Answer
$$c = 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^3} + x - 4;{\text{ }}\left[ { - 1,2} \right],{\text{ }}a = - 1,{\text{ }}b = 2 \cr
& {\text{Calculate }}f'\left( c \right) \cr
& f'\left( x \right) = 3{x^2} + 1 \cr
& f'\left( c \right) = 3{c^2} + 1 \cr
& f\left( x \right){\text{ is continuous and differentiable on the interval }}\left[ { - 1,2} \right], \cr
& {\text{Using the Mean - Value Theorem}} \cr
& f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr
& 3{c^2} + 1 = \frac{{f\left( 2 \right) - f\left( { - 1} \right)}}{{2 - \left( { - 1} \right)}} \cr
& 3{c^2} + 1 = \frac{{\left[ {{{\left( 2 \right)}^3} + \left( 2 \right) - 4} \right] - \left[ {{{\left( { - 1} \right)}^3} + \left( { - 1} \right) - 4} \right]}}{{2 - \left( { - 1} \right)}} \cr
& 3{c^2} + 1 = \frac{{6 + 6}}{3} \cr
& 3{c^2} + 1 = 4 \cr
& 3{c^2} = 3 \cr
& c = \pm 1,{\text{ but only }}c = 1{\text{ is in the interval }}\left( { - 1,2} \right),{\text{ then}} \cr
& c = 1 \cr} $$
