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Discuss the applicability of Rolle's theorem on the function given by $f(x)=\left \{\begin{array}{1 1}x^2+1, & if\;0\leq x\leq1\\3\;x, & if\;1<x<2\end{array}\right.$

1 Answer

  • 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:
$f(x)=\left\{\begin{array}{1 1}x^2+1,&when\;0\leq x\leq 1\\3-x,&when\;1 < x\leq 2\end{array}\right.$
A polynomial function is continuous everywhere and also differentiable.
Therefore $f(x)$ is continuous and differentiable at all points except at $x=1$.
Now let us consider the differentiability of $f(x)$ at $x=1$.
The LHD at $x=1=\lim\limits_{\large x\to 1^-}\large\frac{f(x)-f(1)}{x-1}$
$\Rightarrow$ LHD at $x=1=\lim\limits_{\large x\to 1}\large\frac{(x^2+1)-(1-1)}{x-1}$
$\Rightarrow \lim\limits_{\large x\to 1}\large\frac{x^2-1}{x-1}$
$\Rightarrow \lim\limits_{\large x\to 1}\large\frac{(x+1)(x-1)}{(x-1)}$
On applying limits we get,
Step 2:
RHD at $x=1=\lim\limits_{\large x\to 1^+}\large\frac{f(x)-f(1)}{x-1}$
$\qquad\quad=\lim\limits_{\large x\to 1^+}\large\frac{(3-x)-(1+1)}{x-1}$
RHD at $x=1=\lim\limits_{\large x\to 1}\large\frac{-(x-1)}{(x-1)}$$=-1$
Therefore LHD at $x=1\neq$ RHD at $x=1$
So $f(x)$ is not differentiable at $x=1$
Since the condition of differentiability at each point of the given interval is not satisfied.
Hence Rolle's theorem is not applicable to the given function on the interval $[0,2]$
answered Jul 3, 2013 by sreemathi.v

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