Single Variable Calculus: Concepts and Contexts, Enhanced Edition

4th Edition

ISBN: 9781337687805

Author: James Stewart

Publisher: Cengage Learning

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Question

Chapter 4.3, Problem 62E

(a)

To determine

To prove: The equation for $x≥0$ .

(a)

Expert Solution

Answer to Problem 62E

The equation is proved.

Explanation of Solution

Given information:

The equation is $ex≥1+x$ .

Calculation:

Let $f(0)=ex−1−x$

Compute the value $f(0)$ as follows:

$f(0)=e0−1−0=1−1=0$

Derivative of the function $f(x)=ex−1−x:$

$f(x)=ddx(ex−1−x)=ddx(ex)−ddx(1)−ddx(x)=ex−0−1=ex−1$

So, $f(x)=ex−1$

For $x≥0$ , we have $f(x)=ex−1≥0$ .

Thus, the function $f(x)=ex−1−x$ is increasing on $(0,∞)$ .

Since, $f(0)=0$ and $f$ is increasing on $(0,∞)$ , $f(x)≥0$ for $x≥0$ .

It follows that

$ex−1−x≥0ex≥1+x$

Therefore, the statement $ex≥1+x$ is proved for $x≥0$ .

(b)

To determine

To prove: The equation for $x≥0$ .

(b)

Expert Solution

Answer to Problem 62E

The equation $ex≥1+x+12x2$ is proved for $x≥0$ .

Explanation of Solution

Given information:

The given equation is $ex≥1+x+12x2$ .

Calculation:

Let $f(x)=ex−1−x−12x2.$

Compute the value $f(0)=e0−1−x−12(02)=1−1=0$

Calculate the derivative of the function $f(x)=ex−1−x−12x2.$

$f(x)=ddx(ex−1−x−12x2)=ddx(ex)−ddx(1)−ddx(x)−12ddx(x2)=ex−0−1−12(2x)=ex−1−x$

Thus, $f'(x)=ex−1−x$ .

Since. $f'(x)=ex−1≥0$ for $x≥0$ , so the function $f'(x)=ex−1−x$ is increasing on $(0,∞)$ . And since $f'(0)=0,$ therefore, $f'(x)=ex−1−x$ is positive for $x≥0$ .

So, the function $f(x)=ex−1−x−12x2.$ is increasing on $(0,∞)$ .

Since, $f'(0)=0,$ and $f$ is increasing on $(0,∞)$ , $f(x)≥0$ for $x≥0$ .

It follows that

$ex≥1+x+12x2$

Therefore, the statement $ex≥1+x+12x2$ is proved for $x≥0$ .

(c)

To determine

To prove: The equation for $x≥0$ and positive integer n.

(c)

Expert Solution

Answer to Problem 62E

The equation is proved shown below.

Explanation of Solution

Given information:

The given equation is $ex≥1+x+x22!+.....+xnn!$

Calculation:

Let the function $f$ be $f(x)=ex−1−x−x22!−.....−xnn!$

First, prove that the statement is true for $n=1$ .

Let $f(x)=ex−1−x$ .

compute the value $f(0)$ as follows:

$f(0)=e0−1−0=1−1=0$

Calculate the derivative of the function $f(x)=ex−1−x$ :

$f(x)=ddx(ex−1−x)=ddx(ex)−ddx(1)−ddx(x)=ex−0−1=ex−1$

So, $f'(x)=ex−1$

For $x≥0$ , we have $f'(x)=ex−1≥0$ .

Thus, the function $f(x)=ex−1−x$ is increasing on $(0,∞)$ .

Since, $f(0)=0$ and $f$ is increasing on $(0,∞)$ , $f(x)≥0$ for $x≥0$ .

It follows that

$ex−1−x≥0ex≥1+x$

Thus, $ex≥1+x$ for $x≥0$ .

Therefore, the statement is proved for $n=1$ .

Now, let $f(x)=ex−1−x−x22!−.....−xkk!$

Assume that the statement is true for $n=k$ .

That means, $ex≥1+x+x22!+.....+xkk!$ for $x≥0$ .

That implies $f(x)=ex−1−x−x22!−.....−xkk!≥0$ for $x≥0$ .

Now. Let $f(x)=ex−1−x−x22!−.....−xkk!−xk+1k+1!$ .

Then, $f(x)=ex−1−x−x22!−.....−xkk!≥0$ by assumption, so,

$f(x)=ex−1−x−x22!−.....−xkk!−xk+1k+1!$ is increasing on $(0,∞)$ .

Thus, $ex≥1+x+x22!+.....+xkk!+xk+1k+1!$ .

Therefore, for $x≥0$ , $ex≥1+x+x22!+.....+xnn!$ for every positive integer $n$ , by mathematical induction.

Therefore, the statement $ex≥1+x+x22!+.....+xnn!$ is proved for $x≥0$ and for any positive integer $n$ , by using mathematical induction.

Chapter 4.3, Problem 61E

Chapter 4.3, Problem 63E

Chapter 4 Solutions

Single Variable Calculus: Concepts and Contexts, Enhanced Edition

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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