Question: Given \( x > 0 \), find the maximum of the expression \(\frac{4}{x + 1} - \frac{1}{x}\).

Given \( x > 0 \), find the maximum of the expression \(\frac{4}{x + 1} - \frac{1}{x}\).

Solution

To find the maximum of the expression given by \[ f(x) = \frac{4}{x+1} - \frac{1}{x} \] we first need to find the derivative of \( f(x) \) and set it equal to zero to find critical points. The derivative of \( f(x) \) is: \[ f’(x) = -\frac{4}{(x+1)^2} + \frac{1}{x^2} \] Set \( f’(x) = 0 \) to find critical points: \[ -\frac{4}{(x+1)^2} + \frac{1}{x^2} = 0 \] Simplify and solve for \( x \): \[ \frac{1}{x^2} = \frac{4}{(x+1)^2} \] Cross-multiply to get: \[ (x+1)^2 = 4x^2 \] Expand and rearrange the equation: \[ x^2 + 2x + 1 = 4x^2 \] \[ 3x^2 - 2x - 1 = 0 \] Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 3 \), \( b = -2 \), and \( c = -1 \). \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \] \[ x = \frac{2 \pm \sqrt{4 + 12}}{6} \] \[ x = \frac{2 \pm \sqrt{16}}{6} \] \[ x = \frac{2 \pm 4}{6} \] This gives us \[ x = 1 \quad \text{and} \quad x = -\frac{1}{3} \] Since \( x > 0 \), we only consider \( x = 1 \). Check the second derivative to ensure it’s a maximum: \[ f”(x) = \frac{8}{(x+1)^3} - \frac{2}{x^3} \] Plug \( x = 1 \) into \( f”(x) \): \[ f”(1) = \frac{8}{(1+1)^3} - \frac{2}{1^3} = \frac{8}{8} - 2 = 1 - 2 = -1 \] Since \( f”(1) < 0 \), there is a local maximum at \( x = 1 \). Thus, evaluate \( f(x) \) at \( x = 1 \): \[ f(1) = \frac{4}{1+1} - \frac{1}{1} = \frac{4}{2} - 1 = 2 - 1 = 1 \] The maximum value of the expression is 1.