Let f(x) = abs(x^2-1), so
f(x) = { (x^2-1," "" "x < 1),(-x^2+1," "-1 <=x <= 1),(x^2-1," "" "1>x) :}
At all x other than 1 and -1, we can quickly see that f'(x) = 2x or -2x
Now, lim_(xrarr1^+)(f(x)-f(1))/(x-1) = lim_(xrarr1^+)(x^2-1)/(x-1) = 2,
but lim_(xrarr1^-)(f(x)-f(1))/(x-1) = lim_(xrarr1^-)(-x^2+1)/(x-1) = -2.
We conclude that f has no derivative at 1.
A similar argument will show that there is no derivative at -1.
f(x) = { (2x," "" "x < 1),(-2x, " " -1 < x < 1),(2x," "" "1>x) :}
Because both 1 and -1 are in the domain of f and f' does not exists at either, these are both critical numbers for f.
In addition, f'(x) = 0 at x=0, so 0 is another critical number for f.
