Answer
$$C\left( x \right) = \frac{{3{x^{5/3}} + 114}}{5} + 2x$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = {x^{2/3}} + 2;{\text{ 8 units cost }}\$ 58 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}{x^{2/3}} + 2{\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {\left( {{x^{2/3}} + 2} \right)} dx \cr
& {\text{integrating }} \cr
& C\left( x \right) = \frac{{{x^{2/3 + 1}}}}{{2/3 + 1}} + 2x + K{\text{ }} \cr
& C\left( x \right) = \frac{{{x^{5/3}}}}{{5/3}} + 2x + K{\text{ }} \cr
& C\left( x \right) = \frac{{3{x^{5/3}}}}{5} + 2x + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the 8 units cost }}\$ 58{\text{ then }}C\left( 8 \right) = 58 \cr
& 58 = \frac{{3{{\left( 8 \right)}^{5/3}}}}{5} + 2\left( 8 \right) + K \cr
& 58 = \frac{{96}}{5} + 16 + K \cr
& K = 58 - \frac{{96}}{5} - 16 \cr
& K = \frac{{114}}{5} \cr
& {\text{then substituting }}K = \frac{{114}}{5}{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr
& C\left( x \right) = \frac{{3{x^{5/3}}}}{5} + 2x + \frac{{114}}{5} \cr
& C\left( x \right) = \frac{{3{x^{5/3}} + 114}}{5} + 2x \cr} $$
