Question

How do you find the first three iterate of the function `f(x)=3x^2-4` for the given initial value `x_0=1`?

Answer 1

`-1, -1, -1`

Explanation:

Applying the function, we find:

`x_1 = f(x_0) = 3x_0^2-4 = 3(1)^2-4 = 3-4 = -1`

`x_2 = f(x_1) = 3x_1^2-4 = 3(-1)^2-4 = 3-4 = -1`

`x_3 = f(x_2) = 3x_2^2-4 = 3(-1)^2-4 = 3-4 = -1`

In fact, for any `n >= 1` we have `x_n = -1`, since:

`f(-1) = 3(-1)^2-4 = -1`

is a fixed point.