Alan

First look at the prime factors of 33275

$${factorize}{\left({\mathtt{33\,275}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{25}} = {{\mathtt{5}}}^{{\mathtt{2}}}\\
{\mathtt{1\,331}} = {{\mathtt{11}}}^{{\mathtt{3}}}\\
\end{array} \right\}$$

i.e. $$33275=5^2*11^3$$

This should give you a clue.

g(7) = 7² - 5 = 49 - 5 = 44

f(44) = 44 + 6 = 50

Replace the t by 2x

f(2x) = 3*(2x)² - 2 

f(2x) = 3*4x² - 2

f(2x) = 12x² - 2

Perhaps I've misunderstood, but if c1 and c2 are initial values (n=0) for x and y then (for n = 1):

These don't look the same to me!

I get  $$-\lambda^3+8\lambda^2-19\lambda+12=0$$

Multiply through by -1 (not essential - I just find it easier that way!)

$$\lambda^3-8\lambda^2+19\lambda-12=0$$

If we assume that, because you are doing it by hand, the solutions are integers, then consider

$$(\lambda-a)(\lambda-b)(\lambda-c)$$

The three constants must multiply together such that $$abc=12$$

If they are each integers then we have either 1, 2 and 6 or 1, 3 and 4 (ok a couple of them might have negative signs attached!).

The sum of the three must also be 8, and since 1+2+6 is not 8, and 1+3+4 is, there is a good chance that the three eigenvalues are 1, 3 and 4.

Plugging each of these into the LHS of the first equation above does indeed give zero.

In the figure below start at A and follow the edges in alphabetical order.  Every edge is visited once and you finish where you started.  Note that there is an even number of edges at every node.

-w² is -(w²) not (-w)² because exponentiation (raising to a power) normally takes priority over the unary minus.

This would mean -w²(w - 15) = w²(-w + 15) = -w³ + 15w

To ensure there is no confusion you could use brackets!

f(0.5) = 16^0.5 = √(16) = 4

If n = 9 then (-1/3)^n-1 = (-1/3)⁸ = (-1)⁸/3⁸ = 1/3⁸ = 1⁸/3⁸ = (1/3)⁸