Answer
$f(x)$ is increasing on $\left( 0.5,\infty\right )$ and decreasing on $\left(0, 0.5\right )$. $f(x)$ has a local minimum at $x= 0.5$.
Work Step by Step
Given $$y=x^{-2}-4 x^{-1} \quad(x>0)$$
Since
$$f'(x) = -2x^{-3}+4x^{-2} =\frac{4x-2}{x^3} $$
Then $f'(x)=0$ for $x= 0 $ and $x= 0.5$. Since $x=0 $ is not in the interval, choose $x=0.25 $ and $x=1$. Then
\begin{align*}
f'(0.25)& <0\\
f'(1)&>0
\end{align*}
Then, $f(x)$ is increasing on $\left( 0.5,\infty\right )$ and decreasing on $\left(0, 0.5\right )$. Hence, $f(x)$ has a local minimum at $x= 0.5$.
