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: 24


$f(x)$ has local minima at $x= 2$ and local maxima at $x= -2$

Work Step by Step

Given $$f(x)=x^{3}-12 x-4$$ Since $$f'(x) = 3x^2-12 $$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 3x^2-12 &=0\\ x^2-4&=0 \end{align*} Then $x= -2$ and $x=2$ are critical points. To find the intervals where the function is increasing and decreasing, choose $x=-3,\ x= 0$ and $x=3$ \begin{align*} f'( -3)&>0 \\ f'(0)& <0\\ f'(3)&>0 \end{align*} Hence, by using the first derivative test, $f(x)$ has a local minima at $x= 2$ and a local maxima at $x= -2$.
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