Explanation Given: The function is given by x 3 − x − 1 , 1 ≤ x ≤ 2 Graph: Formula Used: According to the intermediate value property if there are two point ( x 1 , y 1 ) a n d ( x 2 , y 2 ) on a graph of the continuous function f , then on the given interval x 1 ≤ x ≤ x 2 f assumes every value between y 1 a n d y 2 at least once. Calculation: First, we have to evaluate f ( x 1 ) a n d f ( x 2 ) . First, we have to evaluate f ( 1 ) f ( 1 ) = 1 3 − 1 − 1 = 1 − 1 − 1 = − 1 Now, we have to evaluate f ( 2 ) f ( 2 ) = 2 3 −

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ISBN: 9780321845092

Author: Rockswold

Publisher: YUZU

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Chapter 4.4, Problem 104E

To determine

a) To evaluate $f (1)$ and $f (2)$

b) To use the intermediate value property, show that $f (x) = 0$ for some $x$ on the given interval.

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Chapter 4.4, Problem 103E

Chapter 4.4, Problem 105E

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Chapter 4 Solutions

EBK COLLEGE ALGEBRA WITH MODELING & VIS

Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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a) To evaluate f ( 1 ) and f ( 2 ) b) To use the intermediate value property, show that f ( x ) = 0 for some x on the given interval.

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a) To evaluate f(1) and f(2) b) To use the intermediate value property, show that f(x)=0 for some x on the given interval.

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Chapter 4 Solutions

a) To evaluate $f (1)$ and $f (2)$

b) To use the intermediate value property, show that $f (x) = 0$ for some $x$ on the given interval.