#### Answer

$f(x)$ increasing on $\left( \dfrac{16}{25} ,\infty\right )$ and decreasing on $\left(0, \dfrac{16}{25}\right )$. $f(x)$ has a local minimum at $x= \dfrac{16}{25}$.

#### Work Step by Step

Given $$y=x^{5 / 2}-x^{2} \quad(x>0)$$
Since
$$f'(x) = \frac{5}{2}x^{3/2}-2 x = x ((5/2)x^{1/2}-2)$$
Then $f'(x)=0$ for $x= 0 $ and $x= \dfrac{16}{25}$. Since $x=0$ is not in the interval, choose $x=0.5 $ and $x=1$. Then
\begin{align*}
f'(0.5)& <0\\
f'(1)&>0
\end{align*}
Thus $f(x)$ is increasing on $\left( \dfrac{16}{25} ,\infty\right )$ and decreasing on $\left(0, \dfrac{16}{25}\right )$. Hence, $f(x)$ has a local minimum at $x= \dfrac{16}{25}$.