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 \[ \frac{3x - 1}{x(x + 1)} \] under the condition \( x > 0 \), we can use calculus and try to analyze the behavior of the function. First, let’s denote the function as \[ f(x) = \frac{3x - 1}{x(x + 1)} \] We start by finding the derivative of the function. Find the derivative \( f’(x) \) using the quotient rule: \[ f’(x) = \frac{(x(x + 1))(3) - (3x - 1)(2x + 1)}{(x(x + 1))^2} \] Simplify the numerator: - Multiply out each term: \[ 3x(x + 1) = 3x^2 + 3x \] \[ (3x - 1)(2x + 1) = 6x^2 + 3x - 2x - 1 = 6x^2 + x - 1 \] - Subtract the terms: \[ 3x^2 + 3x - (6x^2 + x - 1) = 3x^2 + 3x - 6x^2 - x + 1 = -3x^2 + 2x + 1 \] Now, the derivative is: \[ f’(x) = \frac{-3x^2 + 2x + 1}{(x(x + 1))^2} \] Find critical points by setting \( f’(x) = 0 \): \[ -3x^2 + 2x + 1 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -3 \), \( b = 2 \), \( c = 1 \). \[ x = \frac{-2 \pm \sqrt{2^2 - 4(-3)(1)}}{2(-3)} \] \[ x = \frac{-2 \pm \sqrt{4 + 12}}{-6} \] \[ x = \frac{-2 \pm \sqrt{16}}{-6} \] \[ x = \frac{-2 \pm 4}{-6} \] \[ x_1 = \frac{2}{-6} = -\frac{1}{3}, \quad x_2 = \frac{-6}{-6} = 1 \] Since \( x > 0 \), we only consider \( x = 1 \). Evaluate \( f(x) \) at critical point and endpoints of the domain \( x > 0 \). At \( x = 1 \): \[ f(1) = \frac{3(1) - 1}{1(1 + 1)} = \frac{2}{2} = 1 \] As \( x \rightarrow 0^+ \), the expression tends to negative infinity. As \( x \rightarrow \infty \), the expression tends towards zero. The maximum value of \( f(x) \) is \[ 1 \] Hence, the maximum value is 1.