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

Consider $f(x) = \int\limits_0^x {\left( {t + \frac{1}{t}} \right)\,dt}$ and $g(x) = f'(x)$ for $x \in \left[ {\frac{1}{2}, 3} \right]$. If $P$ is a point on the curve $y = g(x)$ such that the tangent to this curve at $P$ is parallel to a chord joining the points $\left( {\frac{1}{2}, g\left( {\frac{1}{2}} \right)} \right)$ and $(3, g(3))$ of the curve,then the coordinates of the point $P$ are:

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 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:

$f:[2,10] \rightarrow R$ is defined as $f(x) = \begin{cases} \frac{1}{2}(x-6)^2-3, & x \leq 4 \\ x-5, & x > 4 \end{cases}$. Which of the following is true?

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