Question

Determine the interval(s) in which \(f(x)=5x^{3/2}-3x^{5/2}\), \(x>0\) is increasing.

Hint:

Find f'(x) = (15/2) sqrt(x) (1 - x) and see where it is positive for x > 0.

Answer 1

Step 1: Use the test for an increasing function:
A differentiable function is increasing on an interval where \(f^{\prime}(x)>0\). So we find \(f^{\prime}(x)\) and look at its sign for \(x>0\).

Step 2: Differentiate:
\[ f(x)=5x^{3/2}-3x^{5/2} \]
\[ f^{\prime}(x)=5\cdot\dfrac{3}{2}x^{1/2}-3\cdot\dfrac{5}{2}x^{3/2}=\dfrac{15}{2}x^{1/2}-\dfrac{15}{2}x^{3/2} \]

Step 3: Factorise:
Take \(\dfrac{15}{2}x^{1/2}\) common.
\[ f^{\prime}(x)=\dfrac{15}{2}\sqrt{x}\,(1-x) \]

Step 4: Find where the derivative is positive:
For \(x>0\), the factor \(\dfrac{15}{2}\sqrt{x}\) is always positive.
So the sign of \(f^{\prime}(x)\) is the sign of \((1-x)\).
\(1-x>0\) when \(x<1\). Together with \(x>0\), this gives \(0<x<1\).
For \(x>1\), \(f^{\prime}(x)<0\), so the function decreases there.

Final Answer:
\(f\) is increasing on the interval \((0,1)\). \[ \boxed{(0,1)} \]