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

To solve the problem, we need to find the maximum of the function: \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] First, find the derivative of \(f(x)\) with respect to \(x\). \[ f’(x) = \frac{d}{dx}\left(\frac{4}{x+1} - \frac{1}{x}\right) \] Differentiate each term: The derivative of \(\frac{4}{x+1}\) using the quotient rule: \[ \frac{d}{dx}\left(\frac{4}{x+1}\right) = -\frac{4}{(x+1)^2} \] The derivative of \(-\frac{1}{x}\): \[ \frac{d}{dx}\left(-\frac{1}{x}\right) = \frac{1}{x^2} \] Thus, the derivative of \(f(x)\) is: \[ f’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set \(f’(x) = 0\) to find critical points: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] Simplify and solve for \(x\): \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} \] Cross-multiply: \[ (x+1)^2 = 4x^2 \] Expand and rearrange: \[ x^2 + 2x + 1 = 4x^2 \] \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation: The quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For \(3x^2 - 2x - 1 = 0\), \(a = 3\), \(b = -2\), \(c = -1\): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \] \[ x = \frac{2 \pm \sqrt{16}}{6} \] \[ x = \frac{2 \pm 4}{6} \] Calculate the roots: \[ x = 1 \quad \text{or} \quad x = -\frac{1}{3} \] Since \(x > 0\), take \(x = 1\). Evaluate \(f(x)\) at \(x = 1\): \[ f(1) = \frac{4}{1+1} - \frac{1}{1} \] \[ f(1) = \frac{4}{2} - 1 \] \[ f(1) = 2 - 1 \] \[ f(1) = 1 \] Thus, the maximum value is 1.