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 54E

(a)

To determine

To show: The function f(x)=x+ex is one to one.

Expert Solution

Explanation of Solution

Given:

The function is f(x)=x+ex.

Proof:

Obtain the derivative of the function f(x)=x+ex.

Differentiate implicitly with respect to x,

f(x)=ddx(x+ex)=ddx(x)+ddx(ex)=1+ex

Here, the value ex is always positive, then f(x)=1+ex is positive.

Note that, if f(x) is positive then the function f(x) is increasing function.

Thus, f(x)=x+ex is increasing function, every element of the codomain has at most one preimage.

Therefore, the function f(x) is one to one.

Hence the required result is proved.

(b)

To determine

To find: The values of f1(1).

Expert Solution

Answer to Problem 54E

The value f1(1)=0.

Explanation of Solution

Calculation:

Obtain the value of f1(1).

Let x=f1(1). That is, compute f(x)=1.

Substitute x=0 in f(x)=x+ex,

f(0)=0+e0=0+1=1

Thus, the value is f(0)=1.

Since f(x) is one to one,f1(1)=0.

Therefore, the value f1(1)=0.

(c)

To determine

To find: The value of the f1(1) (use formula from ex 77 (a))

Expert Solution

Answer to Problem 54E

The value (f1)(1)=12.

Explanation of Solution

Formula used:

The derivative of inverse function (f1)(x)=1f(f1(x)).

Calculation:

Given function is f(x)=x+ex.

Substitute 1 for x in (f1)(x)=1f(f1(x)),

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

From part (a) and (b), the derivative is f(x)=1+ex and the value of f1(1)=0.

Substitute 0 for x in f(x)=1+ex,

f(0)=1+e0=1+1=2

Thus, the value is f(0)=2.

Substitute f1(1)=0 in (f1)(1),

(f1)(1)=1f(0)=12                [Qf(0)=2]

Therefore, the value (f1)(1)=12.

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