Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 259: 98

Answer

$-a$ is the critical point. There is a local minimum at x = −a.

Work Step by Step

$f'(x) = 4(x + a)^3$, which is $0$ for $x = −a$. Note that $f''(x) = 12(x + a)^2$, which is $0$ at $x = −a$, so the test is inconclusive. The first derivative test shows that there is a local minimum at x = −a.
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