If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

Let $f(x)$ satisfy all the conditions of the Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \le \frac{1}{2}$ for all $x$ in $[0, 2]$,then:

Let $f(x)$ be continuous on $[0,4]$,differentiable on $(0,4)$,$f(0)=4$ and $f(4)=-2$. If $g(x)=\frac{f(x)}{x+2}$,then the value of $g^{\prime}(c)$ for some Lagrange's constant $c \in (0,4)$ is

If the $L.M.V.T.$ holds for the function $f(x) = x + \frac{1}{x}$ on the interval $x \in [1, 3]$,then $c$ is:

Let the function $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x) \leq 2$ for all $x \in (-7,0)$,then for all such functions $f$,$f(-1)+f(0)$ lies in the interval:

Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,