To solve the inequality \( f(x) = |x-1| + |x-2| \geq 4 \), we analyze the function by considering the critical points where the expressions inside the absolute values change their behavior. The critical points are at \( x = 1 \) and \( x = 2 \). We divide the real number line into three intervals based on these points:
1. For \( x < 1 \): \[ f(x) = -(x-1) + -(x-2) = -x + 1 - x + 2 = -2x + 3 \] The inequality becomes: \[ -2x + 3 \geq 4 \implies -2x \geq 1 \implies x \leq -\frac{1}{2} \] So, \( x \in (-\infty, -\frac{1}{2}] \).
2. For \( 1 \leq x < 2 \): \[ f(x) = (x-1) + -(x-2) = x - 1 - x + 2 = 1 \] The inequality \( 1 \geq 4 \) does not hold. Therefore, there are no solutions in this interval.
3. For \( x \geq 2 \): \[ f(x) = (x-1) + (x-2) = x - 1 + x - 2 = 2x - 3 \] The inequality becomes: \[ 2x - 3 \geq 4 \implies 2x \geq 7 \implies x \geq \frac{7}{2} \] So, \( x \in [\frac{7}{2}, \infty) \).
Combining the solutions from the intervals where the inequality holds, we get: \[ x \in (-\infty, -\frac{1}{2}] \cup [\frac{7}{2}, \infty) \] This is the solution set for the inequality \( f(x) \geq 4 \).