Question: Given \( x > 0 \), find the maximum of \[ \left( \frac{3x - 1}{x(x + 1)} \right) \] with the…

Given \( x > 0 \), find the maximum of \[ \left( \frac{3x - 1}{x(x + 1)} \right) \] with the basic inequality.

Solution

To find the maximum of the function \( f(x) = \frac{3x - 1}{x(x + 1)} \) for \( x > 0 \). \[ f(x) = \frac{3x - 1}{x^2 + x} & \] Take the derivative of \( f(x) \): \[ f’(x) = \frac{(3)(x^2 + x) - (3x - 1)(2x + 1)}{(x^2 + x)^2} & \] Set the derivative equal to zero to find critical points: \[ 3(x^2 + x) - (3x - 1)(2x + 1) = 0 & \] Expand and simplify: \[ 3x^2 + 3x - (6x^2 + 3x - 2x - 1) = 0 & \] \[ 3x^2 + 3x - 6x^2 - x + 1 = 0 & \] \[ -3x^2 + 2x + 1 = 0 & \] Multiply both sides by -1: \[ 3x^2 - 2x - 1 = 0 & \] Solve the quadratic equation: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} & \] Since \( x > 0 \), we take the positive solution: \[ x = 1 & \] Evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = \frac{3(1) - 1}{1(1 + 1)} = \frac{2}{2} = 1 & \] Thus, the maximum value of the function is \( 1 \).