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: 69

Answer

$g$ is concave down on $(−∞, 0)$ and on $(2, 4)$, and $g$ is concave up on $(0, 2)$ and on $(4,∞)$. There are inflection points at $t = 0, 2$, and $4$.

Work Step by Step

$g'(t) = 15t^4 − 120t^3 + 240t^2$, and $g''(t) = 60t^3 − 360t^2 + 480t = 60t(t − 2)(t − 4)$. Note that $g$ is $0$ for $t = 0, 2$, and $4$. Note also that $g' < 0$ on $(−∞, 0)$ and on $(2, 4)$, so $g$ is concave down on those intervals, while $g' > 0$ on $(0, 2)$ and on $(4,∞)$, so $g$ is concave up there. There are inflection points at $t = 0, 2$, and $4$.
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