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Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse of Roller's Theorem from these examples?
$f(x)=x^{2}-1$ for $x \in[1,2]$
Solution
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 $x \in[1,2].$
