$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ is

Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,

The value of $c$ for which Rolle's theorem holds for the function $f(x)=x^3-3x^2+2x$ in the interval $[0,2]$ is:

If $c$ is a point at which Rolle's theorem holds for the function $f(x) = \log_{e}\left(\frac{x^{2}+\alpha}{7x}\right)$ in the interval $[3, 4]$,where $\alpha \in R$,then $f''(c)$ is equal to

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:

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to: