Question: Given \( x > 0 \), find the maximum of \(\left( \frac{4}{x + 1} - \frac{1}{x} \right)\).

Given \( x > 0 \), find the maximum of \(\left( \frac{4}{x + 1} - \frac{1}{x} \right)\).

Solution

Find the maximum value of the expression \(\left(\frac{4}{x+1} - \frac{1}{x}\right)\) for \(x > 0\). \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] Calculate the derivative of \(f(x)\): \[ f’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set the derivative equal to zero to find critical points: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] Rearrange the equation: \[ \frac{4}{(x+1)^2} = \frac{1}{x^2} \] Cross-multiply to solve for \(x\): \[ 4x^2 = (x+1)^2 \] 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: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \] This gives two solutions: \[ x = 1 \quad \text{or} \quad x = -\frac{1}{3} \] Since \(x > 0\), the valid solution is: \[ x = 1 \] Evaluate \(f(x)\) at \(x = 1\): \[ f(1) = \frac{4}{1+1} - \frac{1}{1} = \frac{4}{2} - 1 = 2 - 1 = 1 \] The maximum value of the expression is \(1\).