## Calculus (3rd Edition)

$f(x)$ is increasing on $\left( 0.5,\infty\right )$ and decreasing on $\left(0, 0.5\right )$. $f(x)$ has a local minimum at $x= 0.5$.
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 $x= 0.5$.