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- Lagrange's Mean Value Theorem :
- Let $f(x) $ be a real valued function that satisfies the following conditions.
- (i) $f(x)$ is continuous on the closed interval $[a,b]$
- (ii) $f(x)$ is differentiable in the open interval $(a,b)$
- (iii) $f(a)=f(b)$
- Then there exists atleast one value $c \in (a,b)$ such that $f'(c)=0$

$f(x)=x^{\large\frac{2}{3}}$ is $[-2,2]$ $f(x)$ is continuous

$f'(x)=\large\frac{2}{3}x^{\large\frac{-1}{3}}. $$f'(x) $ is not defined at $x=0$

$\therefore\; f'(x)$ does not exists at $0 \in (-2,2)$.

The conditions for Lagrange's theorem are not satisfied.

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