Step 1: Plan:
We will find the critical point of $f$ and then test the sign of the slope on each side of it.
Step 2: Slope function:
Using the power rule on each term,
\[ f^{\prime}(x)=\dfrac{15}{2}x^{1/2}-\dfrac{15}{2}x^{3/2}=\dfrac{15}{2}x^{1/2}(1-x) \]
Step 3: Critical point:
Put $f^{\prime}(x)=0$. Since $x>0$, the factor $x^{1/2}$ is not zero, so $1-x=0$ and $x=1$.
This point splits the domain $(0,\infty)$ into $(0,1)$ and $(1,\infty)$.
Step 4: Test points:
Take $x=\dfrac{1}{4}$ in $(0,1)$: $f^{\prime}\left(\dfrac{1}{4}\right)=\dfrac{15}{2}\cdot\dfrac{1}{2}\cdot\dfrac{3}{4}=\dfrac{45}{16}>0$. So $f$ rises here.
Take $x=4$ in $(1,\infty)$: $f^{\prime}(4)=\dfrac{15}{2}\cdot 2\cdot(-3)=-45<0$. So $f$ falls here.
Final Answer:
The function increases only on the interval from 0 to 1. \[ \boxed{0<x<1} \]