Question: Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{3x - 1}{x(x + 1)}…

Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{3x - 1}{x(x + 1)} \right)_{\text{max}} \]

Solution

To find the maximum value of the expression \(\frac{3x - 1}{x(x + 1)}\) for \(x > 0\), we can use calculus. First, let \(f(x) = \frac{3x - 1}{x(x + 1)}\). Find the derivative \(f’(x)\) using the quotient rule, which states that if \(f(x) = \frac{u(x)}{v(x)}\), then \[ f’(x) = \frac{u’(x)v(x) - u(x)v’(x)}{v(x)^2}. \] Identify \(u(x) = 3x - 1\) and \(v(x) = x(x + 1)\). Calculate \(u’(x)\) and \(v’(x)\): \(u’(x) = 3\). \(v(x) = x^2 + x\). \(v’(x) = 2x + 1\). Apply the quotient rule: \[ f’(x) = \frac{3(x^2 + x) - (3x - 1)(2x + 1)}{(x^2 + x)^2} \] Simplify the numerator: \[ = \frac{3x^2 + 3x - (6x^2 + 5x - 1)}{(x^2 + x)^2} \] \[ = \frac{3x^2 + 3x - 6x^2 - 5x + 1}{(x^2 + x)^2} \] \[ = \frac{-3x^2 - 2x + 1}{(x^2 + x)^2} \] Set the derivative \(f’(x)\) to zero to find critical points: \[ -3x^2 - 2x + 1 = 0 \] Solve this quadratic equation: \[ 3x^2 + 2x - 1 = 0 \] Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 2\), \(c = -1\). \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{6} \] \[ x = \frac{-2 \pm \sqrt{16}}{6} \] \[ x = \frac{-2 \pm 4}{6} \] \[ x = \frac{2}{6} \text{ or } x = \frac{-6}{6} \] \[ x = \frac{1}{3} \text{ or } x = -1 \] Since \(x > 0\), the only critical point is \(x = \frac{1}{3}\). Evaluate \(f(x)\) at \(x = \frac{1}{3}\): \[ f\left(\frac{1}{3}\right) = \frac{3 \cdot \frac{1}{3} - 1}{\frac{1}{3}\left(\frac{1}{3} + 1\right)} \] \[ = \frac{1 - 1}{\frac{1}{3} \cdot \frac{4}{3}} \] \[ = \frac{0}{\frac{4}{9}} \] \[ = 0 \] Therefore, the maximum value is 0.