## Calculus (3rd Edition)

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