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I need help checking to make sure my work is correct. 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 x 1 = 2f^2 (1) = 2 x 2 x 1 = 4f^3 (1) = 2 x 2 x 2 x 1 = 8The question Part 1: If you continue the 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. I put: f^4 (1) = 2 x 2 x 2 x 2 x 1 = 16f^5 (1) = 2 x 2 x 2 x 2 x 2 x 1 = 32f^6 (1) = 2 x 2 x 2 x 2 x 2 x 2 x 1 = 64f^7 (1) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 1 = 128f^8 (1) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 1 = 256f^9 (1) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 1 = 512f^10 (1) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 1 = 1024Part 2: Repeat the process with an initial value of -1. What happens as the number of iterations grows?

Respuesta :

Given: The function below:

[tex]f(x)=2x[/tex]

To Determine: The interation with initial value of 1

When the initial value is 1, it means that x = 1

If x =1, we can determine f(1) by the substituting for x in the function as shown below:

[tex]\begin{gathered} f(x)=2x \\ x=1 \\ f(1)=2(1)=2\times1=2 \end{gathered}[/tex][tex]f^2(1)=2^2\times1=2\times2\times1=4[/tex]

Part 1:

It can be observed that as the number of iterations grow, the number increase in powers of 2

This can be modelled as

[tex]f^n=2^n\times1=2^n[/tex][tex]f^{10}=2^{10}\times1=1024[/tex]

Part 2:

If we repeat the process with an initial value of -1. As the number of iterations grows, the number can be modelled as

[tex]\begin{gathered} f^{-n}=2^{-n}\times1 \\ f^{-1}=2^{-1}\times1=\frac{1}{2}\times1=\frac{1}{2} \\ \text{For initial value of -2, we would have} \\ f^{-2}=2^{-2}\times1=\frac{1}{2^2}\times1=\frac{1}{4} \end{gathered}[/tex]

So, as the initial value decreases, it can be observed by the above calculations that the number would be decreasing by the the reciprocal of the power of 2.

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