Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.2 Mathematical Models: A Catalog of Essential Functions - 1.2 Exercises - Page 34: 11



Work Step by Step

It is given that $f(-1)=f(0)=f(2)=0$. Therefore, there must be a zero at x = -1, 0 and 2. Now, we know that the equation is a $cubic function$. We already have 3 roots, and hence we can write the equation as $f(x)=ax(x+1)(x-2)$. The last step is to find $a$ in order to find the final expression. This is done through using $f(1)=6$. Like so: $f(1) = a(1)(1+1)(1-2)$ $f(1)=-2a$ But$f(1)=6$ so, $-2a=6$ which leads to $a=-3$ Which then leads to our final answer: $f(x)=-3x(x+1)(x-2)$
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