   Chapter 2.5, Problem 51E

Chapter
Section
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 .

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

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)=9614.0404957

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

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Domain Find the domain of the function. 17. f(x)=x+3x

Precalculus: Mathematics for Calculus (Standalone Book)

Find y if x6 + y6 = 1.

Single Variable Calculus: Early Transcendentals, Volume I

Rationalize the numerator: x1x1.

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

Solve the equations in Exercises 112 for x (mentally, if possible). ax+b=c(a0)

Finite Mathematics and Applied Calculus (MindTap Course List)

54. If

Mathematical Applications for the Management, Life, and Social Sciences 