Let $f(x) = x + 2$ and $g(x) = 1/f(x)$. What is $g(f(-3))$?
To find $g(f(-3))$, I need to first evaluate $f(-3)$ and then plug that into $g(x)$. I can use the definition of $f(x)$ to get $f(-3) = -3 + 2 = -1$. Now I have $g(f(-3)) = g(-1)$. I can use the definition of $g(x)$ to get $g(-1) = 1/f(-1)$. I can use the definition of $f(x)$ again to get $f(-1) = -1 + 2 = 1$. So $g(-1) = 1/1 = 1$. Therefore, $g(f(-3)) = 1$. 