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

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{{x^2} + 1,}&{{\rm{ if }}0 \le x \le 1}\\{3 - x,}&{{\rm{ if }}1 \le x \le 2}\end{array}} \right.$

We know that, polynomial function is everywhere continuous and differentiability.

So, $f(x)$ is continuous and differentiable at all points except possibly at $x = 1$.

Now, check the differentiability at $x = 1$,
At $x = 1$,
${\rm{LDH}} = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{f(x) - f(1)}}{{x - 1}}$

$= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^2} + 1} \right) - (1 + 1)}}{{x - 1}}$

$= \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x + 1)(x - 1)}}{{x - 1}}$
$= 2$

and ${\rm{RDH}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(3 - x)f(1 + 1)}}{{(x - 1)}}$

$= \mathop {\lim }\limits_{x \to 1} \frac{{3 - x - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{ - (x - 1)}}{{x - 1}} = - 1$

therefore,${\rm{LHD}} \ne {\rm{RHD}}$

Hence we can say that $f(x)$ is not differentiable at $x = 1$.

Therefore, Rolle's theorem is not applicable on the interval [0,2] .