Answer
$$C\left( x \right) = \frac{{2{x^{3/2}} + 7}}{3}$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = {x^{1/2}};{\text{ 16 units cost }}\$ 45 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}{x^{1/2}}{\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {{x^{1/2}}} dx \cr
& {\text{integrating }} \cr
& C\left( x \right) = \frac{{{x^{1/2 + 1}}}}{{1/2 + 1}} + K{\text{ }} \cr
& C\left( x \right) = \frac{{{x^{3/2}}}}{{3/2}} + K{\text{ }} \cr
& C\left( x \right) = \frac{{2{x^{3/2}}}}{3} + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the 16 units cost }}\$ 45{\text{ then }}C\left( {16} \right) = 45 \cr
& 45 = \frac{{2{{\left( {16} \right)}^{3/2}}}}{3} + K \cr
& 45 = \frac{{128}}{3} + K \cr
& K = 45 - \frac{{128}}{3} \cr
& K = \frac{7}{3} \cr
& {\text{then substituting }}K = \frac{7}{3}{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr
& C\left( x \right) = \frac{{2{x^{3/2}}}}{3} + \frac{7}{3} \cr
& C\left( x \right) = \frac{{2{x^{3/2}} + 7}}{3} \cr} $$
