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Verify the Rolle's theorem for the function $f(x)=x(x-1)^2\;in\;[0,1]$

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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:
$f(x)=x(x-1)^2$ in $[0,1]$
On expanding $x(x-1)^2=x(x^2-2x+1)$
$\Rightarrow x^3-2x^2+x$
$f(x)=x^3-2x^2+x$
We know that a polynomial function is differentiable every where and hence continuous also.Hence $f(x)$ being continuous on $[0,1]$ and differentiable on $(0,1)$.
Also $f(0)=f(1)=0$
Then all the condition of Rolle's theorem are verified.
Step 2:
Now let us show that $c\in (0,1)$ such that $f'(c)=0$
$f(x)=x^3-2x^2+x$
$f'(x)=3x^2-4x+1$
Therefore $f'(x)=0=3x^2-4x+1$
$\Rightarrow 3x^2-4x-1=0$
Step 3:
On factorising we get,
$3x^2-3x-x+1=0$
$\Rightarrow 3x(x-1)-1(x-1)=0$
$\Rightarrow (3x-1)(x-1)=0$
Therefore $x=\large\frac{1}{3},$$1$
$\large\frac{1}{3},$$1$ lies within the interval $[0,1]$.
Hence Rolle's theorem is verified.
answered Jul 2, 2013 by sreemathi.v
 

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