Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.4 The Shape of a Graph - Exercises - Page 194: 4

Answer

The inflection point is at $t=2$. Concave up on $t\in (2, \infty)$ Concave down on $t\in (-\infty, -2)$.

Work Step by Step

We have $$y=t^3-6t^2+4, \quad y'=3t^2-12t, \quad y''=6t-12$$ The inflection point is when $y''=6t-12=0$, that is $t=2$. Concave up when $y''\gt 0$, which occurs at $t\in (2, \infty)$ Concave down when $y''\lt 0$, which occurs at $t\in (-\infty, -2)$.
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