For the Mean Value Theorem $f(b) - f(a) = (b - a) f'(x_1)$ where $a < x_1 < b$,if $f(x) = 1/x$,then $x_1 = ?$

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:

Suppose $f(x)$ is twice differentiable in the interval $[1, 3]$ and $f(1)=f(3)$. If $|f^{\prime \prime}(x)| \leq 2$,then for all $x$ in $[1, 3]$,which one of the following is true?

If the function $f(x) = x(x+3)e^{-x/2}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$, then a root of $f'(x) = 0$ is

If the function $f(x)=x^3+ax^2+bx+40$ satisfies the conditions of Rolle's theorem on the interval $[-5,4]$ and $-5,4$ are two roots of the equation $f(x)=0$, then one of the values of $c$ as stated in that theorem is

Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.