## Calculus (3rd Edition)

Published by W. H. Freeman

# 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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