A critical number for #f# is a number #c#, in the domain of #f# at which #f'(c)# does not exist or #f'(c)=0#.

Therefore, we begin by finding #f'(x)# for #f(x) = abs(x^2-1)#.

NOte that the domain of #f# is #(-oo,oo)#.

#f(x) = abs(x^2-1) = {(x^2-1,"if",x^2-1 >= 0),(-(x^2-1),"if",x^2-1 < 0) :}#.

Investigating the sign of #x^2-1# shows that it is positive if #x < 1# or #x > 1# and it is negative if #-1 < x < 1#.

Therefore,

#f(x) = {(x^2-1,"if",x <= -1),(-x^2+1,"if",-1 < x < 1),(x^2-1,"if",x >= 1) :}#.

Differentiating each piece gets us

#f'(x) = {(2x,"if",x < -1),(-2x,"if",-1 < x < 1),(2x,"if",x > 1) :}#.

At the 'joints' of #x=-1# and #x=1# the left and right derivatives do not agree, so the derivative does not exist.

#-1# and #1# are critical numbers for #f#.

Furthemore, #f'(x) = 0# at #x=0#, so #0# is also a critical number for #f#.

The critical numbers are #-1, 0, 1#