BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 53E

(a)

To determine

To show: The derivative of the inverse function (f1)(x)=1f(f1(x)).

Expert Solution

Explanation of Solution

Given:

The function f(x) is one to one and differentiable and its inverse f1 also differentiable.

Derivative rules:

Chain rule: dydx=dydududx

Proof:

Obtain the derivative of the function f1(x).

Since the function f(x) is one to one function, f(f1(x))=x.

Let u=f1(x).

f(u)=x

Differentiate implicitly with respect to x.

ddx(f(u))=ddx(x)ddx(f(u))=1

Apply the chain rule and simplify the terms,

ddu(f(u))dudx=1f(u)dudx=1

Substitute u=f1(x),

f(f1(x))ddx(f1(x))=1

Divided by f(f1(x)) on both sides,

ddx(f1(x))=1f(f1(x))(f1)(x)=1f(f1(x))

Hence the required result is proved.

(b)

To determine

To find: The value f1(5).

Expert Solution

Answer to Problem 53E

The value of (f1)(5) is 32.

Explanation of Solution

Given:

The values f(4)=5 and  f(4)=23

Calculation:

Obtain the value of (f1)(5).

From part (a), (f1)(x)=1f(f1(x)).

Substitute 5 for x,

(f1)(5)=1f(f1(5))

Since the function is one to one, f(4)=5  implies that f1(5 )=4.

Substitute f1(5 )=4 and f(4)=23,

(f1)(5)=1f(4)=123=32

Therefore, the value of (f1)(5) is 32.

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