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![\bf \begin{cases} f(x)=&x^2+3\\\\ g(x)=&\cfrac{x+2}{x}\\\\ (f\circ g)(x)=&f(~~g(x)~~) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ g(2)=\cfrac{(2)+2}{(2)}\implies g(2)=\cfrac{4}{2}\implies g(2)=\boxed{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{x=2}{f(~~g(2)~~)}=\left( \boxed{2} \right)^2+3\implies f(~~g(2)~~)=4+3\implies \stackrel{(f\circ g)(2)}{f(~~g(2)~~)}=7](/tpl/images/0041/0444/eabb2.png)
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