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Math
CalculusSingle Variable Calculus: Early Transcendentals, Volume I8th Edition

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Chapter 2.5, Problem 62E
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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Problem 1E
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Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
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Problem 10E
Problem 11E
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Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
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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
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
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=x2−3+1x has at least two x-intercepts in (1,2)

Explanation

Result Used:

Intermediate value theorem:

Let f:[x,y]→ℝ is a continuous function such that min{f(a),f(b)}≤k≤max{f(a),f(b)}, then there exists c∈(a,b) satisfying f(c)=k

Proof:

Consider the function f(x)=x2−3+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)2−3+10.2=(15)2−3+10.2≈2

f(b)=(1)2−3+11=−1

Thus, f(a)f(b)=−2<1.

Also, −1≤0≤2 which implies that min{f(15),f(1)}≤0≤max{f(1),f(32)}

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