Question

Rolle's theorem cannot be applied to the function f(x) = x1/3 on the interval [-1, 1] because

Answer 1

Answer with Step-by-step explanation:

We are given that a function

$f(x)=x^{\frac{1}{3}}$ 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

$f'(x)=0$ for some x $a\leq x\leq b$

$a=-1 , b=1$

$f(1)=1$

$f(-1)=(-1)^{\frac{1}{3})=-1$

$f(-1)\neq f(1)$

$f'(x)=\frac{1}{3x^{\frac{2}{3}}}$

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.