## Calculus (3rd Edition)

$f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-1$
Given $$f(x)=x^{3}+x^{-3}$$ Since $$f'(x) =3 x^{2}-3 x^{-4}$$ Then $f(x)$ has critical points when \begin{align*} f'(x)&=0\\ 3 x^{2}-3 x^{-4} &=0\\ \frac{3}{x^{4}}\left(x^{6}-1\right)&=0\\ x^{6}-1&=0 \end{align*} Then $x= -1$ and $x=1$ are critical points. To find the intervals where the function is increasing and decreasing, choose $x=-2,\ x= 0$ and $x=2$ \begin{align*} f'( -2)&>0 \\ f'(0)& <0\\ f'(2)&>0 \end{align*} Hence, by using the first derivative test, $f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-1$.