My Polynomial is $f(x)=14315.7-14158.8x+4668.3x^2-513x^3$

$\Rightarrow F_c(x)=14315.7-14158.8x+4668.3x^2-513x^3+c$

I begin by defining my functions in Mathematica:

```
f[x_] := 14315.7 - 14158.8 x + 4668.3 x^2 - 513 x^3
F[c_][x_] := f[x] + c
```

I next solve for super attractive points and test if the point is fixed:

`Solve[f'[x] == 0, x]`

Which gives us the output {{x -> 3.}, {x -> 3.06667}}, which we test with

```
f[3];
f[3.0666666666666753`];
```

The output of $f(3)=3$, but $ f(3.0666666666666753)\neq 3.0666666666666753$,

and so we have $3$ as a super-attractive fixed point when $c=0$.

Next, we seek to find a value of $c$ such that $F_c(x)$ has a super attractive orbit of period $2$.

We already know $3$ is a super attractive point for $F$, so we again turn to mathematica, define our twice iterated function, and solve for a c which produces the required behavior:

```
FF[c_][x_] = Nest[F[c], x, 2];
Solve[FF[c][3] == 3, c]
```

Which gives out the output {{c -> -0.0167032}, {c -> 0.}, {c -> 0.116703}}.

We already know the solution for $c=0$ corresponds to the fixed point solution, so we need only test the behavior of the two non-zero solutions.

```
NestList[F[-0.01670320638112171`], 3, 4]
NestList[F[0.11670320634838026`], 3, 4]
```

Which produces the outputs {3, 2.9833, 3., 2.9833, 3.} and {3, 3.1167, 3., 3.1167, 3.} respectively. These are both cycles of period $2$, and in both cases $F'_c(3)$=0, so $c=-0.01670320638112171$ and $c=0.11670320634838026$ are super-attractive orbits of period $2$.