Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 32

Given the function f(x)=|x+3|-1:
Rewrite as a piecewise function:
f(x)={ [[x+2,x > - 3],[-1,x=-3],[-x-4,x<-3]]
The function is continuous on the reals.
The first derivative fails to exist at x=-3, and is nonzero for all other x's. (f'(x)=1 on (-oo,-3) , and f'(x)=-1 on (-3,oo) .)
Thus the only critical value is at x=-3. The function increases for x>-3 (since the first derivative is positive on this interval) and decreases for x<-3 (since the first derivative is negative on this interval.)
There is an absolute minimum at x=-3.
The graph:

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