Answer with Step-by-step explanation:
We are given that a function
[tex]f(x)=x^{\frac{1}{3}}[/tex] on the interval [-1,1]
Rolle's theorem : It states that function is continuous on close interval [a,b] and differentiable on open interval (a,b) such that f(a)=f(b) , then
[tex]f'(x)=0[/tex] for some x [tex]a\leq x\leq b[/tex]
[tex]a=-1 , b=1[/tex]
[tex]f(1)=1[/tex]
[tex]f(-1)=(-1)^{\frac{1}{3})=-1[/tex]
[tex]f(-1)\neq f(1)[/tex]
[tex]f'(x)=\frac{1}{3x^{\frac{2}{3}}}[/tex]
f is not differentiable at x=0.Therefore , f(x) is not diffrentiable on interval (-1,1)
Hence, Rolle's threorem cannot be applied for given function because it does not satisfied the condition of rolle's theorem.
