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Math
CalculusSingle Variable Calculus: Early Transcendentals8th Edition

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Chapter 2.5, Problem 51E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
Problem 56E
Problem 57E
Problem 58E
Problem 59E
Problem 60E
Problem 61E
Problem 62E
Problem 63E
Problem 64E
Problem 65E
Problem 66E
Problem 67E
Problem 68E
Problem 69E
Problem 70E
Problem 71E
Problem 72E
Problem 73E
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

If f(x) = x2 + 10 sin x, show that there is a number c such that f(c) = 1000.

To determine

To show: There is a number c such that f(c)=1000 whenever f(x)=x2+10sinx.

Explanation

Theorem used: The Intermediate value Theorem

Suppose that if f is continuous on the closed interval [a, b] and let N be any number between f(a) and f(b), where f(a)≠f(b). Then, there exists a number c in (a, b) such that f(c)=N.

Proof:

Consider the function f(x)=x2+10sinx.

Here, f(x) is a combination of polynomial and trigonometric function. Both are continuous everywhere. Therefore, f(x) also continuous on ℝ.

Without loss of generality, take the subinterval [31,32]⊆ℝ.

Take a=31, b=32 and N=1000.

Substitute 31 for x in f(x),

f(31)=(31)2+10sin(31)=961+10(−0.40404)=961−4.0404≈957

This implies that, f(31)=957<1000

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