Answer:
See the attached.
Explanation:
A graph of f' is a graph of the slope of the function. Your function f(x) is piecewise linear, so different sections of its graph have different constant values of slope.
In the intervals (-5, -2) and (0, 2), the slope is -1. (The graph has a "rise" of -1 for each "run" of 1.) So, in those intervals, the graph of f' looks like a graph of y=-1.
In the interval (-2, 0), the rise is 2 for a run of 2, so the slope is 2/2 = 1. The graph of f' in that interval will look like a graph of y=1.
In the interval (2, 5), the rise of f(x) is 1 for a run of 3, so the slope in that interval is 1/3. There, the graph of f' will look like a graph of y=1/3.
If you want to get technical about it, the slope is undefined at x=-2, x=0, and x=2. Therefore, the line segments that make up the graph of f' ought to have open circles at those points, indicating that f' is not defined.