How can we show that there is a zero of $f$

1. Consider the solutions to $ f(x) = x$. These are $ x_1, x_2 = frac{ - 2019 pm sqrt{ 2019^2 - 4 } } { 2 } < 0 $.
2. We have $ f(x_i) = x_i$, so $ F_n(x_i) = x_i$.
3. In particular, $ F_n(x_i) < 0$.
4. Observe that for $x > 0$ $ f(x) > 0$ , so $ F_n(x) > 0 $ for $ x > 0 $.
5. Hence by IVT, there exists a $x'$ such that $F_n(x') = 0$.

Note: $ frac{ - 2019 + sqrt{ 2019^2 - 4 } } { 2 } approx -0.0005$, and so we can conclude that there is a root in $(-0.0005, 0)$.

First, note that for each $n$, $F_n(x)$ is a polynomial of degree $2n$ (with leading coefficient $1$). This is easy to show by induction as

$$F_1(x)=f(x)=x^2+2020x+1$$

Then if $r(x)$ is the remaining polynomial such that $F_{n-1}(x)=x^{2(n-1)}+r(x)$ (note then that the degree of $r(x)$ is less than $2(n-1)$) we have

$$F_n(x)=f(F_{n-1}(x))=(x^{2(n-1)}+r(x))^2+2020 (x^{2(n-1)}+r(x))+1$$

$$=x^{2n}+2x^{2(n-1)}r(x)+r(x)^2+2020x^{2(n-1)}+2020r(x)+1$$

Since this clearly has degree $2n$ the proposition is proved. This implies that

$$lim_{xtoinfty}F_n(x)=infty$$

or that there exists $Xinmathbb{R}$ such that $xgeq X$ implies $F_n(x)geq 1$. Of course, this implies that $F_n(X)=1$. Now, let $x_0$ be the lesser fixed point of $f(x)$. That is

$$f(x_0)=x_0$$

$$x_0^2+2020x_0+1=x_0$$

$$Rightarrow x_0=frac{-2019pmsqrt{4076357}}{2}$$

Since we want the lesser fixed point, we may conclude that

$$x_0=frac{-2019-sqrt{4076357}}{2}$$

But for this $x_0$ we have

$$F_1(x_0)=f(x_0)=x_0$$

$$F_2(x_0)=f(F_1(x_0))=f(x_0)=x_0$$

$$vdots$$

$$F_n(x_0)=x_0=frac{-2019-sqrt{4076357}}{2}<0$$

Thus, $F_n(x_0)<0$ and $F_n(X)=1>0$. By the mean value theorem, there exists some $xin (x_0,X)$ such that $F_n(x)=0$.