Let $f$ be a polynomial function defined on $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$,then the maximum possible value attained by $f$ at $x=7$ is

The function $f(x) = (x - 3)^2$ satisfies all the conditions of the Mean Value Theorem in $[3, 4]$. $A$ point on $y = (x - 3)^2$,where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$,is

The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

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:

Let $f:[a, b] \rightarrow R$ be continuous in $[a, b]$, differentiable in $(a, b)$ and $f(a)=0=f(b)$. Then