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

To find the maximum value of the expression \( \frac{4}{x+1} - \frac{1}{x} \) for \( x > 0 \), follow these steps: \[ f(x) = \frac{4}{x + 1} - \frac{1}{x} \] Take the derivative of \( f(x) \) with respect to \( x \): \[ f’(x) = -\frac{4}{(x + 1)^2} + \frac{1}{x^2} \] Set the derivative equal to zero to find critical points: \[ -\frac{4}{(x + 1)^2} + \frac{1}{x^2} = 0 \] Simplify the equation: \[ \frac{4}{(x + 1)^2} = \frac{1}{x^2} \] Cross-multiply to eliminate the denominators: \[ 4x^2 = (x + 1)^2 \] Expand the right side of the equation: \[ 4x^2 = x^2 + 2x + 1 \] Move all terms to one side to form a quadratic equation: \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} \] \[ x = \frac{2 + 4}{6} = 1 \quad (\text{since } x > 0) \] Calculate \( f(1) \) to find the maximum value: \[ f(1) = \frac{4}{1 + 1} - \frac{1}{1} = 2 - 1 = 1 \] Thus, the maximum value of the expression is \( 1 \) when \( x = 1 \).