The value of c in the Lagrange's mean value theorem for the function

Since, $f(x)$ is a polynomial function.

$\therefore \quad$ It is continuous and differentiable in $[0,1]$

Here, $f(0)=11, f(1)=1-4+8+11=16$

$f^{\prime}(x)=3 x^{2}-8 x+8$

$\therefore \quad f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}=\frac{16-11}{1}$

$=3 c^{2}-8 c+8$

$\Rightarrow \quad 3 c^{2}-8 c+3=0$

$\Rightarrow \quad c=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}$

$\therefore \quad c=\frac{4-\sqrt{7}}{3} \in(0,1)$