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Verify Lagranges theorem for the following function;$f(x)=\large\frac{1}{x}$$\quad[1,2]$

Note: This is part 2nd of a 5 part question, split as 5 separate questions here.

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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 $f'(c)=0$
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

 

answered Jul 25, 2013 by meena.p
 

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