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

To find the maximum of the expression \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] we need to take the derivative and find the critical points. First, compute the derivative \( f’(x) \). The derivative of \( \frac{4}{x+1} \) is \[ -\frac{4}{(x+1)^2} \] The derivative of \( -\frac{1}{x} \) is \[ \frac{1}{x^2} \] So, the derivative \( f’(x) \) is \[ f’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set \( f’(x) = 0 \) to find critical points: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} \] Cross-multiply to solve: \[ (x+1)^2 = 4x^2 \] Expand the left side: \[ x^2 + 2x + 1 = 4x^2 \] Rearrange the terms: \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = -2 \), \( c = -1 \). \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 3 \times (-1)}}{2 \times 3} \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \] \[ x = \frac{2 \pm \sqrt{16}}{6} \] \[ x = \frac{2 \pm 4}{6} \] \[ x = 1 \quad \text{or} \quad x = -\frac{1}{3} \] Since \( x > 0 \), we only consider \( x = 1 \). Evaluate \( f(x) \) at the critical point \( x = 1 \): \[ f(1) = \frac{4}{1+1} - \frac{1}{1} \] \[ f(1) = \frac{4}{2} - 1 \] \[ f(1) = 2 - 1 = 1 \] Thus, the maximum value of the expression is 1.