If the function $f(x)=x^3+b x^2+c x-6$ satisfies all the conditions of Rolle's theorem in $[1,3]$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$,then $b c=$

The value $c$ of Lagrange's mean value theorem for $f(x)=e^{x}+24$ in $[0,1]$ is

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because

For the Mean Value Theorem $f(b) - f(a) = (b - a) f'(x_1)$ where $a < x_1 < b$,if $f(x) = 1/x$,then $x_1 = ?$

Let $y = f(x)$ and $y = g(x)$ be two differentiable functions in $[0, 2]$ such that $f(0) = 3$,$f(2) = 5$,$g(0) = 1$,and $g(2) = 2$. If there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,then $k$ must be:

The constant $c$ of Lagrange's mean value theorem for $f(x)=\cos x-\sin 2x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is