Inverse Functions

A function takes an input and gives you an output. Its inverse runs the process backwards: give it the output, and it hands you back the original input. Inverses let you “undo” a formula to answer questions the other way around, and later on they’re how logarithms are built from exponential functions.

Key ideas

What an inverse does

If $f$ sends $a$ to $b$, then the inverse $f^{-1}$ sends $b$ back to $a$:

$$f(a) = b \quad \Longleftrightarrow \quad f^{-1}(b) = a$$

For example, if $f(2) = 9$, then $f^{-1}(9) = 2$.

Careful: the $-1$ in $f^{-1}$ is not an exponent. $f^{-1}(x)$ means “the inverse of $f$”, not $\dfrac{1}{f(x)}$.

Swap the inputs and outputs

To find the inverse of a relation, swap the coordinates in every ordered pair: the point $(a, b)$ becomes $(b, a)$.

Because inputs and outputs trade places, the domain and range trade places too:

• the domain of $f^{-1}$ is the range of $f$
• the range of $f^{-1}$ is the domain of $f$

Finding the inverse from an equation

1. Write $y$ in place of $f(x)$.
2. Swap $x$ and $y$.
3. Solve for $y$.
4. Write $f^{-1}(x)$ in place of $y$.

Graphing: reflect in the line y = x

Swapping the coordinates of a point is the same as reflecting it in the line $y = x$. So the graph of $y = f^{-1}(x)$ is the mirror image of the graph of $y = f(x)$ in that line. To sketch it, pick a few key points on $y = f(x)$, swap their coordinates, plot the new points, and join them. Example 4 shows this.

Is the inverse a function?

Every function has an inverse, but the inverse is only a function if $f$ never gives the same output for two different inputs. Check with the horizontal line test: if some horizontal line crosses the graph of $f$ more than once, the inverse is not a function.

For example, $f(x) = x^2$ fails the test: $f(-2) = 4$ and $f(2) = 4$, so the inverse would have to send $4$ to both $-2$ and $2$. The fix is to restrict the domain of $f$, keeping only a part of the graph that passes the test. If you keep $x \ge 0$, the inverse is $f^{-1}(x) = \sqrt{x}$.

Checking your answer

• Quick check: pick a number, put it through $f$, then put the result through $f^{-1}$. You should get your number back.
• Full check: show that both of these are true for every $x$ in the right domain:

$$f\big(f^{-1}(x)\big) = x \qquad \text{and} \qquad f^{-1}\big(f(x)\big) = x$$

Worked examples

Example 1: Ordered pairs

Find the inverse of $f = \{(-2, 7), (0, 3), (1, 1), (4, -5)\}$ and state its domain and range. Is $f^{-1}$ a function?

Solution. Swap the coordinates in each pair:

$$f^{-1} = \{(7, -2), (3, 0), (1, 1), (-5, 4)\}$$

The domain of $f^{-1}$ is the range of $f$: $\{-5, 1, 3, 7\}$. The range of $f^{-1}$ is the domain of $f$: $\{-2, 0, 1, 4\}$.

$f^{-1}$ is a function, because each of its inputs ($7$, $3$, $1$ and $-5$) appears only once.

Notice that $(1, 1)$ didn’t change. It sits on the line $y = x$, so reflecting it leaves it where it is.

Example 2: A linear function

Find $f^{-1}(x)$ for $f(x) = 2x + 6$, then check your answer.

Solution.

1. Write $y$ for $f(x)$: $\quad y = 2x + 6$
2. Swap $x$ and $y$: $\quad x = 2y + 6$
3. Solve for $y$: $\quad x - 6 = 2y$, so $y = \dfrac{x - 6}{2}$
4. Write $f^{-1}(x)$ in place of $y$:

$$f^{-1}(x) = \frac{x - 6}{2} \quad \left(\text{or } \frac{x}{2} - 3\right)$$

Quick check: $f(1) = 2(1) + 6 = 8$, and $f^{-1}(8) = \dfrac{8 - 6}{2} = 1$. We got $1$ back.

Full check:

$$f\big(f^{-1}(x)\big) = 2\left(\frac{x - 6}{2}\right) + 6 = (x - 6) + 6 = x$$ $$f^{-1}\big(f(x)\big) = \frac{(2x + 6) - 6}{2} = \frac{2x}{2} = x$$

Look at the order of the steps. $f$ multiplies by $2$ and then adds $6$. The inverse undoes them in reverse order: subtract $6$ first, then divide by $2$. It’s like socks and shoes: you put socks on first, but you take shoes off first.

Example 3: Temperature and altitude

On a day when it’s $15\,^\circ\text{C}$ at sea level, a simple model for the air temperature $T$ (in $^\circ\text{C}$) at a height of $h$ kilometres is

$$T = 15 - 6.5h$$

Find the inverse, and use it to find the height where the temperature is $-11\,^\circ\text{C}$.

Solution. Here the letters have meanings: $T$ is always a temperature and $h$ is always a height. So don’t swap them. Just solve for the other variable, which gives the same inverse:

$$\begin{aligned} T &= 15 - 6.5h \\ 6.5h &= 15 - T \\ h &= \frac{15 - T}{6.5} \end{aligned}$$

This inverse takes a temperature and gives the height. For $T = -11$:

$$h = \frac{15 - (-11)}{6.5} = \frac{26}{6.5} = 4$$

The temperature is $-11\,^\circ\text{C}$ at a height of $4$ km.

Check: $T = 15 - 6.5(4) = 15 - 26 = -11$. ✓

Example 4: A quadratic with a restricted domain

Let $f(x) = (x - 3)^2 - 4$ for $x \ge 3$. Find $f^{-1}(x)$, state the domain and range of both $f$ and $f^{-1}$, and sketch both graphs.

Solution.

Domain and range of $f$. The domain is given: $\{x \in \mathbb{R} \mid x \ge 3\}$. The graph is the right half of a parabola that opens up from its vertex $(3, -4)$, so the range is $\{y \in \mathbb{R} \mid y \ge -4\}$. Keeping only the right half is what makes $f$ pass the horizontal line test.

Find the inverse.

$$\begin{aligned} y &= (x - 3)^2 - 4 \\ x &= (y - 3)^2 - 4 && \text{swap } x \text{ and } y \\ x + 4 &= (y - 3)^2 \\ y - 3 &= \pm\sqrt{x + 4} \\ y &= 3 \pm \sqrt{x + 4} \end{aligned}$$

Which sign? The range of $f^{-1}$ has to be the domain of $f$, which is $y \ge 3$. The $+$ sign gives values that are at least $3$; the $-$ sign gives values that are at most $3$. So we keep the $+$:

$$f^{-1}(x) = 3 + \sqrt{x + 4}$$

Domain and range of $f^{-1}$. Swap them from $f$: the domain is $\{x \in \mathbb{R} \mid x \ge -4\}$ and the range is $\{y \in \mathbb{R} \mid y \ge 3\}$.

Quick check: $f(5) = (5 - 3)^2 - 4 = 0$, and $f^{-1}(0) = 3 + \sqrt{4} = 5$. ✓

Sketch. Take key points on $y = f(x)$ and swap their coordinates:

On $y = f(x)$	$(3, -4)$	$(4, -3)$	$(5, 0)$	$(6, 5)$
On $y = f^{-1}(x)$	$(-4, 3)$	$(-3, 4)$	$(0, 5)$	$(5, 6)$

Plot the swapped points and join them with a smooth curve. The two graphs are reflections of each other in the line $y = x$.

$y = f(x)$ (blue) and $y = f^{-1}(x)$ (orange) are mirror images in the dashed line $y = x$.

Common mistakes

Reading $f^{-1}(x)$ as $\dfrac{1}{f(x)}$. The $-1$ means “inverse”, not “reciprocal”. For $f(x) = 2x + 6$, the inverse is $f^{-1}(x) = \dfrac{x - 6}{2}$, but $\dfrac{1}{f(x)} = \dfrac{1}{2x + 6}$. Those are completely different functions.

Undoing the steps in the wrong order. For $f(x) = 2x + 6$, a common wrong answer is $\dfrac{x}{2} - 6$: it undoes “times $2$” and “plus $6$” in the same order instead of the reverse order. A quick check catches it: $f(1) = 8$, but $\dfrac{8}{2} - 6 = -2$, not $1$.

Leaving the $\pm$ in (or picking the wrong sign). $y = 3 \pm \sqrt{x + 4}$ isn’t a function, because it gives two outputs. Use the domain of the original function to decide: the range of $f^{-1}$ must match the domain of $f$.

Forgetting the inverse’s domain. The domain of $f^{-1}$ is the range of $f$, not automatically “all real numbers”. In Example 4, $f^{-1}(x) = 3 + \sqrt{x + 4}$ only works for $x \ge -4$.

Assuming every inverse is a function. Use the horizontal line test first. If $f$ fails it, restrict the domain before you find the inverse.

Swapping letters that have meanings. In a word problem like Example 3, solve for the other variable instead of swapping. Otherwise $T$ ends up standing for a height, which gets confusing fast.

Practice

1. (Warm-up)

• (a) If $f(3) = 10$, find $f^{-1}(10)$.
• (b) If $f^{-1}(4) = -1$, find $f(-1)$.
• (c) The point $(6, -1)$ is on the graph of $y = f(x)$. Which point must be on the graph of $y = f^{-1}(x)$?

Solution

(a) $f$ sends $3$ to $10$, so $f^{-1}$ sends $10$ back to $3$: $f^{-1}(10) = 3$.

(b) $f^{-1}$ sends $4$ to $-1$, so $f$ sends $-1$ to $4$: $f(-1) = 4$.

(c) Swap the coordinates: $(-1, 6)$.

2. (Warm-up) Find the inverse of $g = \{(-3, 0), (-1, 2), (1, 6), (4, 8)\}$, and state its domain and range.

Solution

Swap the coordinates in each pair:

$$g^{-1} = \{(0, -3), (2, -1), (6, 1), (8, 4)\}$$

Domain of $g^{-1}$: $\{0, 2, 6, 8\}$. Range of $g^{-1}$: $\{-3, -1, 1, 4\}$.

3. (Core) Find $f^{-1}(x)$ for $f(x) = \dfrac{x}{4} + 7$.

Solution

Swap $x$ and $y$: $x = \dfrac{y}{4} + 7$. Then $x - 7 = \dfrac{y}{4}$, so $y = 4(x - 7)$.

$$f^{-1}(x) = 4x - 28$$

Check: $f(8) = 2 + 7 = 9$, and $f^{-1}(9) = 36 - 28 = 8$. ✓

4. (Core) Find $f^{-1}(x)$ for $f(x) = \dfrac{2x - 1}{3}$.

Solution

Swap $x$ and $y$: $x = \dfrac{2y - 1}{3}$.

Multiply both sides by $3$: $3x = 2y - 1$. Add $1$: $3x + 1 = 2y$. Divide by $2$:

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

Check: $f(2) = \dfrac{3}{3} = 1$, and $f^{-1}(1) = \dfrac{4}{2} = 2$. ✓

5. (Core) Show that $g(x) = 5 - 2x$ and $h(x) = \dfrac{5 - x}{2}$ are inverses of each other.

Solution

Check both compositions:

$$g\big(h(x)\big) = 5 - 2\left(\frac{5 - x}{2}\right) = 5 - (5 - x) = x$$$$h\big(g(x)\big) = \frac{5 - (5 - 2x)}{2} = \frac{2x}{2} = x$$

Both give $x$, so $g$ and $h$ are inverses.

6. (Core) Let $f(x) = (x + 1)^2$.

• (a) Explain why the inverse of $f$ is not a function.
• (b) Restrict the domain of $f$ to $x \ge -1$. Find $f^{-1}(x)$, and state its domain and range.

Solution

(a) Two different inputs give the same output: $f(1) = 4$ and $f(-3) = 4$. So the inverse would send $4$ to both $1$ and $-3$. (On the graph, the horizontal line $y = 4$ crosses the parabola twice.)

(b) Swap $x$ and $y$: $x = (y + 1)^2$, so $y + 1 = \pm\sqrt{x}$. The range of $f^{-1}$ must be the restricted domain of $f$, $y \ge -1$, so take the $+$ sign:

$$f^{-1}(x) = \sqrt{x} - 1$$

Domain of $f^{-1}$: $\{x \in \mathbb{R} \mid x \ge 0\}$. Range of $f^{-1}$: $\{y \in \mathbb{R} \mid y \ge -1\}$.

Check: $f(2) = 9$, and $f^{-1}(9) = 3 - 1 = 2$. ✓

7. (Core) A taxi charges a $3.50 flat fee plus $1.75 per kilometre, so a ride of $d$ kilometres costs $C = 1.75d + 3.50$ dollars. Find the inverse, and use it to find how far you can ride for $24.50.

Solution

$C$ and $d$ have meanings, so solve for $d$ instead of swapping:

$$d = \frac{C - 3.50}{1.75}$$

For $C = 24.50$:

$$d = \frac{24.50 - 3.50}{1.75} = \frac{21}{1.75} = 12$$

You can ride $12$ km. Check: $1.75(12) + 3.50 = 21 + 3.50 = 24.50$. ✓

8. (Challenge) Find $f^{-1}(x)$ for $f(x) = \dfrac{x + 3}{x - 2}$, where $x \ne 2$. What is the domain of $f^{-1}$?

Solution

Swap $x$ and $y$, then collect the $y$ terms on one side:

$$\begin{aligned} x &= \frac{y + 3}{y - 2} \\ x(y - 2) &= y + 3 \\ xy - 2x &= y + 3 \\ xy - y &= 2x + 3 \\ y(x - 1) &= 2x + 3 \\ y &= \frac{2x + 3}{x - 1} \end{aligned}$$$$f^{-1}(x) = \frac{2x + 3}{x - 1}, \qquad \text{domain } \{x \in \mathbb{R} \mid x \ne 1\}$$

(This also tells you that $f$ never outputs $1$.)

Check: $f(3) = \dfrac{6}{1} = 6$, and $f^{-1}(6) = \dfrac{15}{5} = 3$. ✓

9. (Challenge) Show that $f(x) = 6 - x$ is its own inverse. Then use its graph to explain why that makes sense.

Solution

Swap $x$ and $y$: $x = 6 - y$, so $y = 6 - x$. That means $f^{-1}(x) = 6 - x$, which is the same as $f(x)$.

The graph of $y = 6 - x$ is a line with slope $-1$ through $(0, 6)$ and $(6, 0)$, so it’s perpendicular to $y = x$. Reflect any point $(a, 6 - a)$ on it in $y = x$ and you get $(6 - a, a)$. That point is on the same line, because $6 - (6 - a) = a$. So reflecting the graph in $y = x$ gives back the same graph, and the inverse is the function itself.