Let's solve the inequality: \[ |x+3| - |x-2| - |x-1| \leq 4 \] To solve this, consider the critical points where the expressions inside the absolute values are zero: \(x = -3\), \(x = 1\), and \(x = 2\). These points break the number line into intervals. Let's analyze each interval: ### 1. \(x < -3\) - \(x+3 < 0 \implies |x+3| = -(x+3)\) - \(x-2 < 0 \implies |x-2| = -(x-2)\) - \(x-1 < 0 \implies |x-1| = -(x-1)\) Plug these into the inequality: \[ -(x+3) - (-(x-2)) - (-(x-1)) \leq 4 \\ -(x+3) + (x-2) + (x-1) \leq 4 \\ -x-3 + x-2 + x-1 \leq 4 \\ (-x + x + x) + (-3 - 2 - 1) \leq 4 \\ x - 6 \leq 4 \\ x \leq 10 \] But, in this interval, \(x < -3\), so the solution here is \(x < -3\). --- ### 2. \(-3 \leq x < 1\) - \(x+3 \geq 0 \implies |x+3| = x+3\) - \(x-2 < 0 \implies |x-2| = -(x-2)\) - \(x-1 < 0 \implies |x-1| = -(x-1)\) Plug these into the inequality: \[ (x+3) - (-(x-2)) - (-(x-1)) \leq 4 \\ (x+3) + (x-2) + (x-1) \leq 4 \\ x+3 + x-2 + x-1 \leq 4 \\ (x + x + x) + (3 - 2 - 1) \leq 4 \\ 3x + 0 \leq 4 \\ 3x \leq 4 \\ x \leq \frac{4}{3} \] But since \(x < 1\) in this interval, the solution is \(-3 \leq x < 1\). --- ### 3. \(1 \leq x < 2\) - \(x+3 \geq 0 \implies |x+3| = x+3\) - \(x-2 < 0 \implies |x-2| = -(x-2)\) - \(x-1 \geq 0 \implies |x-1| = x-1\) Plug these into the inequality: \[ (x+3) - (-(x-2)) - (x-1) \leq 4 \\ (x+3) + (x-2) - (x-1) \leq 4 \\ x+3 + x-2 - x + 1 \leq 4 \\ (x + x - x) + (3 - 2 + 1) \leq 4 \\ x + 2 \leq 4 \\ x \leq 2 \] But, in this interval, \(1 \leq x < 2\), so the solution here is \(1 \leq x < 2\). --- ### 4. \(x \geq 2\) - \(x+3 \geq 0 \implies |x+3| = x+3\) - \(x-2 \geq 0 \implies |x-2| = x-2\) - \(x-1 \geq 0 \implies |x-1| = x-1\) Plug these into the inequality: \[ (x+3) - (x-2) - (x-1) \leq 4 \\ x+3 - x + 2 - x + 1 \leq 4 \\ (x - x - x) + (3 + 2 + 1) \leq 4 \\ -x + 6 \leq 4 \\ -x \leq -2 \\ x \geq 2 \] But in this interval, \(x \geq 2\), so the solution is \(x \geq 2\). --- ### Combine All Intervals - For \(x < -3\): always true - For \(-3 \leq x < 1\): always true - For \(1 \leq x < 2\): always true - For \(x \geq 2\): always true **Therefore, the inequality is satisfied for all real numbers**: \[ \boxed{(-\infty, \infty)} \] **All real values of \(x\) satisfy the inequality.**