Answer
$$c = 1$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{2}x - \sqrt x ;{\text{ }}\left[ {0,4} \right],{\text{ }}a = 0,{\text{ }}b = 4 \cr
& {\text{Calculating }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{1}{2}\left( 1 \right) - \frac{1}{2}{x^{ - 1/2}} \cr
& f'\left( x \right) = \frac{1}{2} - \frac{1}{{2\sqrt x }} \cr
& f'\left( c \right) = \frac{1}{2} - \frac{1}{{2\sqrt c }} \cr
& {\text{The function is continuous becuase it is a polynomial, then}} \cr
& f\left( a \right) = f\left( 0 \right) = \frac{1}{2}\left( 0 \right) - \sqrt 0 = 0 \cr
& f\left( b \right) = f\left( 4 \right) = \frac{1}{2}\left( 4 \right) - \sqrt 4 = 0 \cr
& {\text{then there is at least one point }}c{\text{ in the interval}}\left( {a,b} \right){\text{ such that }} \cr
& f'\left( c \right) = 0 \cr
& \frac{1}{2} - \frac{1}{{2\sqrt c }} = 0 \cr
& \frac{1}{{2\sqrt c }} = \frac{1}{2} \cr
& \frac{1}{{\sqrt c }} = 1 \cr
& c = 1 \cr
& {\text{Therefore, }} \cr
& f'\left( 1 \right) = 0 \cr} $$
