Stupid fence

The derivative at a maximum is zero. $\frac{d}{dx}(f_{n+1}(x)+f_n(x))=0$

$f_0^\prime(f_n(x))f_n^\prime(x)+f_n^\prime(x)=0$

$f_n^\prime(x)(f_0^\prime(f_n(x))+1)=0$

$f_n(x)=\frac12$ or $f_n^\prime(x)=0$

If $f_n^\prime(x)=0$ , $f_0^\prime(f_{n-1}(x))f_{n-1}^\prime(x)=0$

$f_{n-1}(x)=0$ or $f_{n-1}^\prime(x)=0$

Repeating the process shows it is possible for any $a< n$ that $f_a(x)=0$ , or, when $a=0$ is reached, $f_0^\prime(x)=0$ where $x=0$

$f_a(x)=0\implies f_n(x)=0$ or $f_n(x)=1$ , because $f_0(0)=1$ and $f_0(1)=0$

The maximum is either $1.25$ if $f_n(x)=\frac{1}{2}$ , or $1$ If $f_n(x)=1$ or $f_n(x)=0$

Goal: maximize $L = \lim\limits_{n \to \infty} f_{n + 1}(x) + f_{n}(x)$ .

$\frac{d}{dx} [f_{n + 1}(x) + f_{n}(x)] = 0$

$\frac{d}{dx} [f_{n + 1}(x)] + \frac{d}{dx} [f_{n}(x)] = 0$

Since $f_{n+1}(x) = f_{0}(f_{n}(x))$ by definition, it follows that $\frac{d}{dx} [f_{n + 1}(x)] = f_{0}'(f_{n}(x)) f_{n}'(x)$ .

Thus, $\frac{d}{dx} [f_{n + 1}(x) + f_{n}(x)] = f_{0}'(f_{n}(x)) f_{n}'(x) + f_{n}'(x) = f_{n}'(x)[ f_{0}'(f_{n}(x)) + 1] = 0$ .

Now we can solve $f_{0}'(f_{n}(x)) + 1 = 0$ to get that $f_{0}'(f_{n}(x)) = -1$ .

Since $f_{0}(x) = 1 - x^2$ , we can use $f_{0}'(x) = -2x$ to solve for $f_{n}(x)$ .

$f_{0}'(f_{n}(x)) = -2(f_{n}(x)) = -1 \Rightarrow f_{n}(x) = \frac{1}{2}$ .

Now let's solve for $f_{n+1}(x)$ .

$f_{n+1}(x) = f_{0}(f_{n}(x)) = f_{0}(\frac{1}{2}) = 1 - (\frac{1}{2})^2 = \frac{3}{4}$ .

Therefore, $L = \lim\limits_{n \to \infty} f_{n + 1}(x) + f_{n}(x) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ , or $1.25$ .