Question: Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{4}{x + 1} - \frac{1}{x}…

Given \( x > 0 \), use the basic inequality to find \[ \left( \frac{4}{x + 1} - \frac{1}{x} \right)_{\text{max}} \]

Solution

To find the maximum value of the expression \( \frac{4}{x+1} - \frac{1}{x} \) for \( x > 0 \) using the basic inequality method, follow these steps: Consider the expression: \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] Combine the terms over a common denominator: \[ f(x) = \frac{4x - (x + 1)}{x(x + 1)} = \frac{3x - 1}{x(x + 1)} \] To find the maximum, analyze the behavior of the function as \( x \) varies. Take the derivative of \( f(x) \) with respect to \( x \): \[ f’(x) = \frac{(3)(x)(x + 1) - (3x - 1)(2x + 1)}{(x(x + 1))^2} \] Set the derivative equal to zero to find critical points: \[ 3x(x + 1) - (3x - 1)(2x + 1) = 0 \] 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 through by -1: \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation: \[ x = \frac{2 \pm \sqrt{(2)^2 - 4(3)(-1)}}{2(3)} = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6} \] This gives two solutions: \[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \] Since \( x > 0 \), consider \( x = 1 \). Evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = \frac{4}{1 + 1} - \frac{1}{1} = \frac{4}{2} - 1 = 2 - 1 = 1 \] Check the behavior as \( x \) approaches 0 and infinity: \[ \lim_{x \to 0^+} f(x) = \frac{4}{1} - \infty = -\infty \] \[ \lim_{x \to \infty} f(x) = 0 - 0 = 0 \] Therefore, the maximum value of \( f(x) \) is \( 1 \) at \( x = 1 \).