   Chapter 2.5, Problem 61E

Chapter
Section
Textbook Problem

Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval.y = sin x3, (1, 2)

To determine

To Prove: The function y=sinx3 has at least two x-intercepts in (1,2)

Explanation

Result Used: Intermediate value theorem:

Let f:[x,y] is a continuous function such that

min{f(a),f(b)}kmax{f(a),f(b)}, then there exists c(a,b) satisfying f(c)=k

Proof:

y=sinx3 is a composite function of sin x and x3

Both sin x and x3  are continuous function on the interval (1,2)

Thus, y=sinx3 is continuous function on the interval (1,2).

Bisect the interval (1,2)

Thus, (1,2)=(1,32][32,2)

Take the interval (1,32](1,32)

Let a=1 and b=32

Calculate the value of f at these points.

f(a)=sin(1)3=sin(1)0.841

f(b)=sin(32)3=sin(278)0.231

Thus,f(a)f(b)<1

Also,0.23100.841 which implies min{f(1),f(32)}0max{f(1),f(32)}

Hence all the conditions of intermediate value theorem are satisfied

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