Imagine that you live on an infinitely long and straight street called Infinite Drive. The addresses on Infinite Drive are given by real numbers. Your address on the street is π while your friend Patrick’s is √3 and Karen’s is sin60°. There are two moving companies, f and g. Those companies move people along Infinite Drive from address x to new addresses f(x) or g(x), depending on which company people choose.
1.Assume that the functions f(x) = 1-x and g(x) = 1/x. Describe how the moving companies move people from one address to another.
a.What are all possible addresses that you can move to when taking repeated compositions of the functions f and g?
b.How many possible addresses can you relocate to when using those moving companies? Are they finite? If so, make a list of all the possible new addresses, if your original address is π.
c.If your parents live on address 0, what are all the possible addresses that they can move to by repeatedly composing the two functions?
f(x) = 1-x
f(f(x)) = 1 - (1-x) = x
g(x) = 1/x
g(g(x)) = 1/(1/x) = x
So, in both cases, there are only two possible addresses available for each real number. Alas, g(0) is undefined, so it looks like the mover gets lost.
a. To determine all possible addresses that can be reached by repeated compositions of the functions f(x) = 1-x and g(x) = 1/x, we can apply the functions multiple times.
Starting with an initial address x, applying the function f(x) = 1-x will result in a new address 1-x. Applying the function again on the new address will yield 1-(1-x) = x.
For the function g(x) = 1/x, applying it to an address x will give 1/x as the new address. Applying it again to the new address will result in 1/(1/x) = x.
Therefore, when taking repeated compositions of f and g, all possible addresses that can be reached are the original address x and its reciprocal 1/x.
b. If the original address is π, plugging π into the functions f(x) = 1-x and g(x) = 1/x gives:
f(π) = 1 - π ≈ -0.1416
g(π) = 1/π ≈ 0.3183
Applying f again to the address -0.1416 gives:
f(-0.1416) = 1 - (-0.1416) ≈ 1.1416
Applying g again to the address 0.3183 gives:
g(0.3183) = 1/0.3183 ≈ 3.1421
Therefore, when starting with an original address of π and repeatedly applying f and g, the possible new addresses are approximately -0.1416, 1.1416, 0.3183, and 3.1421. Hence, there are four possible addresses.
c. If your parents live at address 0, applying f and g to the address 0 gives:
f(0) = 1 - 0 = 1
g(0) = 1/0 = undefined
Since division by zero is undefined, applying g to 0 is not possible. Therefore, the only possible address that your parents can move to by repeatedly composing f and g is 1.