Answer
$$c \approx 3.7640$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^{2/3}},{\text{ }}\underbrace {\left[ {1,8} \right]}_{\left[ {a,b} \right]} \to a = 1,b = 8 \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{2/3}}} \right] \cr
& f'\left( x \right) = \frac{2}{3}{x^{ - 1/3}} \cr
& {\text{Thus:}} \cr
& f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} \cr
& \frac{2}{3}{c^{ - 1/3}} = \frac{{f\left( 8 \right) - f\left( 1 \right)}}{{8 - 1}} \cr
& \frac{2}{{3\root 3 \of c }} = \frac{{{{\left( 8 \right)}^{2/3}} - {{\left( 1 \right)}^{2/3}}}}{7} \cr
& {\text{Simplify and solve for }}c \cr
& \frac{2}{{3\root 3 \of c }} = \frac{3}{7} \cr
& \root 3 \of c = \frac{{14}}{9} \cr
& c = {\left( {\frac{{14}}{9}} \right)^3} \cr
& c \approx 3.7640 \cr} $$
