Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$,then for some Lagrange's constant $c \in(0,6)$,$g^{\prime}(c)=$

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

Let $f :[0,1] \rightarrow R$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$,then the least number of points $x \in(0,1)$,at which $f^{\prime\prime}(x)=0$,is $......$

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

Consider the function $f(x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the Mean Value Theorem is:

Let $f(x)$ be a differentiable function in $[0, 2]$,$f(0) = 0$ and $f'(x) \le \frac{1}{2}$ for all $x \in [0, 2]$. Then: