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Verify Lagrange's theorem for the following function;\[\]$f(x)=x^{\large\frac{2}{3}}[-2,2]$

Note: This is part 4th of a 5 part question, split as 5 separate questions here.

1 Answer

Toolbox:
  • 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.
answered Jul 25, 2013 by meena.p
 

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