Answer
$f(x)$ increasing on $\left( 1,\infty\right )$ and decreasing on $\left(0, 1\right )$. $f(x)$ has a local minimum at $x= 1$
Work Step by Step
Given $$y=x+x^{-1} \quad(x>0)$$
Since
$$f'(x) =1- \frac{1}{x^2} =\frac{x^2-1}{x^2} $$
Then $f'(x)=0$ for $x= 0 $ and $x=\pm1$. Since $x=0,\ x=-1$ is not in the interval, choose $x=0.5 $ and $x=2$. Then:
\begin{align*}
f'(0.5)& <0\\
f'(2)&>0
\end{align*}
Thus $f(x)$ is increasing on $\left( 1,\infty\right )$ and decreasing on $\left(0, 1\right )$. Hence, $f(x)$ has a local minimum at $x=1$.
