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Verify Lagrange's mean value theorem for the following function: $ f(x)=x^2-2x+3,\: for \: [4,6]$

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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 $f'(c)=0$
  • $f'(c)=\large\frac{f(b)-f(a)}{b-a}$
Step 1:
Given :$f(x)=x^2-2x+3$ in the interval [4,6]
We know that a polynomial function is continuous everywhere and also differentiable.
So $f(x)$ being a polynomial is continuous and differentiable on (4,6)
So there must exist at least one real number $c\in (4,6)$ such that
Step 2:
$\therefore 2c-2=\large\frac{27-11}{2}$
$c\in (4,6)$
Hence Lagrange's Mean Value theorem is verified.
answered Sep 24, 2013 by sreemathi.v

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