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 257: 70

Answer

$f$ is concave up on $(−∞,∞)$. $x=-1$ is the only point of inflection.

Work Step by Step

$f'(x) = 8x^3 + 24x^2 + 24x − 1$, and $f''(x) = 24x^2 + 48x + 24 = 24(x + 1)^2$. Note that this quantity is always greater than $0$ for $x \neq −1$, and is $0$ only at $x = −1$. Thus $f$ is concave up on $(−∞,−1)$ and on $(−1,∞)$, and since $f$ and $f'$ are continuous at $−1$, we can say that $f$ is concave up on $(−∞,∞)$.
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