Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then, the value of Lagrange's constant $C$ in $(-1,2)$ is

For the function $f(x) = x + \frac{1}{x}$,$x \in [1, 3]$,the value of $c$ for the Mean Value Theorem is:

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

The number of values of $C$ that satisfy the conclusion of Rolle's theorem for the function $f(x) = \sin(2 \pi x)$ on the interval $x \in [-1, 1]$ 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

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