The function $f(x)$ satisfies
\[f(x + y) = f(x) f(y)\]for all real numbers $x$ and $y.$  If $f(2) = 3,$ find $f(6).$
I notice that this is a functional equation, where the value of $f(x + y)$ depends only on the values of $f(x)$ and $f(y).$ For example, if I plug in $x = 2$ and $y = 2,$ I get $f(4) = f(2) f(2) = 3 \cdot 3 = 9.$ Now I want to find $f(6),$ so I need to express $6$ as a sum of two numbers whose function values I already know. One way to do that is to write $6 = 4 + 2.$ Then I can use the functional equation again and plug in $x = 4$ and $y = 2.$ I get $f(6) = f(4) f(2) = 9 \cdot 3 = 27.$

