Answer
$$c = 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt x - 2x,{\text{ }}\underbrace {\left[ {0,4} \right]}_{\left[ {a,b} \right]} \to a = 0,b = 4 \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sqrt x - 2x} \right] \cr
& f'\left( x \right) = \frac{1}{{2\sqrt x }} - 2 \cr
& {\text{Thus:}} \cr
& f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr
& \frac{1}{{2\sqrt c }} - 2 = \frac{{\left[ {\sqrt 4 - 2\left( 4 \right)} \right] - \left[ {\sqrt 0 - 2\left( 0 \right)} \right]}}{{4 - 0}} \cr
& \frac{1}{{2\sqrt c }} - 2 = \frac{{ - 6 - 0}}{4} \cr
& \frac{1}{{2\sqrt c }} = - \frac{3}{2} + 2 \cr
& {\text{Simplify and solve for }}c \cr
& \frac{1}{{2\sqrt c }} = \frac{1}{2} \cr
& \frac{1}{{\sqrt c }} = 1 \cr
& c = 1 \cr} $$
