Rolle's Theorem:
Let f(x) = 1-x^(2/3). Show that f(-1) = f(1) but there is no number c in(-1,1) such that f'(c)=0. Why does this not contradict Rolle's Theorem?

Question

Rolle's Theorem:
Let f(x) = 1-x^(2/3).  Show that f(-1) = f(1) but there is no number c in(-1,1) such that f'(c)=0.  Why does this not contradict Rolle's Theorem?

anonymous · Answer

I see that f is continuous on [-1,1]
And I see that f(-1)= f(1)
So this contradicts Rolle's theorem because it is not differentiable on (-1,1) but I don't understand why it isn't differentiable

phi · Answer

First, what is f' ?  i.e. what is df/dx ?

anonymous · Answer

2/3x^(-1/3)

anonymous · Answer

So since 0 cannot be raised to -1/3 then it isn't differentiable on [-1,1]?

anonymous · Answer

Wups, forgot the negative: f'(x) = -2/3x^(-1/3)

phi · Answer

The short answer is f(x) is not differentiable at x=0
one way to see this is evaluate f'(0) to get infinity.

anonymous · Answer

Got it, thanks!