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: 26


Minimum value is 5 at $ x=0$ and maximum value is 21 at $ x=2$

Work Step by Step

$ f'(x)= 4x+4$ $ f $ is differentiable everywhere. So, the only critical point we get is $ c $ such that $ f'(c)=0$ i.e., $4c+4=0\implies 4c=-4\implies c=-1$ But, this point is outside the given interval. To find the maximum and minimum value, we compare values at the end points only (since there are no critical points inside the given interval). $ f(0)= 2(0)^{2}+4(0)+5=5$ $ f(2)=2(2)^{2}+4(2)+5=21$ The minimum value is 5 at $ x=0$ and maximum value is 21 at $ x=2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.