Lagrange's Mean Value Theorem (LMVT)

Let $f^{\text{'}}\left(x\right)=e^{x^2}$ and $f\left(0\right)=10 .$ If $A<f\left(1\right)<B$ can be concluded from the mean value theorem, then the largest value of $(A-B)$ equals

If $f\left(x\right)$ and $g\left(x\right)$ are differentiable functions for $0\le x\le 1$ such that $f\left(0\right)=10, g\left(0\right)=2, f\left(1\right)=2, g\left(1\right)=4,$ then in the interval $(0,1)$

Suppose that $f$ is differentiable for all $x$ and that $f^{\text{'}}\left(x\right)\le 2$ for all $x.$ If $f\left(1\right)=2$ and $f\left(4\right)=8,$ then $f\left(2\right)$ has the value equal to

A value of $c$ for which the conclusion of mean value theorem holds for the function $f\left(x\right)={\log }_{e}x$ on the interval$[1,3]$ is

If $f\left(x\right)=x^3+7x-1$, then $f\left(x\right)$ has a zero between $x=0$ and $x=1.$ The theorem that best describes this is