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 $x \in[1,2]$
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 $x \in[1,2].$
The number of points, where the curve $y=x^5-20 x^3+50 x+2$ crosses the $x$-axis, is $............$.
If $L.M.V.$ theorem is true for $f(x) = x(x-1)(x-2);\, x \in [0,\, 1/2]$ , then $C =$ ?
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?
$f(x)=[x]$ for $x \in[5,9]$
Let $f(x)$ satisfy all the conditions of mean value theorem in $[0, 2]. $ If $ f (0) = 0 $ and $|f'(x)|\, \le {1 \over 2}$ for all $x$ in $[0, 2]$ then
In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$ is