The function $f(x)$ satisfies
\[f(x + f(x)) = 4f(x)\]for all $x,$ and $f(1) = 4.$  Find $f(21).$
I notice that this function has some kind of recursive property, where knowing the value of $f(x)$ helps me find the value of $f(x + f(x)).$ Maybe I can use this property to find the value of $f(21)$ by starting from $f(1)$ and repeatedly applying the rule. I see that $f(1) = 4,$ so $f(1 + f(1)) = f(5) = 4f(1) = 16.$ Now I have $f(5) = 16,$ so $f(5 + f(5)) = f(21) = 4f(5) = 64.$ So I have found $f(21) = 64.$ 