How do you find the critical points for #abs(x^2 -1)#?

1 Answer
Nov 26, 2015

Answer:

They are #0, 1, "and "-1#.

Explanation:

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#.