7 Inverse Functions and L’Hôpital’s Rule Chapter Introduction 7.2 Derivatives of Inverse Functions

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 $f$ .

Watch the video:
Finding the Inverse of a Function or Showing One Does not Exist, Ex 3 from https://youtu.be/BmjbDINGZGg

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:

(a)

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

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

Solution

(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.
(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 $-\pi/2<y<\pi/2$ . 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

(a)

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

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

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

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

T/F: Every function has an inverse.
2.

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

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 $f^{-1}$ .

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

7.

$f(x)=2x+6$ and $g(x)=\frac{1}{2}x-3$
8.

$f(x)=x^{2}+6x+11$ , $x\geq-3$ and $g(x)=\sqrt{x-2}-3$ , $x\geq 2$
9.

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

11.

$f(x)=\sqrt{16-x^{2}}$
12.

$g(x)=\sqrt{x^{2}-16}$
13.

$r(t)=t^{2}-6t+9$
14.

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

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

15.

$f(x)=\dfrac{x+1}{x-2}$
16.

$f(x)=x^{2}+4$
17.

$f(x)=e^{x+3}-2$
18.

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

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

19.

$\tan^{-1}(0)$
20.

$\tan^{-1}(\tan(\pi/7))$
21.

$\cos(\cos^{-1}(-1/5))$
22.

$\sin^{-1}(\sin(8\pi/3))$
23.

$\sin(\tan^{-1}(1))$
24.

$\sec(\sin^{-1}(-3/5))$
25.

$\cos(\tan^{-1}(3/7))$
26.

$\sin^{-1}(-\sqrt{3}/2)$
27.

$\cos^{-1}(-\sqrt{2}/2)$
28.

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

In Exercises 29–32., simplify the expression.

29.

$\sin\Bigl{(}\tan^{-1}\frac{x}{\sqrt{4-x^{2}}}\Bigr{)}$
30.

$\tan\Bigl{(}\sin^{-1}\frac{x}{\sqrt{x^{2}+4}}\Bigr{)}$
31.

$\cos\Bigl{(}\sin^{-1}\frac{5}{\sqrt{x^{2}+25}}\Bigr{)}$
32.

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

33.

Show that for any $x$ in the domain of $\sec^{-1}$ we have $\sec^{-1}x=\cos^{-1}\frac{1}{x}$ .
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 $\theta$ .
35.

Show that for any $x$ we have $\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x$ .
36.

Show that for $\left\lvert x\right\rvert\geq 1$ we have $\csc^{-1}x=\frac{\pi}{2}-\sec^{-1}x$ .
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 $y(t)=0$ .