Calc 5.3

Verify that one function is the inverse of the other.Determine whether a function has an inverse.

Find the derivative of an inverse function.

This section is a lot review and a little new calculus concepts. Recall that a function f can be represented by a set of ordered pairs in which each element of the domain is mapped to only one element in the range. To find the inverse, you switch x- and y-values in the ordered pairs, or in the rule to get ordered pairs, and solve for y. This is called the inverse of f and if it too is a function, we denote this with f-1. When you switch x- and y- variables, you also switch the domains and ranges of the original function f. f and f-1 have the effect of “undoing” each other. This means that when you form the composition of the two, in other words,

1 1( ( )) and ( ( ))f f x x f f x x

the input becomes the output!

Warning:

11

1( )

( )f x

f x

1. If g is the inverse function of f, then f is the inverse function of g.

2. The domain of f-1 is the range of f. The range of f-1 is the domain of f.

3. A function need not have an inverse function, (could be just a relation), but if it does, the inverse function is unique.

Ex 1 p. 342 Verifying inverse functions.Show that the functions are inverses of each other.

3( ) 8 1f x x 3 1

( )2

xg x

33 31 1

( ( )) 8 12 2

x xf g x f

1

8 1 1 18

xx x

333 (8 1) 1

( ( )) (8 1)2

xg f x g x

3 38 2

2 2

x xx

Since they both equal x, they are inverses.

Important observations about inverse functions:

If you will verbally express what is happening in each function, you can more easily see the “undoing” nature of the inverse function.

y = xIt is a good idea to graph each function and the line y=x. Notice the reflection of each graph across the line y=x results in the other function. Make sure you zoom square to make it look symmetric!

3( ) 8 1f x x

31

( )8

xg x

The vertical line test is done to determine if a graph represents a function. If the vertical line can only intersect the graph once, it is a function. The horizontal line test determines whether the inverse of a function will be a function. A function f has an inverse function if and only if every horizontal line intersects the graph at most once.

Inverse of f(x) not a function since it fails the horizontal line test.

Strictly monotonic means that it is either increasing on its entire domain, or decreasing on its entire domain. It can have a slope of zero at some point, but then must keep with either increasing or decreasing again, whichever applies.

Ex 2 p. 343 The Existence of an Inverse Function

Do any of the following have an inverse function?

3. ( ) 1a f x x x

From the graph you can see that it appears that f is increasing on it entire domain. Looking at its derivative,

you can see that it is positive for all real values of x. So f is strictly monotonic and must . have an inverse function

3. g( ) 1b x x x

2'( ) 3 1f x x

From the graph, you can see that g fails the horizontal line test, or in other words, is not one-to-one. For instance, x = -1, 0, or 1 all result in the same y-value. g does not have an inverse function, only an inverse relation.

Often it is easier to decide a function has an inverse than to find the inverse function, as with f(x).

Guidelines for Finding an Inverse Function1.Use Thm 5.7 to determine whether the function y=f(x) has an inverse. You should look at graph of f(x).

2.Interchange x and y. Then solve for y. The resulting equation is y=f-1(x). If you can’t solve for y, you can graph it by inputting f(x) into y1, go to DRAW, DrawInv then select VARS, Y-VARS, Function, Y1 and enter

3.Define the domain of y=f-1(x) as the range of f(x).

4.Verify that f(f-1(x))=x and f-1(f(x)) =x

Ex 3 p.344 Finding an inverse function.

Find the inverse of ( ) 3 4f x x

3 4x y 2 3 4x y 2 4 3x y

21 4

3 3y x

1 21 4( )

3 3f x x

Square both sides

Add 4 to both sides

Divide both sides by 3

Switch x and y

The domain of f, which is x≥ 4/3, becomes the range, y ≥ 4/3 of the inverse. The range of f, which is y ≥0, becomes the domain, x ≥0 of the inverse. The darker graph is the inverse function, and the line of symmetry, y = x, is shown also.

f-1

f

Sometimes you are given a function that is not one-to-one in its domain. By restricting the domain to one in which the function is strictly monotonic, you can get the inverse to be a function as well.

Ex 4, p.345 Testing whether a function is one-to-one.Show that the cosine function is not one-to-one. Then show that if we restrict the domain from [0, π] we can make the cosine function one-to-one and therefore its inverse exists and is a function as well.

( ) cosf x x Many different values of x give the same y value in this function, for example, cos (-π) = cos (π) = -1.

πIf we restrict the domain to [0, π] it is decreasing through the entire interval, so it is one-to-one and monotonic, and f-1(x) = cos-1 x, which is also called the arccosine x, is a function as well.

The next two theorems discuss the derivative of an inverse function. Proofs are given in Appendix A.Remember, sometimes we can’t write what the inverse function is, but we can know the following things about it.

The two graphs shown are f(x) = x3 and f-1(x) = x1/3

The two graphs shown are f(x) = x3 and g(x) = x1/3

Calculate the slopes of f at (1, 1), (2, 8), (3, 27).Then calculate the slopes for g at (1,1), (8,2), and (27, 3). What do you observe? What happens at (0, 0)?

f ‘(x) =3x2 When x = 1, the slope is 3. When x = 2, the slope is 12. When x = 3, the slope is 27.

When x = 1 the slope is 1/3. When x = 8, the slope is 1/12. When x = 27, the slope is 1/272

3

1'( )

3g x

x

When you put an x-value from an ordered pair that is part of the graph of f(x) into f ‘, and the y-value from that same ordered pair into g’, the slopes are reciprocals!

Ex 5 p. 346 Evaluating the Derivative of an Inverse Function

31Let ( ) 1

4f x x x

1

1

a. What is the value of ( ) when x = 3?

b. What is the value of ( ) '( ) when x = 3?

f x

f x

a. We can’t find f-1. But since we know that f(2) = 3, we know that f-1(3) = 2 (Hint: graph f(x) and y=3 and find intersection.)

b. Using and theorem 5.9 we can write 23'( ) 1

4f x x

11

1( ) '(3)

'( (3))f

f f

1

'(2)f 2

1 13 4(2) 14

Got your head around that? The point (2, 3) is on the graph fAt that point where x = 2, the slope of f is 4. (Recall that ). The slope of f -1 is ¼. Whatever value of x you are given for f-1 you must find the x value of f(x) given this y-value!This reciprocal relationship which follows from Theorem 5.9 can be written as shown below.

23'( ) 14f x x

1If ( ) ( ), then ( ) and '( )dx

y g x f x f y x f ydy

1 1 1'( )

'( ( )) '( )

dyg x

dxdx f g x f ydy

1dydxdxdy

f(x)

f-1(x)

Ex 6 p. 346 Graphs of Inverse Function have Reciprocal Slopes

2Let ( ) for x 0f x x 1and then ( )f x x

Show that the slopes of the two functions are reciprocals for the following points:a. (2, 4) and (4, 2) b. (3, 9) and (9, 3)

The derivatives of f and f -1 are given by

'( ) 2f x x 1 1( ) '( )

2f x

x

a. '(2) 2(2) 4f 1 1 1, ( ) '(4)

42 4f

b. '(3) 2(3) 6,f 1 1 1( ) '(9)

62 9f

5.3 p. 347/ 1-7 odd, 9-12 all, 15, 17-41 every other odd, 53,63,65, 71,73, 83-89 odd