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

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

Solution

Define the function: \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] Find 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: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 & \] Solve for \( x \): \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} & \] Cross-multiply the equation: \[ (x+1)^2 = 4x^2 & \] Expand and simplify: \[ x^2 + 2x + 1 = 4x^2 & \] \[ 3x^2 - 2x - 1 = 0 & \] Use the quadratic formula to solve for \( x \): \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6} & \] Since \( x > 0 \), the valid solution is: \[ x = 1 & \] Evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = \frac{4}{2} - \frac{1}{1} = 2 - 1 = 1 & \] Therefore, the maximum value of the expression is \( 1 \) when \( x = 1 \).