is 0.
Shift left 2.
Shift down 3. (pp. 90–91)

Writing equations and reading monotonic behavior

To write an equation from a transformed toolkit graph, identify the reflection, the stretch and the shifts. Then use a known point to find any unknown stretch factor. A downward parabola with vertex $(3, 1)$ has the form $g(x) = -a(x-3)^2 + 1$ . The point $(2, -1)$ gives $-1 = -a(1) + 1$ , so $a = 2$ . The result is $g(x) = -2(x-3)^2 + 1$ . (pp. 91–92)

To find where a transformed function increases or decreases, start with the toolkit function's intervals. Take $g(x) = -\dfrac{2}{(x-1)^2} + 3$ . The function $1/x^2$ increases on $(-\infty, 0)$ and decreases on $(0, \infty)$ . The negative sign flips this behavior. The shift right by 1 moves the dividing point to $x = 1$ . So $g$ decreases on $(-\infty, 1)$ and increases on $(1, \infty)$ . Vertical shifts and stretches do not move these intervals. (pp. 92–93)

Inverse functions

An inverse function swaps the roles of input and output. If $f(a) = b$ , then $f^{-1}(b) = a$ . So $f(2) = 4$ means $f^{-1}(4) = 2$ . If $f(90) = 70$ miles, then $f^{-1}(70) = 90$ means it took 90 minutes to drive 70 miles. (pp. 103–105)

The raised $-1$ is only notation. It is not an exponent, and $f^{-1}(x)$ does not mean $\dfrac{1}{f(x)}$ . (p. 103)

A function and its inverse undo each other: $f^{-1}(f(a)) = a$ and $f(f^{-1}(b)) = b$ . For $f(x) = 3x - 5$ and $g(x) = \dfrac{x+5}{3}$ , composing in either order gives $x$ . (p. 104)

The domain and range also swap. The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . For $f(x) = 2^x$ , the range is $(0, \infty)$ , so $f^{-1}$ has domain $(0, \infty)$ . This is also why $x^2$ is the inverse of $\sqrt{x}$ only on $[0, \infty)$ . (p. 104)

Tables and graphs. To find $f^{-1}(70)$ , find 70 in the output column, or at height 70 on the vertical axis, and read the matching input. On a graph, for $g^{-1}(3)$ find height 3. That is the point $(5, 3)$ , so $g^{-1}(3) = 5$ . (p. 105)

Formulas. Solve for the input variable. From $C = \tfrac{5}{9}(F - 32)$ , multiply by $\tfrac{9}{5}$ and add 32 to get $F = \tfrac{9}{5}C + 32$ . Evaluating this is easier than solving each time. (pp. 103, 106)

Existence. An inverse is a function only if $f$ is one-to-one. Otherwise one input of $f^{-1}$ would need two different outputs. (p. 106)

Restricting the domain. The function $h(x) = x^2$ fails the horizontal line test. On $[0, \infty)$ it is one-to-one. Solve $y = x^2$ for $x$ and keep the positive root. This gives $h^{-1}(x) = \sqrt{x}$ . Using $(-\infty, 0]$ instead would give $-\sqrt{x}$ . (p. 107)

Graph symmetry. The graph of $f^{-1}$ is the graph of $f$ reflected over the line $y = x$ . Coordinates swap, so $(1, 0)$ becomes $(0, 1)$ and $(4, 2)$ becomes $(2, 4)$ . (pp. 107–108)

Check yourself

Decide whether $f(x) = x^3 + x$ and $g(x) = x^2 + x$ are even, odd or neither. Show your work.
The graph of $f$ passes through $(4, 6)$ . Give the matching point on the graph of $h(x) = 3f\left(\tfrac{x}{2}\right) - 1$ . Then describe the steps in order.
Why does $f(x) = x^2$ have no inverse function on $(-\infty, \infty)$ ? Name one domain restriction that fixes this and give the inverse.
For $f(x) = \sqrt{x - 3}$ , find the domain and range. Then state the domain and range of $f^{-1}$ .
Let $f(x) = 2x + 1$ and $g(x) = x^2$ . Write a formula for $f(g(x))$ . Then find the average rate of change of that composition on $[0, 2]$ .