Let F(X)= So^x F(T)dt Where F Is the Function Graphed Below

1. Functions and Graphs

1.4 Inverse Functions

Learning Objectives

Determine the conditions for when a function has an inverse.
Use the horizontal line test to recognize when a function is one-to-one.
Find the inverse of a given function.
Draw the graph of an inverse function.
Evaluate inverse trigonometric functions.

An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric functions.

Existence of an Inverse Function

We begin with an example. Given a function and an output , we are often interested in finding what value or values were mapped to by . For example, consider the function . Since any output , we can solve this equation for to find that the input is $x=\sqrt[3]{y-4}$ . This equation defines as a function of . Denoting this function as $f^{-1}$ , and writing $x=f^{-1}(y)=\sqrt[3]{y-4}$ , we see that for any in the domain of $f, \, f^{-1}(f(x))=f^{-1}(x^3+4)=x$ . Thus, this new function, $f^{-1}$ , "undid" what the original function did. A function with this property is called the inverse function of the original function.

Note that $f^{-1}$ is read as "f inverse." Here, the -1 is not used as an exponent and $f^{-1}(x) \ne 1/f(x)$ . (Figure) shows the relationship between the domain and range of and the domain and range of $f^{-1}$ .

Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each input to exactly one output. For example, let's try to find the inverse function for . Solving the equation for , we arrive at the equation $x= \pm \sqrt{y}$ . This equation does not describe as a function of because there are two solutions to this equation for every . The problem with trying to find an inverse function for is that two inputs are sent to the same output for each output . The function discussed earlier did not have this problem. For that function, each input was sent to a different output. A function that sends each input to a different output is called a one-to-one function.

Definition

We say a is a one-to-one function if $f(x_1) \ne f(x_2)$ when $x_1 \ne x_2$ .

One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the -plane, according to the horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different from the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test determines whether a function is one-to-one ((Figure)).

Rule: Horizontal Line Test

A function is one-to-one if and only if every horizontal line intersects the graph of no more than once.

Determining Whether a Function Is One-to-One

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

Is the function graphed in the following image one-to-one?

An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of the function

Inverse Trigonometric Functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ((Figure)). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ . By doing so, we define the inverse sine function on the domain such that for any in the interval , the inverse sine function tells us which angle $\theta$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ satisfies $\sin \theta =x$ . Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.

Definition

The inverse sine function, denoted $\sin^{-1}$ or arcsin, and the inverse cosine function, denoted $\cos^{-1}$ or arccos, are defined on the domain $D=\{x|-1 \le x \le 1\}$ as follows:

$\begin{array}{c}\sin^{-1}(x)=y \,\, \text{if and only if} \, \sin (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2};\hfill \\ \cos^{-1}(x)=y \,\, \text{if and only if} \, \cos (y)=x \, \text{and} \, 0 \le y \le \pi \hfill \end{array}$

The inverse tangent function, denoted $\tan^{-1}$ or arctan, and inverse cotangent function, denoted $\cot^{-1}$ or arccot, are defined on the domain $D=\{x|-\infty <x<\infty \}$ as follows:

$\begin{array}{c}\tan^{-1}(x)=y \,\, \text{if and only if} \, \tan (y)=x \, \text{and} \, -\frac{\pi}{2}<y<\frac{\pi}{2};\hfill \\ \cot^{-1}(x)=y \,\, \text{if and only if} \, \cot (y)=x \, \text{and} \, 0<y<\pi \hfill \end{array}$

The inverse cosecant function, denoted $\csc^{-1}$ or arccsc, and inverse secant function, denoted $\sec^{-1}$ or arcsec, are defined on the domain $D=\{x| \, |x| \ge 1\}$ as follows:

$\begin{array}{c}\csc^{-1}(x)=y \,\, \text{if and only if} \, \csc (y)=x \, \text{and} \, -\frac{\pi}{2} \le y \le \frac{\pi}{2}, \, y\ne 0;\hfill \\ \sec^{-1}(x)=y \,\, \text{if and only if} \, \sec (y)=x \, \text{and} \, 0 \le y \le \pi, \, y \ne \pi/2\hfill \end{array}$

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line ((Figure)).

Go to the following site for more comparisons of functions and their inverses.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate $\cos^{-1}(\frac{1}{2})$ , we need to find an angle $\theta$ such that $\cos \theta =\frac{1}{2}$ . Clearly, many angles have this property. However, given the definition of $\cos^{-1}$ , we need the angle $\theta$ that not only solves this equation, but also lies in the interval $[0,\pi]$ . We conclude that $\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}$ .

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions $\sin (\sin^{-1}(\frac{\sqrt{2}}{2}))$ and $\sin^{-1}(\sin(\pi))$ . For the first one, we simplify as follows:

$\sin (\sin^{-1}(\frac{\sqrt{2}}{2}))= \sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ .

For the second one, we have

$\sin^{-1}( \sin (\pi))=\sin^{-1}(0)=0$ .

The inverse function is supposed to "undo" the original function, so why isn't $\sin^{-1}(\sin (\pi))=\pi$ ? Recalling our definition of inverse functions, a function and its inverse $f^{-1}$ satisfy the conditions $f(f^{-1}(y))=y$ for all in the domain of $f^{-1}$ and $f^{-1}(f(x))=x$ for all in the domain of , so what happened here? The issue is that the inverse sine function, $\sin^{-1}$ , is the inverse of the restricted sine function defined on the domain $[-\frac{\pi}{2},\frac{\pi}{2}]$ . Therefore, for in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ , it is true that $\sin^{-1}(\sin x)=x$ . However, for values of outside this interval, the equation does not hold, even though $\sin^{-1}(\sin x)$ is defined for all real numbers .

What about $\sin (\sin^{-1}y)$ ? Does that have a similar issue? The answer is no. Since the domain of $\sin^{-1}$ is the interval , we conclude that $\sin (\sin^{-1}y)=y$ if $-1 \le y \le 1$ and the expression is not defined for other values of . To summarize,

$\sin (\sin^{-1}y)=y \, \text{if} \, -1 \le y \le 1$

and

$\sin^{-1}( \sin x)=x \, \text{if} \, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ .

Similarly, for the cosine function,

$\cos (\cos^{-1}y)=y \, \text{if} \, -1 \le y \le 1$

and

$\cos^{-1}( \cos x)=x \, \text{if} 0 \le x \le \pi$ .

Similar properties hold for the other trigonometric functions and their inverses.

Evaluating Expressions Involving Inverse Trigonometric Functions

The Maximum Value of a Function

In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain, even if we don't know its exact value at a given instant. For instance, if we have a function describing the strength of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails. Safe design often depends on knowing maximum values.

This project describes a simple example of a function with a maximum value that depends on two equation coefficients. We will see that maximum values can depend on several factors other than the independent variable .

Consider the graph in (Figure) of the function $y= \sin x + \cos x$ . Describe its overall shape. Is it periodic? How do you know?
Using a graphing calculator or other graphing device, estimate the – and -values of the maximum point for the graph (the first such point where ). It may be helpful to express the -value as a multiple of $\pi$ .
Now consider other graphs of the form $y=A \sin x + B \cos x$ for various values of and . Sketch the graph when and , and find the – and -values for the maximum point. (Remember to express the -value as a multiple of $\pi$ , if possible.) Has it moved?
Repeat for $A = 1, \, B = 2$ . Is there any relationship to what you found in part (2)?

Complete the following table, adding a few choices of your own for

and


0	1	$\sqrt{3}$	1
1	0	1	$\sqrt{3}$
1	1	12	5
1	2	5	12
2	1
2	2
3	4
4	3

Try to figure out the formula for the -values.
The formula for the -values is a little harder. The most helpful points from the table are $(1,1), \, (1,\sqrt{3}), \, (\sqrt{3},1)$ . (Hint: Consider inverse trigonometric functions.)
If you found formulas for parts (5) and (6), show that they work together. That is, substitute the -value formula you found into $y=A \sin x + B \cos x$ and simplify it to arrive at the -value formula you found.

Key Concepts

Key Equations

For the following exercises, use the horizontal line test to determine whether each of the given graphs is one-to-one.

1.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a function that decreases in a straight in until the origin, where it begins to increase in a straight line. The x intercept and y intercept are both at the origin.

2.
An image of a graph. The x axis runs from 0 to 7 and the y axis runs from -4 to 4. The graph is of a function that is always increasing. There is an approximate x intercept at the point (1, 0) and no y intercept shown.

3.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a function that resembles a semi-circle, the top half of a circle. The function starts at the point (-3, 0) and increases until the point (0, 3), where it begins decreasing until it ends at the point (3, 0). The x intercepts are at (-3, 0) and (3, 0). The y intercept is at (0, 3).

4.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a curved function. The function increases until it hits the origin, then decreases until it hits the point (2, -4), where it begins to increase again. There are x intercepts at the origin and the point (3, 0). The y intercept is at the origin.

5.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a curved function that is always increasing. The x intercept and y intercept are both at the origin.

6.
An image of a graph. The x axis runs from -4 to 7 and the y axis runs from -4 to 4. The graph is of a function that increases in a straight line until the approximate point (, 3). After this point, the function becomes a horizontal straight line. The x intercept and y intercept are both at the origin.

For the following exercises, a. find the inverse function, and b. find the domain and range of the inverse function.

7. $f(x)=x^2-4, \, x \ge 0$

8. $f(x)=\sqrt[3]{x-4}$

Solution

a. $f^{-1}(x)=\sqrt[3]{x-1}$ b. Domain: all real numbers, Range: all real numbers

10. $f(x)=(x-1)^2, \, x \le 1$

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

12. $f(x)=\frac{1}{x+2}$

For the following exercises, use the graph of to sketch the graph of its inverse function.

13.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of an increasing straight line function labeled

Solution

An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of two functions. The first function is an increasing straight line function labeled

14.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a curved decreasing function labeled

15.
An image of a graph. The x axis runs from -8 to 8 and the y axis runs from -8 to 8. The graph is of an increasing straight line function labeled

Solution

An image of a graph. The x axis runs from 0 to 8 and the y axis runs from 0 to 8. The graph is of two function. The first function is an increasing straight line function labeled

16.
An image of a graph. The x axis runs from -4 to 4 and the y axis runs from -4 to 4. The graph is of a decreasing curved function labeled

For the following exercises, use composition to determine which pairs of functions are inverses.

17. $f(x)=8x, \, g(x)=\frac{x}{8}$

Solution

These are inverses.

18. $f(x)=8x+3, \, g(x)=\frac{x-3}{8}$

19. $f(x)=5x-7, \, g(x)=\frac{x+5}{7}$

Solution

These are not inverses.

20. $f(x)=\frac{2}{3}x+2, \, g(x)=\frac{3}{2}x+3$

21. $f(x)=\frac{1}{x-1}, \, x \ne 1, \, g(x)=\frac{1}{x}+1, \, x \ne 0$

Solution

These are inverses.

22. $f(x)=x^3+1, \, g(x)=(x-1)^{1/3}$

23. $f(x)=x^2+2x+1, \, x \ge -1,\, g(x)=-1+\sqrt{x}, \, x \ge 0$

Solution

These are inverses.

24. $f(x)=\sqrt{4-x^2}, \, 0 \le x \le 2, \, g(x)=\sqrt{4-x^2}, \, 0 \le x \le 2$

For the following exercises, evaluate the functions. Give the exact value.

25. $\tan^{-1}(\frac{\sqrt{3}}{3})$

Solution

$\frac{\pi}{6}$

26. $\cos^{-1}(-\frac{\sqrt{2}}{2})$

27. $\cot^{-1}(1)$

Solution

$\frac{\pi}{4}$

28. $\sin^{-1}(-1)$

29. $\cos^{-1}(\frac{\sqrt{3}}{2})$

Solution

$\frac{\pi}{6}$

30. $\cos (\tan^{-1}(\sqrt{3}))$

31. $\sin (\cos^{-1}(\frac{\sqrt{2}}{2}))$

[reveal-answer q="461959″]Show Answer[/reveal-answer]
[hidden-answer a="461959″] $\frac{\sqrt{2}}{2}$

32. $\sin^{-1}( \sin (\frac{\pi}{3}))$

33. $\tan^{-1}( \tan (-\frac{\pi}{6}))$

Solution

$-\frac{\pi}{6}$

34.The function converts degrees Fahrenheit to degrees Celsius.

Find the inverse function $F=T^{-1}(C)$
What is the inverse function used for?

Solution

a. $x=f^{-1}(V)=\sqrt{0.04-\frac{V}{500}}$ b. The inverse function determines the distance from the center of the artery at which blood is flowing with velocity . c. 0.1 cm; 0.14 cm; 0.17 cm

36.A function that converts dress sizes in the United States to those in Europe is given by .

Find the European dress sizes that correspond to sizes 6, 8, 10, and 12 in the United States.
Find the function that converts European dress sizes to U.S. dress sizes.
Use part b. to find the dress sizes in the United States that correspond to 46, 52, 62, and 70.

37. [T] The cost to remove a toxin from a lake is modeled by the function

, where is the cost (in thousands of dollars) and is the amount of toxin in a small lake (measured in parts per billion [ppb]). This model is valid only when the amount of toxin is less than 85 ppb.

Find the cost to remove 25 ppb, 40 ppb, and 50 ppb of the toxin from the lake.
Find the inverse function. c. Use part b. to determine how much of the toxin is removed for $50,000.

Solution

a. $31,250, $66,667, $107,143 b. $(p=\frac{85C}{C+75})$ c. 34 ppb

38. [T] A race car is accelerating at a velocity given by

$v(t)=\frac{25}{4}t+54$ ,

where is the velocity (in feet per second) at time .

Find the velocity of the car at 10 sec.
Find the inverse function.
Use part b. to determine how long it takes for the car to reach a speed of 150 ft/sec.

39. [T] An airplane's Mach number is the ratio of its speed to the speed of sound. When a plane is flying at a constant altitude, then its Mach angle is given by $\mu =2\sin^{-1}(\frac{1}{M})$ .

Find the Mach angle (to the nearest degree) for the following Mach numbers.

1.0".">

$\mu =1.4$
$\mu =2.8$
$\mu =4.3$

41. [T] The temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function

$T(x)=5+18 \sin[\frac{\pi}{6}(x-4.6)]$ ,

where is time in months and corresponds to January 1. Determine the month and day when the temperature is $21^{\circ}$ C.

Solution

$x \approx 6.69,8.51$ ; so, the temperature occurs on June 21 and August 15

42. [T] The depth (in feet) of water at a dock changes with the rise and fall of tides. It is modeled by the function

$D(t)=5 \sin (\frac{\pi}{6}t-\frac{7\pi}{6})+8$ ,

where is the number of hours after midnight. Determine the first time after midnight when the depth is 11.75 ft.

43. [T] An object moving in simple harmonic motion is modeled by the function

$s(t)=-6 \cos (\frac{\pi t}{2})$ ,

where is measured in inches and is measured in seconds. Determine the first time when the distance moved is 4.5 ft.

Solution

$~1.5 \sec$

44. [T] A local art gallery has a portrait 3 ft in height that is hung 2.5 ft above the eye level of an average person. The viewing angle $\theta$ can be modeled by the function

$\theta =\tan^{-1}\frac{5.5}{x}-\tan^{-1}\frac{2.5}{x}$ ,

where is the distance (in feet) from the portrait. Find the viewing angle when a person is 4 ft from the portrait.

45. [T] Use a calculator to evaluate $\tan^{-1}( \tan (2.1))$ and $\cos^{-1}( \cos (2.1))$ . Explain the results of each.

46. [T] Use a calculator to evaluate $\sin (\sin^{-1}(-2))$ and $\tan (\tan^{-1}(-2))$ . Explain the results of each.

Let F(X)= So^x F(T)dt Where F Is the Function Graphed Below

Source: https://opentextbc.ca/calculusv1openstax/chapter/inverse-functions/

Let F(X)= So^x F(T)dt Where F Is the Function Graphed Below

1.4 Inverse Functions

Learning Objectives

Existence of an Inverse Function

Definition

Rule: Horizontal Line Test

Determining Whether a Function Is One-to-One

Inverse Trigonometric Functions

Definition

Evaluating Expressions Involving Inverse Trigonometric Functions

The Maximum Value of a Function

Key Concepts

Key Equations

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

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