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Verify the Rolle's theorem for the function $log(x^2+2)-log 3\;in\;[-1,1]$

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  • If $f[a,b]\rightarrow R$ is continuous on $[a,b]$ and differentiable on $(a,b)$ such that $f(a)=f(b)$,then there exists some c in $(a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=\log(x^2+2)-\log 3$
Since logarithmic function is differentiable and continuous and so continuous on its domain.Therefore $f(x)$ is continuous on $[-1,1]$ and differentiable on $[-1,1]$.
$f(-1)=\log((-1)^2+2)-\log 3$
$\quad\;\;\;\;\;=\log 3-\log 3$
$\quad\;\;\;\;\;=0$
$f(1)=\log((1)^2+2)-\log 3$
$\quad\;\;\;\;\;=\log 3-\log 3$
$\quad\;\;\;\;\;=0$
Also $f(0)=0$
Therefore $f(-1)=f(1)$
Thus all the three conditions of Rolle's theorem are satisfied.
Step 2:
Now we have to show that there exists $c\in (-1,1)$ such that $f'(c)=0$
$f(x)=\log(x^2+2)-\log 3$
$f'(x)=\large \frac{1}{x^2+2}$$(2x)=0$
$\qquad=\large\frac{2x}{x^2+2}$
Therefore $f'(x)=0$
$\Rightarrow \large\frac{2x}{x^2+2}$$=0$
$\Rightarrow x=0$
Thus $c=0$ lies within the interval $[-1,1]$.
Hence Rolle's theorem is verified.
answered Jul 2, 2013 by sreemathi.v
 

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