Condition (1):
Since, f(x) = x1/2 is a polynomial and we know every polynomial function is continuous for all x ϵ R.
⇒ f(x) = x1/2 is continuous on [-1,1].
Condition (2):
Here, f'(x) = \(\frac{1}{2x^{1/2}}\) which does not exist at x = 0 in [-1,1].
f(x) = x1/2 is not differentiable on (-1,1).
Condition (2) of Rolle’s theorem is not satisfied.
So,Rolle’s theorem is not applicable.