If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. The values of $a$ and $b$ are:

The function $f(x) = x(x + 3)e^{-\frac{1}{2}x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$. Find the value of $c$ such that $f'(c) = 0$.

In the Mean Value Theorem,$f'(c) = \frac{f(b) - f(a)}{b - a}$. If $a = 0$,$b = \frac{1}{2}$,and $f(x) = x(x - 1)(x - 2)$,then the value of $c$ is:

If $f(x) = x^{\alpha} \log x, x > 0, f(0) = 0$ and $f(x)$ satisfies Rolle's theorem on $[0, 1]$,then what is the value of $\alpha$?

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: