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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 $f'(c)=0$

Given $f(x)=\large\frac{1}{x} $$\;[1,2]$

Step 1:

The function is continuous in $[1,2]$ and differentiable in $(1,2)$. The conditions for lagranges theorem are satisfied .

Step 2:

To find $c \in (1,2)$ where $f'(c)=\large\frac{f(2)-f(1)}{2-1}$

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

$f'(c)=\large\frac{-1}{c^2}$

$\large\frac{f(2)-f(1)}{2-1}=\frac{\Large\frac{1}{2}-1}{2-1}=\frac{-1}{2}$

$\large\frac{-1}{c^2}=\large\frac{-1}{2}$$=>c=\pm \sqrt 2$

Of these values $\sqrt 2 \in (1,2)$ the theorem is verified

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