f (-1)=(((2+1^2/3)3)^1/2

Question

Does Rolle's Theorem apply?

Answer 1

Yes. If 1^2 is the result of substituting for -1 in f(x), then f(1)=f(-1) so there are two distinct values of x for the same value of f(x). This is essentially Rolles theorem. Between -1 and 1 there is a stationary point where the tangent to the curve f(x) is zero.

Answer 2

Given (((2+1^2/3)3)^1/2
√7