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.

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{x^3}{3} + b\frac{x^2}{2} + cx.$
Statement $1:$ The quadratic equation has at least one root in the interval $(0, 1).$
Statement $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1].$

If $f(x) = (x - 1)(x - 2)(x - 3)$ for $x \in [0, 4]$,then the value of $c \in (0, 4)$ satisfying Lagrange's mean value theorem is

Let $a, b, c$ be real numbers such that $2a + 3b + 6c = 0$ and $g(x) = ax^2 + bx + c = 0$ has at least one root in the interval $(1, 2)$. If a function $f: [1, 2] \rightarrow \mathbb{R}$ for which Rolle's Theorem holds is such that $f(x)$ is a primitive of $g(x)$,then $f(x) = $

If the function $f(x)=\sqrt{x^2-4}$ satisfies the Lagrange's mean value theorem on $[2, 4]$,then the value of $C$ is

Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$, $I_2=[0,1]$, $I_3=[1,2]$, $I_4=[-2,-1]$. Then,