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 expression \[ \frac{3x - 1}{x(x + 1)} \] given \(x > 0\), we can use the basic inequality. First, let’s simplify and rewrite the expression for easier handling: \[ y = \frac{3x - 1}{x^2 + x} \] Our goal is to find the critical points by taking the derivative and setting it to zero. Differentiate \(y\) with respect to \(x\): \[ \frac{d}{dx}\left(\frac{3x - 1}{x^2 + x}\right) = \frac{(x^2 + x)(3) - (3x - 1)(2x + 1)}{(x^2 + x)^2} \] Simplify the numerator: \[ (3x^2 + 3x) - (6x^2 + 3x - 2x - 1) = 3x^2 + 3x - 6x^2 - 3x + 2x + 1 \] \[ = -3x^2 + 2x + 1 \] Set the numerator equal to zero to find critical points: \[ -3x^2 + 2x + 1 = 0 \] Solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = -3\), \(b = 2\), \(c = 1\). \[ x = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{-6} \] \[ x = \frac{-2 \pm \sqrt{16}}{-6} \] \[ x = \frac{-2 \pm 4}{-6} \] Calculate the roots: For \(x = \frac{2}{-6} = -\frac{1}{3}\), which is not applicable as \(x > 0\). For \(x = \frac{-6}{-6} = 1\). Since the only relevant critical point in the domain \(x > 0\) is at \(x = 1\), we evaluate the expression at that point: \[ y(1) = \frac{3(1) - 1}{1 \cdot (1 + 1)} = \frac{2}{2} = 1 \] Thus, the maximum value of \(\frac{3x - 1}{x(x + 1)}\) for \(x > 0\) is 1.