   Chapter 2.5, Problem 70E

Chapter
Section
Textbook Problem

If a and b are positive numbers, prove that the equation a x 3 + 2 x 2 − 1 + b x 3 + x − 2 = 0 has at least one solution in the interval (–1, 1).

To determine

To prove: If a and b are the positive numbers then at least one solution of the given equation

ax3+2x21+bx3+x2=0 lies in the interval (1,1).

Explanation

Given:

The equation ax3+2x21+bx3+x2=0, where a and b are positive numbers.

Formula used:

(1) Intermediate Value Theorem.

Calculation:

The given equation is ax3+2x21+bx3+x2=0.

We can express the given equation in the form of polynomial of degree three is as follows.

a(x3+x2)+b(x3+2x21)(x3+2x21)(x3+x2)=0a(x3+x2)+b(x3+2x21)=0[a0,b0]

Assume p(x)=a(x3+x2)+b(x3+2x21)

Since p(x) is continuous in the interval [1,1] because p(x) is a polynomial of degree three and every finite degree polynomial is continuous, therefore p(x) is also continuous.

For x=1

p(1)=a((1)3+(1)2)+b((1)3+2(1)21)=a(4)+b(0)=4a<0

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