Question

Discuss the applicability of Rolle's theorem to the function $f\left(x\right)=\{\begin{array}{cc}{x}^2+1;0\le x<1& \\ 3-x;1<x\le 2& \end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}{x}^2+1,0\le x<1\\ 3-x,1<x\le 2\end{array}$

$f\left(x\right)=x^2+1$ , is a polynomial function. We know that polynomial functions are continuous everywhere.

So, $f\left(x\right)=x^2+1$ is continuous for $0\le x<1$

Again , $f\left(x\right)=3-x$ , is a polynomial function. We know that polynomial functions are continuous everywhere.

So, $f\left(x\right)=3-x$ is continuous for $1<x\le 2$

Now, we will check the continuity at $x=1$

$LHL={lim}_{x\to 1^{-}}f\left(x\right)$

$={lim}_{h\to 0}f(1-h)$

$={lim}_{h\to 0}(1-h{)}^2+1$

$={lim}_{h\to 0}1+h^2-2h+1$

$\Rightarrow LHL=2$

$RHL={lim}_{x\to 1^{+}}f\left(x\right)$

$={lim}_{h\to 0}f(1+h)$

$={lim}_{h\to 0}3-(1+h)$

$={lim}_{h\to 0}3-1-h$

$\Rightarrow RHL=2$

Also, $f\left(1\right)=1^2+1=2$

Hence, $LHL=RHL=f\left(1\right)$

Hence, f(x) is continuous at $x=1$

Hence, f(x) is continuous in [0,2]

Now, we will check differentiability,

$f^{\prime }\left(x\right)=\{\begin{array}{c}2x0\le x<1\\ -1,1<x\le 2\end{array}$

$f\left(x\right)$ is differentiable in [0,2]

Now, $f\left(0\right)=0+1=1$

$f\left(2\right)=3-2=1$

$\Rightarrow f\left(0\right)=f\left(2\right)$

Hence , all the three conditions of Rolle's theorem holds. Hence Rolle's theorem is applicable