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: 35

Answer

$f(x)$ is increasing on $(-\infty,0)\cup (0,\infty)$ and $f(x)$ has neither a local minimum nor a local maximum.

Work Step by Step

Given $$y=x^{5}+x^{3}+1$$ Since $$f'(x) = 5x^4+3x^2 = x^2( 5x^2+3) $$ Then $f'(x)=0$ for $x=0$. Choose $x= -1$ and $x=1$. We get: \begin{align*} f'(-1)&=8>0\\ f'(1)&=8>0 \end{align*} Hence, $f(x)$ is increasing on $(-\infty,0)\cup (0,\infty)$ and $f(x)$ has neither a local minimum nor a local maximum.
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