# 7.1 Inverse Functions

We say that two functions $f$ and $g$ are inverses if $g(f(x))=x$ for all $x$ in the domain of $f$ and $f(g(x))=x$ for all $x$ in the domain of $g$. A function can only have an inverse if it is one-to-one, i.e. if we never have $f(x_{1})=f(x_{2})$ for different elements $x_{1}$ and $x_{2}$ of the domain. This is equivalent to saying that the graph of the function passes the horizontal line test. The inverse of $f$ is denoted $f^{-1}$, which should not be confused with the function $1/f(x)$.

###### Key Idea 7.1.1 Inverse Functions

For a one-to–one function $f$,

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

• $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$.

• $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$.

• The graph of $y=f^{-1}(x)$ is the reflection across $y=x$ of the graph of $y=f(x)$.

• $y=f^{-1}(x)$ if and only if $f(y)=x$ and $y$ is in the domain of $f$.

To determine whether or not $f$ and $g$ are inverses for each other, we check to see whether or not $g(f(x))=x$ for all $x$ in the domain of $f$,and $f(g(x))=x$ for all $x$ in the domain of $g$.

margin: Figure 7.1.1: A function $f$ along with its inverse $f\mkern 1.35mu ^{-1}$. (Note how it does not matter which function we refer to as $f$; the other is $f\mkern 1.35mu ^{-1}$.) Λ
###### Example 7.1.1 Verifying Inverses

Determine whether or not the following pairs of functions are inverses:

1. (a)

$f(x)=3x+1$; $g(x)=\dfrac{x-1}{3}$

2. (b)

$f(x)=x^{3}+1$; $g(x)=x^{1/3}-1$

Solution

1. (a)

To check the composition we plug $f(x)$ in for $x$ in the definition of $g$ as follows:

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

So $g(f(x))=x$ for all $x$ in the domain of $f$. Likewise, you can check that $f(g(x))=x$ for all $x$ in the domain of $g$, so $f$ and $g$ are inverses.

2. (b)

If we try to proceed as before, we find that:

 $g(f(x))=(f(x))^{1/3}-1=(x^{3}+1)^{1/3}-1$

This doesn’t seem to be the same as the identity function $x$. To verify this, we find a number $a$ in the domain of $f$ and show that $g(f(a))\neq a$ for that value. Let’s try $x=1$. Since $f(1)=1^{3}+1=2$, we find that $g(f(1))=g(2)=2^{1/3}-1\approx 0.26$. Since $g(f(1))\neq 1$, these functions are not inverses.

## Functions that are not one-to-one.

margin: Figure 7.1.2: The function $f(x)=x^{2}$ is not one-to-one. Λ

Unfortunately, not every function we would like to find an inverse for is one-to-one. For example, the function $f(x)=x^{2}$ is not one-to-one because $f(-2)=f(2)=4$. If $f^{-1}$ is an inverse for $f$, then $f^{-1}(f(-2))=-2$ implies that $f^{-1}(4)=-2$. On the other hand, $f^{-1}(f(2))=2$, so $f^{-1}(4)=2$. We cannot have it both ways if $f^{-1}$ is a function, so no such inverse exists. We can find a partial solution to this dilemma by restricting the domain of $f$. There are many possible choices, but traditionally we restrict the domain to the interval $[0,\infty)$. The function $f^{-1}(x)=\sqrt{x}$ is now an inverse for this restricted version of $f$.

## The inverse sine function

We consider the function $f(x)=\sin x$, which is not one-to-one. A piece of the graph of $f$ is in Figure 7.1.3(a). In order to find an appropriate restriction of the domain of $f$, we look for consecutive critical points where $f$ takes on its minimum and maximum values. In this case, we use the interval $[-\pi/2,\pi/2]$. We define the inverse of $f$ on this restricted range by $y=\sin^{-1}x$ if and only if $\sin y=x$ and $-\pi/2\leq y\leq\pi/2$. The graph is a reflection of the graph of $g$ across the line $y=x$, as seen in Figure 7.1.3(b).

(a) (b) Figure 7.1.3: (a) A portion of $y=\sin x$. (b) A one-to-one portion of $y=\sin x$ along with $y=\sin^{-1}x$.

## The inverse tangent function

Next we consider the function $f(x)=\tan x$, which is also not one-to-one. A piece of the graph of $f$ is given in Figure 7.1.4(a). In order to find an interval on which the function is one-to-one and on which the function takes on all values in the range, we use an interval between consecutive vertical asymptotes. Traditionally, the interval $(-\pi/2,\pi/2)$ is chosen. Note that we choose the open interval in this case because the function $f$ is not defined at the endpoints. So we define $y=\tan^{-1}x$ if and only if $\tan y=x$ and $-\pi/2. The graph of $y=\tan^{-1}x$ is shown in Figure 7.1.4(b). Also note that the vertical asymptotes of the original function are reflected to become horizontal asymptotes of the inverse function.

(a) (b) Figure 7.1.4: (a) A portion of $y=\tan x$. (b) A one-to-one portion of $y=\tan x$ along with $y=\tan^{-1}$.

The other inverse trigonometric functions are defined in a similar fashion. The resulting domains and ranges are summarized in Figure 7.1.5.

Function Restricted Domain Range Inverse Function Domain Range $\sin x$ $[-\pi/2,\pi/2]$ $[-1,1]$ $\sin^{-1}x$ $[-1,1]$ $[-\pi/2,\pi/2]$ $\cos x$ $[0,\pi]$ $[-1,1]$ $\cos^{-1}x$ $[-1,1]$ $[0,\pi]$ $\tan x$ $(-\pi/2,\pi/2)$ $(-\infty,\infty)$ $\tan^{-1}x$ $(-\infty,\infty)$ $(-\pi/2,\pi/2)$ $\csc x$ $[-\pi/2,0)\cup(0,\pi/2]$ $(-\infty,-1]\cup[1,\infty)$ $\csc^{-1}x$ $(-\infty,-1]\cup[1,\infty)$ $[-\pi/2,0)\cup(0,\pi/2]$ $\sec x$ $[0,\pi/2)\cup(\pi/2,\pi]$ $(-\infty,-1]\cup[1,\infty)$ $\sec^{-1}x$ $(-\infty,-1]\cup[1,\infty)$ $[0,\pi/2)\cup(\pi/2,\pi]$ $\cot x$ $(0,\pi)$ $(-\infty,\infty)$ $\cot^{-1}x$ $(-\infty,\infty)$ $(0,\pi)$ Figure 7.1.5: Domains and ranges of the trigonometric and inverse trigonometric functions.
###### Example 7.1.2 Evaluating Inverse Trigonometric Functions

Find exact values for the following: margin: Sometimes, $\arcsin$ is used instead of $\sin^{-1}$. Similar “arc” functions are used for the other inverse trigonometric functions as well. Λ
(a) $\tan^{-1}(1)$ (b) $\cos(\sin^{-1}(\sqrt{3}/2))$ (c) $\sin^{-1}(\sin(7\pi/6))$ (d) $\tan(\cos^{-1}(11/15))$

Solution

1. (a)

$\tan^{-1}(1)=\pi/4$

2. (b)

$\cos(\sin^{-1}(\sqrt{3}/2))=\cos(\pi/3)=1/2$

3. (c)

Since $7\pi/6$ is not in the range of the inverse sine function, we should be careful with this one.

 $\sin^{-1}(\sin(7\pi/6))=\sin^{-1}(-1/2)=-\pi/6.$
4. (d)

Since we don’t know the value of $\cos^{-1}(11/15)$, we let $\theta$ stand for this value. We know that $\theta$ is an angle between $0$ and $\pi$ and that $\cos(\theta)=11/15$. In Figure 7.1.6, we use this information to construct a right triangle with angle $\theta$, where the adjacent side over the hypotenuse must equal 11/15. Applying the Pythagorean Theorem we find that margin: Figure 7.1.6: A right triangle for the situation in Example 7.1.2 (4). Λ

 $y=\sqrt{15^{2}-11^{2}}=\sqrt{104}=2\sqrt{26}.$

Finally, we have:

 $\tan(\cos^{-1}(11/15))=\tan(\theta)=\frac{2\sqrt{26}}{11}.$

## Exercises 7.1

### Terms and Concepts

1. 1.

T/F: Every function has an inverse.

2. 2.

In your own words explain what it means for a function to be “one to one.”

3. 3.

If $(1,10)$ lies on the graph of $y=f(x)$, what can be said about the graph of $y=f^{-1}(x)$?

4. 4.

If a function doesn’t have an inverse, what can we do to help it have an inverse?

### Problems

In Exercises 5–6., given the graph of $f$, sketch the graph of $f^{-1}$.

1. 5.
2. 6.

In Exercises 7–10., verify that the given functions are inverses.

1. 7.

$f(x)=2x+6$ and $g(x)=\frac{1}{2}x-3$

2. 8.

$f(x)=x^{2}+6x+11$, $x\geq-3$ and $g(x)=\sqrt{x-2}-3$, $x\geq 2$

3. 9.

$\displaystyle f(x)=\frac{3}{x-5}$, $x\neq 5$ and $\displaystyle g(x)=\frac{3+5x}{x}$, $x\neq 0$

4. 10.

$\displaystyle f(x)=\frac{x+1}{x-1}$, $x\neq 1$ and $g(x)=f(x)$

In Exercises 11–14., find a restriction of the domain of the given function on which the function will have an inverse.

1. 11.

$f(x)=\sqrt{16-x^{2}}$

2. 12.

$g(x)=\sqrt{x^{2}-16}$

3. 13.

$r(t)=t^{2}-6t+9$

4. 14.

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

In Exercises 15–18., find the inverse of the given function.

1. 15.

$f(x)=\dfrac{x+1}{x-2}$

2. 16.

$f(x)=x^{2}+4$

3. 17.

$f(x)=e^{x+3}-2$

4. 18.

$f(x)=\ln(x-5)+1$

In Exercises 19–28., find the exact value.

1. 19.

$\tan^{-1}(0)$

2. 20.

$\tan^{-1}(\tan(\pi/7))$

3. 21.

$\cos(\cos^{-1}(-1/5))$

4. 22.

$\sin^{-1}(\sin(8\pi/3))$

5. 23.

$\sin(\tan^{-1}(1))$

6. 24.

$\sec(\sin^{-1}(-3/5))$

7. 25.

$\cos(\tan^{-1}(3/7))$

8. 26.

$\sin^{-1}(-\sqrt{3}/2)$

9. 27.

$\cos^{-1}(-\sqrt{2}/2)$

10. 28.

$\cos^{-1}(\cos(8\pi/7))$

In Exercises 29–32., simplify the expression.

1. 29.

$\sin\Bigl{(}\tan^{-1}\frac{x}{\sqrt{4-x^{2}}}\Bigr{)}$

2. 30.

$\tan\Bigl{(}\sin^{-1}\frac{x}{\sqrt{x^{2}+4}}\Bigr{)}$

3. 31.

$\cos\Bigl{(}\sin^{-1}\frac{5}{\sqrt{x^{2}+25}}\Bigr{)}$

4. 32.

$\cot\Bigl{(}\cos^{-1}\frac{3}{\sqrt{x}}\Bigr{)}$

1. 33.

Show that for any $x$ in the domain of $\sec^{-1}$ we have $\sec^{-1}x=\cos^{-1}\frac{1}{x}$.

2. 34.
Show that for $\left\lvert x\right\rvert\leq 1$ we have $\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$. Hint: Recall the cofunction identity $\cos\theta=\sin(\frac{\pi}{2}-\theta)$ for all $\theta$.
3. 35.

Show that for any $x$ we have $\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x$.

4. 36.

Show that for $\left\lvert x\right\rvert\geq 1$ we have $\csc^{-1}x=\frac{\pi}{2}-\sec^{-1}x$.

5. 37.

A mass attached to a spring oscillates vertically about the equilibrium position $y=0$ according to the function $y(t)=e^{-t}(\cos(\sqrt{3}t)+\frac{1}{\sqrt{3}}\sin(\sqrt{3}t))$. Find the first positive time $t$ for which $y(t)=0$.