## Calculus (3rd Edition)

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