Explanation Given information: The function f ( x ) = x 1 x is increasing for 0 < x < e and decreasing for x > e , and its maximum value at x = e . The maximum value of the function f ( x ) is e 1 e . Proof: The function f ( x ) has the maximum value of e 1 e , so the function f ( x ) does not go beyond e 1 e . Therefore, no solution exists for c > e 1 e To determine (b) To graph: The function f ( x ) = x 1 x .

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.5, Problem 64E

To determine

(a)

To show:

That x1x=c has a unique solution of 0≤c≤1 or c=e1e, two solutions for 1<c<e1e and no solution for c>e1e.

To determine

(b)

To graph:

The function f(x)=x1x.

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Chapter 4.5, Problem 63E

Chapter 4.5, Problem 65E

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Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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(a) To show: That x 1 x = c has a unique solution of 0 ≤ c ≤ 1 or c = e 1 e , two solutions for 1 < c < e 1 e and no solution for c > e 1 e .

Solution Summary: The author illustrates the function f(x)=x 1e, which has two solutions for 1c

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(a)

To show:

That x1x=c has a unique solution of 0≤c≤1 or c=e1e, two solutions for 1<c<e1e and no solution for c>e1e.

(b)

To graph:

The function f(x)=x1x.

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Chapter 4 Solutions

(a) To show: That x 1 x = c has a unique solution of 0 ≤ c ≤ 1 or c = e 1 e , two solutions for 1 &lt; c &lt; e 1 e and no solution for c &gt; e 1 e .

Solution Summary: The author illustrates the function f(x)=x 1e, which has two solutions for 1c

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(a) To show: That x1x=c has a unique solution of 0≤c≤1 or c=e1e, two solutions for 1<c<e1e and no solution for c>e1e.

(b) To graph: The function f(x)=x1x.

Want to see the full answer?

Chapter 4 Solutions

(a) To show: That x 1 x = c has a unique solution of 0 ≤ c ≤ 1 or c = e 1 e , two solutions for 1 < c < e 1 e and no solution for c > e 1 e .

(a)

To show:

That x1x=c has a unique solution of 0≤c≤1 or c=e1e, two solutions for 1<c<e1e and no solution for c>e1e.

(b)

To graph:

The function f(x)=x1x.