Question

Let $f(x) = \frac{1}{3\sqrt{x}}(\frac{2}{x} - 3), x > 0$. Then $f(x)$ is decreasing in}

• A. $(1,4)$
• B. $(0,3)$
• C. $(1,3)$
• D. $(2,5)$
• E. $(0,2)$

Hint:

Converting rational expressions with square roots into power notation (\( x^n \)) before differentiating avoids the quotient rule and makes the inequality much simpler to solve.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
A function is decreasing in an interval if its first derivative is negative (\( f'(x) < 0 \)) in that interval.

Step 2: Detailed Explanation:
Given \( f(x) = \frac{1}{3\sqrt{x}} \left( \frac{2}{x} - 3 \right) \).
Rewrite the function in power form for easier differentiation:
\[ f(x) = \frac{2}{3x\sqrt{x}} - \frac{3}{3\sqrt{x}} = \frac{2}{3}x^{-3/2} - x^{-1/2} \]
Now, differentiate with respect to $x$:
\[ f'(x) = \frac{2}{3} \left( -\frac{3}{2} \right) x^{-5/2} - \left( -\frac{1}{2} \right) x^{-3/2} \]
\[ f'(x) = -x^{-5/2} + \frac{1}{2} x^{-3/2} \]
For the function to be decreasing, we require \( f'(x) < 0 \):
\[ -x^{-5/2} + \frac{1}{2} x^{-3/2} < 0 \]
\[ \frac{1}{2} x^{-3/2} < x^{-5/2} \]
Multiply both sides by \( x^{5/2} \) (which is positive for $x > 0$):
\[ \frac{1}{2} x^{5/2-3/2} < 1 \]
\[ \frac{1}{2} x^1 < 1 \implies x < 2 \]
Since the domain is given as $x > 0$, the interval is $(0, 2)$.

Step 3: Final Answer:
The function $f(x)$ is decreasing in the interval $(0,2)$.