Unit 2 · Topic 2.8 · about 30 minutes · 7 parts · 16 scored checks, 4 of them multiple choice

Inverse Functions

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The graph of f(x) = (x minus 2) squared plus 1 for x at least 2, rising from (2, 1), and the graph of its inverse, 2 plus the square root of (x minus 1) for x at least 1, rising from (1, 2). A dashed line y = x runs between the two curves. The point (4, 5) on f and the point (5, 4) on the inverse sit across the line from each other at the same distance, joined by a short dashed segment.

Part 1 Think first

A recipe in the wrong units

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A recipe from the United States says to bake at 350 degrees Fahrenheit. Your oven only shows Celsius. To turn Celsius into Fahrenheit, you multiply by 1.8 and then add 32. About what should you set your oven to?

In short. The inverse function $f^{- 1}$ runs $f$ backward: if $f (a) = b$ , then $f^{- 1} (b) = a$ , on a domain where every output of $f$ comes from exactly one input.

A table of $f^{- 1}$ swaps every pair $(a, b)$ of $f$ to $(b, a)$ , and the graph of $f^{- 1}$ is the graph of $f$ reflected over the line $y = x$ . The domain and range trade places.
From a formula, swap $x$ and $y$ in $y = f (x)$ and solve for $y$ . Then $f (f^{- 1} (x)) = f^{- 1} (f (x)) = x$ .
A function that repeats an output has no inverse until its domain is restricted, and a context can limit where the inverse makes sense.

Part 2 Learn it

Four short ideas, each checked as you go

2a. Running a function backward

A function sends each input to an output. Its inverse function, written $f^{- 1}$ , sends each output back to the input it came from. It is a reverse mapping:

if f (a) = b, then f^{- 1} (b) = a .

Here is a table for a function $f$ .

Values of $f$
$x$	0	1	2	3	4
$f (x)$	3	5	9	17	33

Since $f (2) = 9$ , you know $f^{- 1} (9) = 2$ . To get a whole table for $f^{- 1}$ , reverse every input-output pair: $(a, b)$ on $f$ becomes $(b, a)$ on $f^{- 1}$ . The outputs of $f$ become the inputs of $f^{- 1}$ .

Values of $f^{- 1}$ , the same pairs reversed
$x$	3	5	9	17	33
$f^{- 1} (x)$	0	1	2	3	4

Read the notation carefully. The $- 1$ in $f^{- 1}$ is not an exponent. $f^{- 1} (9)$ asks which input $f$ sends to 9. It does not mean $\frac{1}{9}$ , and it does not mean $\frac{1}{f (9)}$ .

Quick check · Fill the statement

No calculator. A function $g$ has $g (- 2) = 6$ , $g (0) = 4$ , $g (1) = - 1$ , $g (3) = 0$ and $g (5) = - 3$ . Tap a blank, then tap an answer.

$g^{- 1} (4)$ = and $g^{- 1} (- 3)$ = . The inputs of $g^{- 1}$ are the of $g$ .

2b. When there is no way back

Some functions cannot be run backward as they stand. Take $q (x) = (x - 3)^{2}$ . Both $q (1)$ and $q (5)$ equal 4. So what would $q^{- 1} (4)$ be? It would have to be 1 and 5 at once, and a function gives each input only one output.

A function has an inverse, or is invertible, on a domain where each output value comes from a unique input value. Being a function is not enough: $q$ gives every input one output, but two inputs share the output 4. On a graph you can see the problem as a horizontal line that crosses the graph twice.

The fix is to restrict the domain. Keep only $x \geq 3$ , the right half of the parabola, where $q$ is always increasing. Now every output from 0 upward comes from one input, and $q^{- 1} (4) = 5$ . Keeping $x \leq 3$ also works, and then $q^{- 1} (4) = 1$ . So does $4 \leq x \leq 10$ . A domain can be restricted in many ways. What matters is that no output repeats on the piece you keep.

That is why a statement such as " $q$ has an inverse" means nothing until you say on which domain.

Quick check · Decide and justify

No calculator. Let $p (x) = (x - 1)^{2} + 2$ on the domain $- 1 \leq x \leq 4$ . Is $p$ invertible on this domain?

Which reason supports your answer?

2c. On a graph, swap the axes

Every point $(a, b)$ on the graph of $f$ gives the point $(b, a)$ on the graph of $f^{- 1}$ . Swapping the coordinates of every point is the same as swapping the roles of the $x$ - and $y$ -axes.

On the page, that swap is a reflection over the line $y = x$ , which is the graph of the identity function $h (x) = x$ . In the diagram, $(4, 5)$ on $f$ lands at $(5, 4)$ on $f^{- 1}$ : the same distance from the line, on the other side. Where the graph of $f$ meets the line $y = x$ , the graph of $f^{- 1}$ meets it at the same point.

The swap carries the domain and range with it. On its invertible domain, the domain of $f$ becomes the range of $f^{- 1}$ , and the range of $f$ becomes the domain of $f^{- 1}$ . In the diagram, $f$ takes inputs $x \geq 2$ to outputs $y \geq 1$ , so $f^{- 1}$ takes inputs $x \geq 1$ to outputs $y \geq 2$ .

Quick check · Multiple choice

No calculator. The graph of an invertible function $f$ passes through $(- 1, 4)$ and $(3, 0)$ . Which two points must be on the graph of $f^{- 1}$ ?

2d. With a formula, undo the steps

When $f$ is given by a formula, write $y = f (x)$ , swap $x$ and $y$ , and solve for $y$ . Swapping the letters is the algebra version of reversing every pair. Take $f (x) = 3 x - 12$ .

$y = 3 x - 12$
Swap: $x = 3 y - 12$
Add 12: $x + 12 = 3 y$
Divide by 3: $f^{- 1} (x) = \frac{x + 12}{3}$

Look at the order. $f$ multiplies by 3 and then subtracts 12. The inverse uses the inverse operations in reverse order: add 12 first, then divide by 3. You put on socks before shoes, and you take off the shoes first.

Then check. A function and its inverse undo each other, so composing them gives the identity function: $f (f^{- 1} (x)) = f^{- 1} (f (x)) = x$ on the invertible domain. Here $f (f^{- 1} (x)) = 3 \cdot \frac{x + 12}{3} - 12 = x$ . If your composition simplifies to anything other than $x$ , the inverse is wrong.

One trap: $\frac{1}{3 x - 12}$ is the reciprocal of $f (x)$ , not its inverse. Compose it with $f$ and you will not get $x$ .

Quick check · Put it in order

No calculator. Put the steps for finding the inverse of $g (x) = 2 (x - 1)^{3} + 5$ in order. Tap them in sequence.

Part 3 Watch it done

An inverse on a restricted domain

Watch each step, then do the last one yourself.

Let $f (x) = (x - 2)^{2} + 1$ on the domain $x \geq 2$ . Find an expression for $f^{- 1} (x)$ and its domain, then find $f^{- 1} (10)$ .

Step 1 · Check that it is invertible
On $x \geq 2$ , $(x - 2)^{2}$ grows as $x$ grows, so $f$ is always increasing there and no output repeats. The outputs start at $f (2) = 1$ and go up, so the range of $f$ is $y \geq 1$ .
Step 2 · Swap x and y
$y = (x - 2)^{2} + 1$ becomes $x = (y - 2)^{2} + 1$ .
Step 3 · Undo in reverse order
$f$ subtracted 2, squared, then added 1. So subtract 1, take the square root, then add 2. Subtracting 1 gives $x - 1 = (y - 2)^{2}$ , and the square root gives $y - 2 = \pm \sqrt{x - 1}$ .
Step 4 · Choose the sign
Keep the plus: $f^{- 1} (x) = 2 + \sqrt{x - 1}$ , with domain $x \geq 1$ , the range of $f$ . Check: $f (f^{- 1} (x)) = (\sqrt{x - 1})^{2} + 1 = x$ for $x \geq 1$ . These are the two curves in the diagram.
Why this step?
Why the plus sign and not the minus?
The outputs of $f^{- 1}$ are the inputs of $f$ , and those were restricted to $x \geq 2$ . $2 + \sqrt{x - 1}$ is never less than 2. $2 - \sqrt{x - 1}$ is never more than 2, which would give back the left half of the parabola, the half the restriction removed.
Your step · Calculate

No calculator. Finish it. What is $f^{- 1} (10)$ ?

Your answer

Part 4 Try it

A rocket on the way up

A model rocket is launched straight up. Its height in meters $t$ seconds after launch is $h (t) = - 5 t^{2} + 40 t$ , for $0 \leq t \leq 8$ . It reaches its highest point, 80 meters, at $t = 4$ , and it lands at $t = 8$ .

Over the whole flight, $h$ is not invertible: the rocket passes most heights twice, once climbing and once falling. Restrict $h$ to the climb, $0 \leq t \leq 4$ , and it is invertible. That inverse takes a height and gives the time the rocket reaches it on the way up.

Try it · Calculate

No calculator. Using $h$ restricted to $0 \leq t \leq 4$ , find $h^{- 1} (60)$ .

Your answer seconds

Try it · Multiple choice

No calculator. On the climb, $h^{- 1} (75) = 3$ . What does this statement mean in this context?

Try it · Sort it

No calculator. Use $h$ restricted to the climb, $0 \leq t \leq 4$ . Tap each request, then whether this inverse can answer it.

$h^{- 1}$ gives a time

Outside what this $h^{- 1}$ can answer

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 1, Function Concepts.

Free-response practice · Question 1 style

Graphing calculator required. The function $f$ is given by $f (x) = \frac{2 x + 1}{x - 3}$ for all real numbers except $x = 3$ , and $f$ is invertible on that domain. (i) Given $f (4) = 9$ , identify $f^{- 1} (9)$ and explain how you know. (ii) Determine an expression for $f^{- 1} (x)$ , showing the steps. (iii) The range of $f$ is all real numbers except 2. Identify the domain of $f^{- 1}$ and give a reason. (iv) Use your expression to find $f^{- 1} (5)$ , and show that $f (f^{- 1} (5)) = 5$ .

Your answer stays on this page. It is never sent or saved.

Model answer

(i) $f^{- 1} (9) = 4$ . The pair $(4, 9)$ is on $f$ , so the reversed pair $(9, 4)$ is on $f^{- 1}$ . (ii) Write $y = \frac{2 x + 1}{x - 3}$ and swap: $x = \frac{2 y + 1}{y - 3}$ . Multiply by $y - 3$ : $x y - 3 x = 2 y + 1$ . Collect the $y$ terms: $x y - 2 y = 3 x + 1$ , so $y (x - 2) = 3 x + 1$ and $f^{- 1} (x) = \frac{3 x + 1}{x - 2}$ . (iii) All real numbers except 2, because the domain of $f^{- 1}$ is the range of $f$ . (iv) $f^{- 1} (5) = \frac{16}{3}$ , and $f (\frac{16}{3}) = \frac{32 / 3 + 1}{16 / 3 - 3} = \frac{35 / 3}{7 / 3} = 5$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Gives $f^{- 1} (9) = 4$ , reasoning that the pair $(4, 9)$ on $f$ reverses to $(9, 4)$ .
Swaps $x$ and $y$ and solves for $y$ , reaching $f^{- 1} (x) = \frac{3 x + 1}{x - 2}$ .
Gives the domain of $f^{- 1}$ as all real numbers except 2, because it is the range of $f$ .
Finds $f^{- 1} (5) = \frac{16}{3}$ and shows that $f (\frac{16}{3}) = 5$ .

Part 6 Check yourself

An inverse from a table

The three case questions use the same table. Answer them from it.

The table gives every value of a function $f$ . Its domain is just these five inputs, and no output repeats, so $f$ is invertible.

Values of $f$
$x$	$f (x)$
−3	11
−1	6
0	4
2	−1
5	−3

Case question 1 · Calculate

No calculator. What is $f^{- 1} (- 1)$ ?

Your answer

Case question 2 · Multiple choice

No calculator. What is the value of $f^{- 1} (f (- 1) - 2)$ ?

Case question 3 · Multiple choice

No calculator. Which list gives the domain of $f^{- 1}$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student works with $p (x) = 4 x - 8$ . Tap the line with the mistake.

Check yourself · Match

No calculator. Tap a function, then its inverse.

Check yourself · Decide and justify

No calculator. Let $p (x) = 2 x + 6$ . A student claims that $q (x) = \frac{x}{2} - 6$ is the inverse of $p$ . Is the student right?

Which reason supports your answer?

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 2.3 Exponential Functions · Calculate

No calculator. A function has $g (0) = 7$ , $g (1) = 13$ , $g (2) = 31$ and $g (3) = 85$ . These outputs are not proportional, but subtracting the same constant from every output makes them proportional. What is the common ratio of those new outputs?

Your answer

From 1.9 Rational Functions and Vertical Asymptotes · Decide and justify

No calculator. Let $r (x) = \frac{(x - 1)^{2} (x + 3)}{(x - 1)^{3}}$ . Does the graph of $r$ have a vertical asymptote at $x = 1$ ?

Which reason supports your answer?

In three lines

If $f (a) = b$ , then $f^{- 1} (b) = a$ : a table swaps every pair, a graph reflects over $y = x$ , and the domain and range trade places.
A function is invertible on a domain where each output comes from one input; restricting the domain can make it so, and a context can limit the inverse further.
To find $f^{- 1}$ , swap $x$ and $y$ and solve for $y$ , undoing the last step first, then check that $f (f^{- 1} (x)) = f^{- 1} (f (x)) = x$ .

Vocabulary

Inverse function: The function $f^{- 1}$ that maps each output of $f$ , on its invertible domain, back to its input: if $f (a) = b$ , then $f^{- 1} (b) = a$ .
Invertible: A function is invertible on a domain when each of its output values there comes from a unique input value.
Input-output pair: A pair $(a, b)$ with $f (a) = b$ ; the inverse function has the reversed pair $(b, a)$ .
Identity function: The function $h (x) = x$ ; a function composed with its inverse gives it, and its graph is the line a graph reflects over to give its inverse's graph.
Invertible domain: A domain on which a function is invertible, often found by restricting the domain to a piece where no output repeats.

Course alignment, for teachers

AP Precalculus topic 2.8, Unit 2: Exponential and Logarithmic Functions.

Skill 1.A: Solve equations and inequalities represented analytically, with and without technology.
Skill 2.B: Construct equivalent graphical, numerical, analytical, and verbal representations of functions that are useful in a given mathematical or applied context, with and without technology.

This lesson maps all 7 essential knowledge statements of the topic to the part that teaches each one: 7 are checked by a scored question. The map is in the lesson file.