Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 31

Answer

The minimum value is $ f(1)=5$ and the maximum value is $ f(2)=28$

Work Step by Step

$ f'(t)=6t^{2}+6t $ is differentiable everywhere. The critical points of the function are the solutions of $6t^{2}+6t=0$ $6t^{2}+6t=0\implies t=0,-1$ Both critical points are outside the given interval. Therefore, we compare the values of $ f(t)$ at the end points only to find the maximum and minimum. $ f(1)=2(1)^{3}+3(1)^{2}=5$ $ f(2)=2(2)^{3}+3(2)^{2}=28$ The minimum value is $ f(1)=5$ and the maximum value is $ f(2)=28$
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