The value of $c$ in Roll's theorem for the function $f\left(x\right)=\frac{x(x+1)}{e^x}$ defined on $[-1,0]$, is

Solution:

We have, $f\left(x\right)=x(x+1)e^{-x},x\in [-1,0]$

$\therefore f^{\mathrm{\prime }}\left(x\right)=-\left(x^2-x-1\right)e^{-x},f(-1)=0,f\left(0\right)=0$

Hence, $f^{\mathrm{\prime }}\left(x\right)=\frac{f\left(0\right)-f(-1)}{0-(-1)}$

$\Rightarrow -\left(x^2-x-1\right)e^{-x}=0\Rightarrow x^2-x-1=0\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$

Hence, $c=\frac{1-\sqrt{5}}{2}$