Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then,the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. The values of $a$ and $b$ are:

Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$ and $f''(x) \geq 4$ for all $x \in (1,6)$,then:

Let $f$ be a function which is differentiable for all real $x$. If $f(2) = -4$ and $f^{\prime}(x) \geq 6$ for all $x \in [2, 4]$, then which of the following is true?

Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.