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 \(f(x)\) is strictly decreasing on an interval if its first derivative \(f'(x)<0\) for all \(x\) in that interval.
Step 2: Key Formula or Approach:
Distribute the terms in \(f(x)\) to turn it into a simple power rule differentiation problem.
Find \(f'(x)\), set up the inequality \(f'(x)<0\), and solve for \(x\).
Step 3: Detailed Explanation:
Simplify \(f(x)\):
\[ f(x) = \frac{1}{3}x^{-1/2} (2x^{-1} - 3) \] \[ f(x) = \frac{2}{3}x^{-3/2} - x^{-1/2} \] Differentiate with respect to \(x\):
\[ f'(x) = \frac{2}{3}\left(-\frac{3}{2}\right)x^{-5/2} - \left(-\frac{1}{2}\right)x^{-3/2} \] \[ f'(x) = -x^{-5/2} + \frac{1}{2}x^{-3/2} \] Factor out the common term with the lowest power (\(x^{-5/2}\)):
\[ f'(x) = x^{-5/2} \left( -1 + \frac{1}{2}x^1 \right) \] \[ f'(x) = \frac{1}{x^{5/2}} \left( \frac{x - 2}{2} \right) = \frac{x - 2}{2x^{5/2}} \] For the function to be decreasing, \(f'(x)<0\):
\[ \frac{x - 2}{2x^{5/2}}<0 \] Since \(x>0\) (from the domain), the denominator \(2x^{5/2}\) is always positive.
Therefore, the numerator must be negative:
\[ x - 2<0 \implies x<2 \] Combining this with the domain restriction \(x>0\), we get:
\[ 0<x<2 \] Step 4: Final Answer:
The function decreases in \((0, 2)\).