Question 73
Let
$$
f(x)=\left(x-F_{1}\right)\left(x-F_{2}\right) \ldots\left(x-F_{3030}\right)
$$
where $\left(F_{n}\right)$ is the Fibonacci sequence, defined by $F_{1}=1, F_{2}=2, F_{n+2}= F_{n+1}+F_{n}, n \geq 1$. Suppose that on the range $\left(F_{1}, F_{3030}\right)$, the function $|f(x)|$ takes on the maximum value at $x=x_{0}$. Prove that $x_{0}\gt 2^{2018}$.
$$
f(x)=\left(x-F_{1}\right)\left(x-F_{2}\right) \ldots\left(x-F_{3030}\right)
$$
where $\left(F_{n}\right)$ is the Fibonacci sequence, defined by $F_{1}=1, F_{2}=2, F_{n+2}= F_{n+1}+F_{n}, n \geq 1$. Suppose that on the range $\left(F_{1}, F_{3030}\right)$, the function $|f(x)|$ takes on the maximum value at $x=x_{0}$. Prove that $x_{0}\gt 2^{2018}$.