Let $f\left(x\right)$ be differentiable function and $g\left(x\right)$ be twice differentiable function. Zeros of $f\left(x\right),g^{\text{'}}\left(x\right)$ be $a,b$ respectively $\left(a<b\right)$. Show that there exists at least one root of equation $f^{\text{'}}\left(x\right)g^{\text{'}}\left(x\right)+f\left(x\right)g^{"}\left(x\right)=0$ on $\left(a,b\right)$.

If $f\left(x\right)=\left|\begin{array}{ccc}{\sin }^{3}x& {\sin }^{3}a& {\sin }^{3}b\\ x{e}^x& a{e}^a& b{e}^b\\ x& a& b\\ 1+x^2& 1+a^2& 1+b^2\end{array}\right|$

where $0<a<b<2\mathrm{\pi }$, then show that the equation $f^{\text{'}}\left(x\right)=0$ has at least one root in the interval $\left(a,b\right)$.

A function $y=f\left(x\right)$ is defined on $\left[0, 6\right]$ as $f\left(x\right)=\left\{\begin{array}{ccc}-8x& ;& 0\le x\le 1\\ {\left(x-3\right)}^3& ;& 1<x<4\\ 2& ;& 4\le x\le 6\end{array}\right.$

Show that for the function $y=f\left(x\right),$ all the three conditions of Rolle's theorem are violated on $\left[0, 6\right]$ but still $f\left(x\right)$ vanishes at a point in $\left(0, 6\right)$