Graph and image of |x-1|-|x+1|

Problem

Sketch qualitatively the graph of $f(x)=|x-1|-|x+1|$ and determine its image.

Solution

The absolute values are removed by reasoning by cases on the critical points $x=-1$ and $x=1$: $f(x)=\begin{cases} (1-x)-(-(x+1)) = 2 & x\le -1, \\ (1-x)-(x+1) = -2x & -1<x<1, \\ (x-1)-(x+1) = -2 & x\ge 1. \end{cases}$ The graph is: a constant piece at height $2$ for $x\le -1$, a segment of slope $-2$ descending from $(-1,2)$ to $(1,-2)$, and a constant piece at height $-2$ for $x\ge 1$. The function is continuous.

The values taken range from $-2$ to $2$: image $=\boxed{[-2;\,2]}$.

Links

Topics: Functions and properties
Concepts: Piecewise function · Image · Absolute value
Functions: Absolute value
Skills: Reasoning by cases · Plotting a graph
Exercise types: Study of a function