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Verify Lagrange's theorem for the following function;\[\] $f(x)=2x^{3}+x^{2}-x-1\quad [0,2]$

Note: This is part 3rd 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 $f'(c)=0$
$f(x)=2x^3+x^2-x-1 \quad [0,2]$
Step 1:
This is a poly nonomial function. It is continuous in $[0,2]$ and differentiable in $(0,2)$
Step 2:
To find $c \in (0,2)$ where $f'(c)=\large\frac{f(2)-f(0)}{2-0}$
$f'(x)=6x^2+2x-1 \qquad f'(c)=6c^2+2c-1$
$ \large\frac{f(2)-f(0)}{2-0}=\large\frac{2(8)+4-2-1(-1)}{2}$$=9$
Therefore $6c^2+2c-1=9$
$6c^2+2c-10=0$
$c=\large\frac{-2 \pm \sqrt {4 +240}}{12}$
$\quad=\large\frac{-2 \pm 2\sqrt {61}}{12}$
$\quad=\large\frac{-1 \pm \sqrt {61}}{6}$
The value $\large\frac{-1 \pm \sqrt {61}}{6}$$ \in (0,2)$
The theorem is verified
answered Jul 25, 2013 by meena.p
 

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