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

Let \( E(x) = \frac{4}{x+1} - \frac{1}{x} \) To find the maximum of \( E(x) \), we first find its derivative. \[ E’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set the derivative equal to zero to find critical points. \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] Rearrange the equation. \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} \] Take the square root of both sides. \[ \frac{1}{x} = \frac{2}{x+1} \] Solve for \( x \). \[ x + 1 = 2x \] \[ 1 = x \] Substitute \( x = 1 \) back into \( E(x) \) to find the maximum value. \[ E(1) = \frac{4}{1+1} - \frac{1}{1} = 2 - 1 = 1 \] The maximum value of the expression is \( 1 \).