Explanation Formula Used: Let us suppose f ( x ) be any polynomial of degree say n ( n ≥ 1 ) and a is any real number then : ( i ) x − a is factor of f ( x ) if f ( a ) = 0 (ii) f ( a ) = 0 , if x − a is factor of f ( x ) Calculation: Since we are given f ( − 1 ) = f ( 0 ) = f ( 2 ) = 0 we know that there are three factors x ( x − 2 ) ( x + 1 ) Since we have to find an expression of a cubic function then there should be third factor , Let the 3 r d factor be ( a x + b ) therefore f ( x ) − x ( x − 2 ) ( x + 1 )

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Chapter 1.2, Problem 9E

To determine

Find an expression for a cubic function f if f(1)=6 and f(−1)=f(0)=f(2)=0 .

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Chapter 1.2, Problem 8E

Chapter 1.2, Problem 10E

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Chapter 1 Solutions

Biocalculus

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Find an expression for a cubic function f if f ( 1 ) = 6 and f ( − 1 ) = f ( 0 ) = f ( 2 ) = 0 .

Solution Summary: The author explains how to find an expression for a cubic function f if it's n(nge 1), and the third factor is (ax+b).

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Find an expression for a cubic function f if f(1)=6 and f(−1)=f(0)=f(2)=0 .

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Chapter 1 Solutions