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.
