Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.2 Relative Extrema - 5.2 Exercises - Page 272: 15

Answer

f(x) is increasing on $(-\infty, -3) \cup (-1,+\infty)$ and decreasing on $(-3,-1)$ . On $(-\infty, -3)$, f(x) achieves a maximum value of -8 at x=−3 and on $(-1,+\infty)$ a minimum value of -12 at x=-1.

Work Step by Step

$f(x) =x^{3}+6x^{2}+9x-8$ $f'(x)=3x^{2}+12x+9$ $f'(x)=0 \rightarrow 3x^{2}+12x+9=0 \rightarrow x=-3, x=-1$ Thus, f(x) is increasing on $(-\infty, -3) \cup (-1,+\infty)$ and decreasing on $(-3,-1)$ On $(-\infty, -3)$, f(x) achieves a maximum value of -8 at x=−3 and on $(-1,+\infty)$ a minimum value of -12 at x=-1.
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