Question: Given \( x > 0 \), find the value of \( \frac{4}{x + 1} - \frac{1}{x} \leq ? \)

Given \( x > 0 \), find the value of \( \frac{4}{x + 1} - \frac{1}{x} \leq ? \)

Solution

To solve this inequality, we need to find the value such that \[ \frac{4}{x+1} - \frac{1}{x} \leq ? \] First, find a common denominator for the two fractions. The common denominator is \(x(x+1)\). Rewriting the fractions with the common denominator: \[ \frac{4}{x+1} = \frac{4x}{x(x+1)} \] \[ \frac{1}{x} = \frac{x+1}{x(x+1)} \] Subtract the second fraction from the first: \[ \frac{4x}{x(x+1)} - \frac{x+1}{x(x+1)} = \frac{4x - (x+1)}{x(x+1)} \] Simplify the numerator: \[ 4x - (x+1) = 4x - x - 1 = 3x - 1 \] So, the expression becomes: \[ \frac{3x - 1}{x(x+1)} \] Now, find for what this expression is less than or equal to a positive number. The expression is linear in the numerator and increasing, and for large \(x\), it approaches 3. Therefore, the expression is less than 3 for all \(x > 0\). Thus, the inequality is: \[ \frac{4}{x+1} - \frac{1}{x} \leq 3 \] for all \(x > 0\). The question mark is replaced by 3.