Question: Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{4}{x + 1} - \frac{1}{x}…

Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{4}{x + 1} - \frac{1}{x} \right)_{\text{max}} \]

Solution

Step 1: Let us consider the function to maximize. \[ f(x) = \frac{4}{x + 1} - \frac{1}{x} \] Step 2: Find the derivative of \( f(x) \) with respect to \( x \). \[ f’(x) = -\frac{4}{(x + 1)^2} + \frac{1}{x^2} \] Step 3: Set the derivative equal to zero to find critical points. \[ -\frac{4}{(x + 1)^2} + \frac{1}{x^2} = 0 \] Step 4: Solve for \( x \). \[ \frac{1}{x^2} = \frac{4}{(x + 1)^2} \] \[ (x + 1)^2 = 4x^2 \] \[ x^2 + 2x + 1 = 4x^2 \] \[ 3x^2 - 2x - 1 = 0 \] Step 5: Solve the quadratic equation. \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \] \[ x = 1 \quad \text{or} \quad x = -\frac{1}{3} \] Since \( x > 0 \), we have \( x = 1 \). Step 6: Substitute \( x = 1 \) into \( f(x) \). \[ f(1) = \frac{4}{2} - \frac{1}{1} = 2 - 1 = 1 \] Hence, the maximum value of the expression is \( 1 \).