# The function f ( x ) = x + e x is one to one. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### 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 f−1(1).

Expert Solution

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 f−1(1) (use formula from ex 77 (a))

Expert Solution

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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