Calculus (3rd Edition)

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

Chapter 4 - Applications of the Derivative - 4.3 The Mean Value Theorem and Monotonicity - Exercises - Page 189: 28


$f(x) $ is decreasing on $(-\infty,-0.6) $ and increasing on $(-0.6,\infty) $ . $x= -0.6$ is a local minimum

Work Step by Step

Given $$y=5 x^{2}+6 x-4$$ Since $$f'(x) = 10x+6$$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 10x+6&=0 \end{align*} Then $x= -0.6$ is a critical point. To find the interval where $f$ is increasing and decreasing, choose $x=0$, $x= -1$ \begin{align*} f'(0)&= 6>0\\ f'(-1)&= -4<0 \end{align*} Hence, $f(x) $ is decreasing on $(-\infty,-0.6) $ and increasing on $(-0.6,\infty) $. Since $f(x)$ changes from decreasing to increasing, then $x= -0.6$ is a local minimum.
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