From the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(x_1)$ where $a < x_1 < b$. If $f(x) = \frac{1}{x}$,then $x_1 = $

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

Let $f: R \rightarrow R$ be a twice continuously differentiable function. Let $f(0)=f(1)=f^{\prime}(0)=0$. Then,

Let $f(x)$ be differentiable on $[1, 6]$ and $f(1) = -2$. If $f(x)$ has only one root in $(1, 6)$, then there exists $c \in (1, 6)$ such that:

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 for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is