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: 34


$f(x) $ is decreasing on $(-\infty,-3/4) $ and increasing on $ ( -3/4,0) \cup (0 ,\infty) $. $f(x)$ has a local minimum at $x= -3/4$.

Work Step by Step

Given $$y=x^{4}+x^{3}$$ Since $$f'(x) =4x^{3}+3x^2 $$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 4x^{3}+3x^2 &=0\\ x^2(4x +3)&=0 \end{align*} Then $x= 0$ and $x=-3/4$ are critical points. To find the intervals where $f$ is increasing and decreasing, choose $x=-1,\ x= -0.5$ and $x= 1$ \begin{align*} f'( -1)&<0 \\ f'(-0.5)& >0\\ f'(1)&>0 \end{align*} Hence, $f(x) $ is decreasing on $(-\infty,-3/4) $ and increasing on $ ( -3/4,0) \cup (0 ,\infty) $. Hence, by using the first derivative test, $f(x)$ has a local minimum at $x= -3/4$. $f(0)$ is a point of inflection.
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