Let $f$ be differentiable for all $x$. If $f(1) = -2$ and $f'(x) \ge 2$ for $x \in [1, 6]$,then

For what value of $c$ does the conclusion of the Mean Value Theorem hold for the function $f(x) = \log_e x$ on the interval $[1, 3]$?

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

If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$,the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval

Suppose that $f(0) = -3$ and $f'(x) \le 5$ for all values of $x$. Then the largest value which $f(2)$ can attain is