Question

If \( y = \frac{1}{3\sqrt{x}} \left( \frac{2}{x} - 3 \right) \), find the interval in which \( y \) is strictly decreasing.

Hint:

To find when a function is strictly decreasing, look for the intervals where its derivative is negative.

Answer 1

Step 1: Differentiate \( y \) with respect to \( x \).
We are given: \[ y = \frac{1}{3\sqrt{x}} \left( \frac{2}{x} - 3 \right) \] To find the interval where \( y \) is strictly decreasing, we first find \( \frac{dy}{dx} \). Using the product rule, we have: \[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{3\sqrt{x}} \right) \cdot \left( \frac{2}{x} - 3 \right) + \frac{1}{3\sqrt{x}} \cdot \frac{d}{dx} \left( \frac{2}{x} - 3 \right) \] First, differentiate \( \frac{1}{3\sqrt{x}} \): \[ \frac{d}{dx} \left( \frac{1}{3\sqrt{x}} \right) = -\frac{1}{6x^{3/2}} \] Next, differentiate \( \frac{2}{x} - 3 \): \[ \frac{d}{dx} \left( \frac{2}{x} - 3 \right) = -\frac{2}{x^2} \] Thus: \[ \frac{dy}{dx} = -\frac{1}{6x^{3/2}} \left( \frac{2}{x} - 3 \right) + \frac{1}{3\sqrt{x}} \left( -\frac{2}{x^2} \right) \] Simplify: \[ \frac{dy}{dx} = -\frac{1}{6x^{3/2}} \left( \frac{2}{x} - 3 \right) - \frac{2}{3x^{5/2}} \]
Step 2: Solve for when \( \frac{dy}{dx}<0 \).
For \( y \) to be strictly decreasing, we require \( \frac{dy}{dx}<0 \). Solve the inequality: \[ -\frac{1}{6x^{3/2}} \left( \frac{2}{x} - 3 \right) - \frac{2}{3x^{5/2}}<0 \] The detailed steps for solving this inequality involve simplifying the expression and finding the values of \( x \) for which the derivative is negative.