The function f(x)=|x^2-1| on [0,2] is continous and has a maximum and minimum.

Find the critical points, i.e. where the derivative does not exists or is zero. The place where the derivative does not exist can only occur at x^2-1=0 that is x=1 on [0,2]. I leave it to thee to show the function is not differenciable there.

Next find the derivative of f(x). To do that use the following theorem.

Theorem:Let f(x)=|x| then if x!=0 the function is differenciable and the derivative is sgn(x). Here.

Thus, the derivative of f(x)=|x^2-1| on (0,2) and x!=1 is:

2x*sgn(x^2-1)

We need that,

2x*sgn(x^2-1)=0

Since sgn(x^2-1)!=0 for x!=0 we must have,

2x=0.

Which has no solutions on (0,2).

Thus, the only critical point is the non-differenciable point we found before x=1.

Compare values with x=0 and x=2. To find maximum and minimum.