Q.Find the intervals in which the function $f(x) = 5x^{\frac{3}{2}} - 3x^{\frac{5}{2}}$ is

Concept: Monotonic Function Analysis — we find $f'(x)$ and check its sign.

Step 1: Domain

Since $x^{3/2}$ and $x^{5/2}$ require $x \ge 0$, the domain is $[0, \infty)$.

Step 2: Differentiate

$f'(x) = 5 \cdot \frac{3}{2} x^{1/2} - 3 \cdot \frac{5}{2} x^{3/2} = \frac{15}{2} x^{1/2} - \frac{15}{2} x^{3/2}$

Factor: $f'(x) = \frac{15}{2} x^{1/2} (1 - x)$.

Step 3: Sign analysis

For $x > 0$, $x^{1/2} > 0$, so sign of $f'(x)$ depends on $(1 - x)$.

• $f'(x) > 0$ when $1 - x > 0 \implies 0 < x < 1$ → increasing. …

We are given $f(x) = 5x^{3/2} - 3x^{5/2}$. The first thing to notice is the domain: because of the half-powers, $x$ cannot be negative. So we work with $x \ge 0$.

To find where a function increases or decreases, we look at its derivative. If $f'(x) > 0$, the function is increasing; if $f'(x) < 0$, it is decreasing. The critical points (where $f'(x) = 0$ or is undefined) split the domain into intervals we can test.

Let’s differentiate carefully.

1. Differentiate $f(x)$. Write $f(x) = 5x^{3/2} - 3x^{5/2}$. Using the power rule $\frac{d}{dx} x^n = n x^{n-1}$:

Factor out the common term $\frac{15}{2} x^{1/2}$:

2. Find critical points. The derivative $f'(x)$ is defined for all $x > 0$ (at $x=0$, $x^{1/2}=0$ so $f'(0)=0$). Set $f'(x) = 0$:

This gives $x^{1/2} = 0$ or $1 - x = 0$, so $x = 0$ or $x = 1$.

Since the domain is $x \ge 0$, our critical points are $x = 0$ and $x = 1$.

3. Test intervals.

   The domain splits into $(0, 1)$ and $(1, \infty)$. (At $x=0$ itself, the function is at the boundary; we can check behaviour just to the right.)

   Pick a test point in each interval:

   • For $x$ in $(0, 1)$, say $x = 0.25$:

     $x^{1/2} > 0$, $(1 - x) > 0$, so $f'(x) > 0$. Hence $f$ is increasing on $(0, 1)$.

   • For $x$ in $(1, \infty)$, say $x = 4$:

     $x^{1/2} > 0$, $(1 - x) < 0$, so $f'(x) < 0$. Hence $f$ is decreasing on $(1, \infty)$. …