Answer
\[\bar c{\,^,}\left( x \right) = \frac{{ - 3x - 4}}{{2{x^2}\sqrt {3x + 2} }}\]
Work Step by Step
\[\begin{gathered}
c\,\left( x \right) = \sqrt {3x + 2} \hfill \\
The\,\,average\,\,cost\,\,is\,\,defined\,\,by \hfill \\
\bar c\,\left( x \right) = \frac{{c\,\left( x \right)}}{x} \hfill \\
\bar c\,\left( x \right) = \frac{{\sqrt {3x + 2} }}{x} \hfill \\
The\,\,marginal\,\,cost\,\,is\,\,given\,\,by \hfill \\
\frac{d}{{dx}}\,\,\left[ {\bar c\,\left( x \right)} \right] = {\left[ {\frac{{\sqrt {3x + 2} }}{x}} \right]^,} \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{x\,{{\left( {\sqrt {3x + 2} } \right)}^,} - \sqrt {3x + 2} \,{{\left( x \right)}^,}}}{{\,{{\left( x \right)}^2}}} \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{x\,\left( {\frac{3}{{2\sqrt {3x + 2} }}} \right) - \sqrt {3x + 2} }}{{{x^2}}} \hfill \\
Simplifying \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{3x - 2\,{{\left( {\sqrt {3x + 2} } \right)}^2}}}{{2{x^2}\sqrt {3x + 2} }} \hfill \\
\bar c{\,^,}\left( x \right) = \frac{{ - 3x - 4}}{{2{x^2}\sqrt {3x + 2} }} \hfill \\
\end{gathered} \]
