For the function $f(x)=|x-2|-|x+3|$, determine its minimum and maximum values.

Step-by-step solution

1. Find points where expressions inside absolute values change sign: $x=2$ and $x=-3$.
2. Split the real line into three intervals: $(-\infty ,-3]$, $(-3,2]$, and $(2,\infty )$.
3. For $x\le -3$: $|x-2|=-(x-2)=-x+2$, $|x+3|=-(x+3)=-x-3$. Thus $f=(-x+2)-(-x-3)=5$.
4. For $-3<x\le 2$: $|x-2|=-(x-2)=-x+2$, $|x+3|=x+3$. Thus $f=(-x+2)-(x+3)=-2x-1$, a decreasing linear function on this interval.
5. For $x>2$: $|x-2|=x-2$, $|x+3|=x+3$. Thus $f=(x-2)-(x+3)=-5$.
6. Evaluate $f$ on each interval: On $(-\infty ,-3]$ and $(2,\infty )$, $f$ is constantly $5$ and $-5$ respectively.
7. On $(-3,2]$, the maximum occurs at $x=-3$ giving $f=5$, and the minimum occurs at $x=2$ giving $f=-5$.
8. Hence the maximum value of $f$ is $5$ and the minimum value is $-5$.

Concept tested: Range of a piecewise absolute‑value function

Published 13 October 2026 · How this solution was prepared and checked