Official solution
f(x) = x3
- 4x2
+ 8x + 11
f(0) = 11
f(1) = 16
Using LMVT
f'(c) = \[{{f\left( 1 \right) - f\left( 0 \right)} \over {1 - 0}}\]
\[ \Rightarrow \] 3c2
-- 8c + 8 = \[{{16 - 11} \over {1 - 0}}\]
\[ \Rightarrow \] 3c2
-- 8c + 3 = 0
\[ \therefore \] c = \[{{8 \pm 2\sqrt 7 } \over 6}\]
\[ \therefore \] c = \[{{4 - \sqrt 7 } \over 3}\] as c \[ \in \] [0, 1]