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 \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] first, take the derivative of \(f(x)\) with respect to \(x\): \[ f’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set the derivative equal to zero to find critical points: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] Solving the equation: \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} \] Cross-multiply: \[ (x+1)^2 = 4x^2 \] Expand and simplify: \[ x^2 + 2x + 1 = 4x^2 \] \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 3\), \(b = -2\), and \(c = -1\). \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \] \[ x = \frac{2 \pm \sqrt{16}}{6} \] \[ x = \frac{2 \pm 4}{6} \] The possible solutions are: \[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \] Since \(x > 0\), take \(x = 1\). To ensure it is a maximum, check the second derivative: \[ f”(x) = \frac{8}{(x+1)^3} - \frac{2}{x^3} \] Evaluate \(f”(x)\) at \(x = 1\): \[ f”(1) = \frac{8}{2^3} - \frac{2}{1^3} \] \[ f”(1) = \frac{8}{8} - 2 \] \[ f”(1) = 1 - 2 \] \[ f”(1) = -1 \] Since \(f”(1) < 0\), \(x = 1\) is a maximum. The maximum value of the expression is \[ f(1) = \frac{4}{1+1} - \frac{1}{1} = 2 - 1 = 1 \]