Answer
$$C\left( x \right) = {1.2^x} + 8$$
Work Step by Step
$$\eqalign{
& C'\left( x \right) = {1.2^x}\left( {\ln 1.2} \right);{\text{ 2 units cost }}\$ 9.44 \cr
& {\text{The marginal function cost is the derivative of the function cost }}C'\left( x \right) \cr
& {\text{then}}{\text{, the function cost }}C\left( x \right){\text{ is }} \cr
& C\left( x \right) = \int {C'\left( x \right)} dx \cr
& {\text{replacing }}{1.2^x}\left( {\ln 1.2} \right){\text{ for }}C'\left( x \right) \cr
& C\left( x \right) = \int {\left( {{{1.2}^x}\left( {\ln 1.2} \right)} \right)} dx \cr
& C\left( x \right) = \left( {\ln 1.2} \right)\int {{{1.2}^x}} dx \cr
& {\text{use the integration formula }}\int {{a^x}} dx = \frac{{{a^x}}}{{\ln a}} + K. \cr
& C\left( x \right) = \left( {\ln 1.2} \right)\left( {\frac{{{{1.2}^x}}}{{\ln 1.2}}} \right) + K \cr
& C\left( x \right) = {1.2^x} + K{\text{ }}\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find }}K,{\text{we know that the 2 units cost }}\$ 9.44{\text{ then }}C\left( 2 \right) = 9.44 \cr
& 9.44 = {1.2^2} + K \cr
& K = 9.44 - {1.2^2} \cr
& K = 8 \cr
& {\text{then substituting }}K = 8{\text{ into the equation }}\left( {\bf{1}} \right){\text{ we obtain}} \cr
& C\left( x \right) = {1.2^x} + 8 \cr} $$
