ચકાસો કે આપેલ વિધેયમાં રોલનું પ્રમેય લગાડી શકાય કે નહિ  : $f(x)=x^{2}-1,$ $x \in[1,2]$

By Rolle's Theorem, for a function $f:[a, b] \rightarrow R,$ if

a) $f$ is continuous on $[a, b]$

b) $f$ is continuous on $(a, b)$

c) $f(a)=f(b)$

Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$

Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f(x)=x^{2}-1$ for $x \in[1,2]$

It is evident that $f$, being a polynomial function, is continuous in $[1,2]$ and is differentiable in $(1,2).$

$f(1)=(1)^{2}-1=0$

$f(2)=(2)^{2}-1=3$

$\therefore f(1) \neq f(2)$

It is observed that $f$ does not satisfy a condition of the hypothesis of Roller's Theorem.

Hence, Roller's Theorem is not applicable for $f(x)=x^{2}-1$ for $x \in[1,2].$