a. Explanation Theorem used: Intermediate theorem: The function f ( x ) is continuous over an integral [ a , b ] , where u is a value of f ( x ) such that f ( a ) < u < f ( b ) then, there exist a number c between a and b , where f ( c ) = u . Calculation: The function is given by f ( x ) = x 3 − x − 1 . Here, the function is polynomial with degree 3. So, the function is continuous on ( − ∞ , ∞ ) . Also, the function is continuous on [ 1 , 2 ] . In order to show that the function f ( x ) = x 3 − x − 1 has a root between x = 1 and x = 2 , compute f ( 1 ) and f ( 2 ) . Compute f ( 1 ) as follows b. To determine The zero of a function f ( x ) = x 3 − x − 1 on the interval ( 1 , 2 ) and approximate the answer to four decimal places.

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

10th Edition

ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Question

Chapter 11.7, Problem 20E

To determine

To show: The function f(x)=x3−x−1=0 has a root between x=1 and x=2.

To determine

The zero of a function f(x)=x3−x−1 on the interval (1,2) and approximate the answer to four decimal places.

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Chapter 11.7, Problem 19E

Chapter 11.7, Problem 21E

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Chapter 11 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

Ch. 11.1 - Prob. 1CQ Ch. 11.1 - Prob. 2CQ Ch. 11.1 - Prob. 1E Ch. 11.1 - In Exercises 110, find the first three Taylor...Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E

Ch. 11.1 - In Exercises 110, find the first three Taylor...Ch. 11.1 - Prob. 10E Ch. 11.1 - Prob. 11E Ch. 11.1 - Prob. 12E Ch. 11.1 - Prob. 13E Ch. 11.1 - Prob. 14E Ch. 11.1 - In Exercises 1120, find the nth Taylor polynomial...Ch. 11.1 - Prob. 16E Ch. 11.1 - Prob. 17E Ch. 11.1 - Prob. 18E Ch. 11.1 - Prob. 19E Ch. 11.1 - Prob. 20E Ch. 11.1 - Prob. 21E Ch. 11.1 - Prob. 22E Ch. 11.1 - Prob. 23E Ch. 11.1 - Prob. 24E Ch. 11.1 - Prob. 25E Ch. 11.1 - Prob. 26E Ch. 11.1 - Prob. 27E Ch. 11.1 - Prob. 28E Ch. 11.1 - Prob. 29E Ch. 11.1 - Prob. 30E Ch. 11.1 - Prob. 31E Ch. 11.1 - Prob. 32E Ch. 11.1 - Prob. 33E Ch. 11.1 - Prob. 34E Ch. 11.1 - Let f(x) = ex. a. 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The function f ( x ) = x 3 − x − 1 = 0 has a root between x = 1 and x = 2 .

Solution Summary: The author explains that the function f(x) is continuous over an integral and is polynomial with degree 3.