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 function \( f(x) = \frac{3x - 1}{x(x + 1)} \) for \( x > 0 \), follow these steps: Step 1: Simplify the function. \[ f(x) = \frac{3x - 1}{x(x + 1)} \] Step 2: Take the derivative of \( f(x) \) with respect to \( x \) to find critical points. \[ f’(x) = \frac{d}{dx} \left( \frac{3x - 1}{x(x + 1)} \right) \] Step 3: Apply the quotient rule for differentiation. \[ f’(x) = \frac{(3)(x^2 + x) - (3x - 1)(2x + 1)}{(x^2 + x)^2} \] Step 4: Simplify the numerator. \[ 3x^2 + 3x - (6x^2 + 3x - 2x - 1) = -3x^2 + 2x + 1 \] Step 5: Set the derivative equal to zero to find critical points. \[ -3x^2 + 2x + 1 = 0 \] Step 6: Multiply both sides by -1 to simplify. \[ 3x^2 - 2x - 1 = 0 \] Step 7: Solve the quadratic equation using the quadratic formula. \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2 \times 3} = \frac{2 \pm \sqrt{16}}{6} \] Step 8: Find the positive solution since \( x > 0 \). \[ x = \frac{2 + 4}{6} = 1 \] Step 9: Substitute \( x = 1 \) back into the original function to find the maximum value. \[ f(1) = \frac{3(1) - 1}{1(1 + 1)} = \frac{2}{2} = 1 \] The maximum value of the function is 1.