#### Answer

$f(x)$ is increasing on $\left( \left( \dfrac{3}{2}\right)^{2/5},\infty\right )$ and decreasing on $\left( -\infty, \left(\dfrac{3}{2}\right)^{2/5}\right )$.
$f(x)$ has local a minimum at $x=\left( \dfrac{3}{2}\right)^{2/5} $

#### Work Step by Step

Given $$y=x^{4}-4 x^{3 / 2} \quad(x>0)$$
Since
$$f'(x) = 4x^{3}-6 x^{1 / 2} =2x^{1/2} (2x^{5/2}-3)$$
Then $f'(x)=0$ for $x= 0 $ and $x= \left( \dfrac{3}{2}\right)^{2/5}$. Since $x=0$ is not in the interval, choose $x=1 $ and $x=2$. Then:
\begin{align*}
f'(1)&= -2<0\\
f'(2)&>0
\end{align*}
Then $f(x)$ is increasing on $\left( \left( \dfrac{3}{2}\right)^{2/5},\infty\right )$ and decreasing on $\left( -\infty, \left(\dfrac{3}{2}\right)^{2/5}\right )$. Hence, $f(x)$ has a local minimum at $x=\left( \dfrac{3}{2}\right)^{2/5} $