## Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations. f(1)=2*1=2 f^2(1)=2

Start with the basic function f(x)=2x. If you have an initial value of 1, then you end up with the following iterations. f(1)=2*1=2 f^2(1)=2*2*1=4 f^3(1)=2*2*2*1=8 1. If you continue this pattern, what do you expect would happen to the numbers as the number of iterations grows? Check your result by conducting at least 10 iterations. 2. Repeat the process with an initial value of −1. What happens as the number of iterations grows?

## Answers ( No )

f(x)=2x

f(1)=2*1=2

f^2(1)=2*2*1=4

f^3(1) =2*2*2*1=8

1. If you

continue this pattern, what do you expect would happen to the numbers as

the number of iterations grows?

I expect the numbers continue growing multiplying each time by 2.

Check your result by conducting at

least 10 iterations.

f^4(1) = f^3(1) * f(1) = 8*2 = 16

f^(5)(1) = f^4(1) * f(1) = 16 * 2 = 32

f^6 (1) = f^5 (1) * f(1) = 32 * 2 = 64

f^7 (1) = f^6 (1) * f(1) = 64 * 2 = 128

f^8 (1) = f^7 (1) * f(1) = 128 * 2 = 256

f^9 (1) = f^8 (1) * f(1) = 256 * 2 = 512

f^10 (1) = f^9 (1) * f(1) = 512 * 2 = 1024

2. Repeat the process with an initial value of −1.

What happens as the number of iterations grows?

f(-1) = 2(-1) = – 2

f^2 (-1) = f(-1) * f(-1) = – 2 * – 2 = 4

f^3 (-1) = f^2 (-1) * f(-1) = 4 * (-2) = – 8

f^4 (-1) = f^3 (-1) * f(-1) = – 8 * (-2) = 16

f^5 (-1) = f^4 (-1) * f(-1) = 16 * (-2) = – 32

As you see the magnitude of the number increases, being multiplied by 2 each time, and the sign is aleternated, negative positive negative positive …