Answer
$c = -\sqrt 5 $
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt {25 - {x^2}} ;{\text{ }}\left[ { - 5,3} \right],{\text{ }}a = - 5,{\text{ }}b = 3 \cr
& {\text{Calculate }}f'\left( c \right) \cr
& f'\left( x \right) = \frac{{ - 2x}}{{2\sqrt {25 - {x^2}} }} = - \frac{x}{{\sqrt {25 - {x^2}} }} \cr
& f'\left( c \right) = - \frac{c}{{\sqrt {25 - {c^2}} }} \cr
& f\left( x \right){\text{ is continuous and differentiable on the interval }}\left[ { - 5,3} \right], \cr
& {\text{Using the Mean - Value Theorem}} \cr
& f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr & - \frac{c}{{\sqrt {25 - {c^2}} }} = \frac{{f\left( 3 \right) - f\left( { - 5} \right)}}{{3 - \left( { - 5} \right)}} \cr & - \frac{c}{{\sqrt {25 - {c^2}} }} = \frac{{\sqrt {25 - {{\left( 3 \right)}^2}} - \sqrt {25 - {{\left( { - 5} \right)}^2}} }}{8} \cr
& - \frac{c}{{\sqrt {25 - {c^2}} }} = \frac{{\sqrt {16} - 0}}{8} \cr
& - \frac{c}{{\sqrt {25 - {c^2}} }} = \frac{1}{2} \cr
& \frac{{{c^2}}}{{25 - {c^2}}} = \frac{1}{4} \cr
& 4{c^2} = 25 - {c^2} \cr
& 5{c^2} - 25 = 0 \cr
& {c^2} = 5 \cr
& c = \pm \sqrt 5 \cr
& {\text{Only }} - \sqrt 5 {\text{ is a solution of }} - \frac{c}{{\sqrt {25 - {c^2}} }} = \frac{1}{2},{\text{ then}} \cr
& c =- \sqrt 5 \cr} $$
