Inverse Functions — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

A weather app shows

20

degrees Celsius, but you think in Fahrenheit. The app's rule turns Fahrenheit into Celsius; you need the rule that runs the other way. A function that undoes another one, sending every output back to the input it came from, is an inverse function.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Let $I(x) = x$ and $f(x) = 2x + 5$ . What is $f(I(3))$ ?

$11$
$3$
$8$
$16$

Show the answer

$I(3) = 3$ , so $f(I(3)) = f(3) = 11$ .

Correct: $11$ :
$3$ :
$8$ :
$16$ :

2. Solve $y = 3x - 6$ for $x$ .

$x = \frac{y + 6}{3}$
$x = 3y - 6$
$x = \frac{y - 6}{3}$
$x = 3y + 6$

Show the answer

Add $6$ , then divide by $3$ .

Correct: $x = \frac{y + 6}{3}$ :
$x = 3y - 6$ :
$x = \frac{y - 6}{3}$ :
$x = 3y + 6$ :

Part 4 · See it

See it first

Graph of f(x) = x squared over 4 for x at least 0 and its inverse 2 times the square root of x, mirror images across the dashed line y = x; (2, 1) on f matches (1, 2) on the inverse. — $f(x) = \frac{x^2}{4}$ for $x \ge 0$ and $f^{-1}(x) = 2\sqrt{x}$ reflect across $y = x$ : the point $(2, 1)$ on $f$ becomes $(1, 2)$ on $f^{-1}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f$ sends the input $a$ to the output $b$ $f^{-1}$ sends $b$ back to $a$ : the point $(a, b)$ becomes $(b, a)$
Swap every input and outputthe graph reflects across the line $y = x$ , and the domain and range trade places
Two inputs share an output, such as $h(1) = h(3)$ the inverse would need two outputs, so $h$ is not invertible
Restrict the domain to a piece where $f$ only increases or only decreasesthat piece has an inverse
Compose a function with its inverse $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ : the identity function

Part 6 · Key ideas

Key ideas

$f^{-1}(b) = a$ means $f(a) = b$ . The $-1$ names the inverse function; $f^{-1}(x)$ is not $\frac{1}{f(x)}$ .
A function is invertible when each output comes from exactly one input (it is one-to-one).
Justify with specific values: " $h(1) = h(3) = 1$ , so $h$ has no inverse", never a bare rule name.
To find a formula, swap $x$ and $y$ in $y = f(x)$ and solve for $y$ .
The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .

Part 7 · Misconception

A common mistake

The wrong idea: $f^{-1}(x)$ means $\frac{1}{f(x)}$ , so the inverse of $f(x) = x - 5$ is $\frac{1}{x - 5}$ .

What actually happens: $f^{-1}$ undoes $f$ : it adds the $5$ back, so $f^{-1}(x) = x + 5$ . Check with composition: $f(f^{-1}(x))$ $= (x + 5) - 5$ $= x$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ and $h$

Values of the functions f and h at x = 0, 1, 2, 3, 4 and 5
$x$	0	1	2	3	4	5
$f(x)$	-3	-1	2	5	7	12
$h(x)$	4	1	0	1	4	9

1. The function $f$ is invertible. Using the table, what is $f^{-1}(5)$ ?

$3$
$12$
$\frac{1}{12}$
$\frac{1}{5}$

Show the answer

$f^{-1}(5)$ is the input of $f$ whose output is $5$ . The table has $f(3) = 5$ , so $f^{-1}(5) = 3$ .

Correct: $3$ : $f(3) = 5$ , so the inverse sends $5$ back to $3$ .
$12$ : This is $f(5) = 12$ . The inverse runs the other way: from the output $5$ back to an input.
$\frac{1}{12}$ : This is $\frac{1}{f(5)}$ . The $-1$ in $f^{-1}$ names the inverse function, not a reciprocal.
$\frac{1}{5}$ : This is $5^{-1}$ . $f^{-1}(5)$ is an input of $f$ , not the reciprocal of $5$ .

2. Which statement correctly decides whether $h$ is invertible, using values from the table?

No: $h(1) = 1$ and $h(3) = 1$ , so two inputs share the output $1$
Yes: each input of $h$ has exactly one output
No: $h(2) = 0$ , and an output of $0$ has no inverse
Yes: $h(5) = 9$ is the largest output in the table

Show the answer

An invertible function gives each output for one input. $h(1) = h(3) = 1$ , so $h^{-1}(1)$ would have to be both $1$ and $3$ : $h$ has no inverse.

Correct: No: $h(1) = 1$ and $h(3) = 1$ , so two inputs share the output $1$ : Two inputs, $1$ and $3$ , share the output $1$ , named with values: this earns the point.
Yes: each input of $h$ has exactly one output: That makes $h$ a function. To be invertible, each output must also come from one input.
No: $h(2) = 0$ , and an output of $0$ has no inverse: An output of $0$ is no problem; $h(2) = 0$ comes from one input.
Yes: $h(5) = 9$ is the largest output in the table: A largest output says nothing about repeated outputs; $h(1) = h(3)$ does.

3. Using the table, which value can be an input of $f^{-1}$ ?

$7$
$4$
$0$
$1$

Show the answer

The inputs of $f^{-1}$ are the outputs of $f$ : $-3, -1, 2, 5, 7, 12$ . Of these choices, only $7$ is one of them.

Correct: $7$ : $7 = f(4)$ is an output of $f$ , so $f^{-1}(7) = 4$ .
$4$ : $4$ is an input of $f$ , but no input of $f$ gives the output $4$ in the table.
$0$ : $0$ is an input of $f$ ; it is not an output of $f$ , so $f^{-1}(0)$ is not in the table.
$1$ : $1$ is an input of $f$ ; $f(1) = -1$ , so $1$ is not an output in the table.

Graph

The graph of $g$

The graph shows

g(x) = \frac{x^3}{8} + 1

, which is increasing.

4. Use the graph to find $g^{-1}(0)$ .

$-2$
$1$
$0$
$2$

Show the answer

$g^{-1}(0)$ is the input of $g$ whose output is $0$ . The graph passes through $(-2, 0)$ , so $g^{-1}(0) = -2$ .

Correct: $-2$ : $g(-2) = 0$ , so the inverse sends $0$ back to $-2$ .
$1$ : This is $g(0) = 1$ , the forward direction.
$0$ : $0$ is the output we start from; the answer is the input that produces it.
$2$ : $g(2) = 2$ ; that point does not have output $0$ .

5. Which point is on the graph of $y = g^{-1}(x)$ ?

$(1, 0)$
$(0, 1)$
$(-2, 0)$
$(2, 0)$

Show the answer

$(0, 1)$ is on the graph of $g$ , so $(1, 0)$ is on the graph of $g^{-1}$ : the coordinates trade places.

Correct: $(1, 0)$ : $g(0) = 1$ , so $g^{-1}(1) = 0$ .
$(0, 1)$ : This point is on the graph of $g$ itself.
$(-2, 0)$ : This point is on $g$ ; its swapped partner $(0, -2)$ is on the inverse.
$(2, 0)$ : $g^{-1}(2) = 2$ , so the point is $(2, 2)$ , not $(2, 0)$ .

6. Let $f(x) = \frac{x - 5}{3}$ . Write $f^{-1}(x)$ .

Type an expression for $f^{-1}(x) =$ .

Show the answer

$f$ subtracts $5$ , then divides by $3$ . Undo in reverse order: multiply by $3$ , then add $5$ . So $f^{-1}(x) = 3x + 5$ .

Answer: $f^{-1}(x) =$ $3x + 5$

7. Let $f(x) = x^2 - 4x$ , defined for every real number. Which statement shows that $f$ has no inverse?

$f(0) = 0$ and $f(4) = 0$
$f(0) = 0$ and $f(2) = -4$
$f(x)$ is a function, so each input has one output
$f(2) = -4$ is a negative output

Show the answer

Two different inputs, $0$ and $4$ , give the same output $0$ , so an inverse could not decide which input to return.

Correct: $f(0) = 0$ and $f(4) = 0$ : Two inputs with one shared output: specific values that prove it.
$f(0) = 0$ and $f(2) = -4$ : Different inputs with different outputs show nothing about repeats.
$f(x)$ is a function, so each input has one output: That is true of every function; it does not decide invertibility.
$f(2) = -4$ is a negative output: A negative output is allowed; repeated outputs are what matter.

Part 9 · Summary

Summary

An inverse function undoes a function:

f^{-1}(b) = a

exactly when

f(a) = b

. It exists when no output repeats, which you justify with specific values, or after restricting the domain. Find it from a table by swapping rows, from a graph by reflecting across

y = x

, and from a formula by swapping

x

and

y

and solving. Composing a function with its inverse gives the identity.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff and hh

The graph of gg

Summary

What comes next

Connections

Values of $f$ and $h$

The graph of $g$