Answer
$$c = 4$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} - 8x + 15;{\text{ }}\left[ {3,5} \right],{\text{ }}a = 3,{\text{ }}b = 5 \cr
& {\text{Calculating }}f'\left( x \right) \cr
& f'\left( x \right) = 2x - 8 \cr
& f'\left( c \right) = 2c - 8 \cr
& {\text{The function is continuous becuase it is a polynomial, then}} \cr
& f\left( a \right) = f\left( 3 \right) = {\left( 3 \right)^2} - 8\left( 3 \right) + 15 = 0 \cr
& f\left( b \right) = f\left( 5 \right) = {\left( 5 \right)^2} - 8\left( 5 \right) + 15 = 0 \cr
& {\text{then there is at least one point }}c{\text{ in the interval}}\left( {a,b} \right){\text{ such that }} \cr
& f'\left( c \right) = 0 \cr
& 2c - 8 = 0 \cr
& c = 4 \cr
& {\text{Therefore, }} \cr
& f'\left( 4 \right) = 0 \cr} $$
