My Initials are WSF. So from last time I have

a = 3

b = 2

c = 0

So my function is f(x) = 3x^2 + 2x.

Let g(x) = x^2 + \gamma and \varphi(x) = \alpha x+\beta.

Then

f(\varphi(x)) = 3(\alpha x +\beta)^2+ 2(\alpha x + \beta) = (3\alpha^2)x^2 +(6\alpha\beta + 2\alpha)x + (3\beta^2 + 2\beta)

\varphi(g(x)) = \alpha(x^2 + \gamma) + \beta = \alpha x^2 + (\alpha\gamma + \beta)

Setting f(\varphi(x)) = \varphi(g(x)), we can construct the following system of equations:

3\alpha^2 = \alpha

6\alpha\beta + 2\alpha = 0

3\beta^2 + 2\beta = \alpha\gamma + \beta

Solving these equations gives us \alpha = 1/3, \beta = -1/3, and \gamma = 0.

So then g(x) = x^2 and \varphi(x) = (1/3)x - 1/3.

Lastly, we’ll check to see if we got the right \alpha, \beta, and \gamma.

f(\varphi(x)) = 3((1/3)x-1/3)^2+2((1/3)x-1/3)
= (1/3)x^2 -1/3

\varphi(g(x)) = (1/3)x^2 - 1/3

We got it!

Now we will compute the first five terms of the critical orbits for f and g.

First, let’s find the critical points for f and g.

Note that f'(x) = 6x + 2 = 0 implies that x = -1/3 and g'(x) = 2x = 0 implies that x = 0.

Using Mathematica and our critical points, we generate the first five terms of the critical orbits for f and g.

For f we have:

```
NestList[3*#^2 + 2*# &, -1/3, 4]
{-(1/3), -(1/3), -(1/3), -(1/3), -(1/3)}
```

For g we have:

```
NestList[#^2 &, 0, 4]
{0, 0, 0, 0, 0}
```

Interestingly, our critical points for f and g were also fixed points.

Now, we will look at the image of the critical orbit of g, \{0,0,0,0,0\}, under \varphi.

Note that \varphi(0) = (1/3)\cdot 0 -1/3 = -1/3.

So then the image of \{0,0,0,0,0\} under \varphi is \{-(1/3),-(1/3),-(1/3),-(1/3),-(1/3)\}, which is the critical orbit of f.