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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Prove, without graphing, that the graph of the function has at least two x-intercepts in the specified interval.

y = x2 3 + 1/x, (0, 2)

To determine

To prove: The function y=x23+1x 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:

Consider the function f(x)=x23+1x.

The given function is a continuous function on the interval (0,2).

Since 1x is discontinuous at 0, consider the interval (15,2)(0,2).

Bisect the interval (15,2) as (15,2)=(15,1][1,2).

Take the interval (15,1](15,1).

Let a=15 and b=1.

Calculate the value of f at these points.

f(a)=(15)23+10.2=(15)23+10.22

f(b)=(1)23+11=1

Thus, f(a)f(b)=2<1.

Also, 102 which implies that min{f(15),f(1)}0max{f(1),f(32)}

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