$f_1(x) = \sqrt{x^2+48} = 2x \Rightarrow 48 = 3x^2 \Rightarrow x = 4$ (negative answer is thrown out because of the definition of square root). Note that $f_{n+1}(4) = \sqrt{4^2+6f_n(4)}$ so if $f_n(4) = 8$, then $f_{n+1}(4) = \sqrt{16+48} = 8$, which means $x = 4$ is a solution for all $n$. Suppose $x < 4$. We know $x$ must be non-negative since we only use the positive square root. Then $f_1(x) = \sqrt{x^2+48}> 2x$. But if $f_n(x) > 2x$, we have $f_{n+1}(x) = \sqrt{x^2+6f_n(x)} > \sqrt{x^2+12x} > 2x$ also (the last step since $x<4$). Now suppose $x > 4$. Then $f_1(x) = \sqrt{x^2+48} < 2x$. But if $f_n(x) < 2x$, we have $f_{n+1}(x) = \sqrt{x^2+6f_n(x)} < \sqrt{x^2+12x} < 2x$ also (the last step since $x>4$). Therefore $x = 4$ is the only solution for all $n$.