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.
Now assume that the moving companies are described by the functions f(x) = 5x and g(x) = 1/x.
a.Find a general form for all the functions that you can get by taking repeated compositions of the two functions.
b.What are all the possible addresses that you can move to? If the list is infinite, list at least three possible addresses.
To find a general form for all the functions that can be obtained by taking repeated compositions of the functions f(x) = 5x and g(x) = 1/x, we can start by considering the possible compositions of these functions.
a. Taking repeated compositions of these functions means applying one function after another. For example, we could have f(g(x)) which means applying function g(x) first and then applying function f(x) to the result. Similarly, we could have g(f(x)) which means applying function f(x) first and then applying function g(x) to the result.
Let's consider the possible compositions:
f(g(x)) = f(1/x) = 5(1/x) = 5/x
g(f(x)) = g(5x) = 1/(5x) = 1/5x
f(f(x)) = f(5x) = 5(5x) = 25x
g(g(x)) = g(1/x) = 1/(1/x) = x
By considering all possible compositions of the functions f(x) and g(x), we have found the following functions:
f(g(x)) = 5/x
g(f(x)) = 1/5x
f(f(x)) = 25x
g(g(x)) = x
So, the general form for all the functions that can be obtained by taking repeated compositions of the functions f(x) = 5x and g(x) = 1/x is:
f(g(x)) = 5/x
g(f(x)) = 1/5x
f(f(x)) = 25x
g(g(x)) = x
b. The possible addresses that can be moved to depend on the input x and the functions applied. Let's consider some examples:
- If we start with address x = π:
- f(g(π)) = f(1/π) = 5/π
- g(f(π)) = g(5π) = 1/(5π)
- f(f(π)) = f(5π) = 25π
- g(g(π)) = g(1/π) = π
- If we start with address x = √3:
- f(g(√3)) = f(1/√3) = 5/√3
- g(f(√3)) = g(5√3) = 1/(5√3)
- f(f(√3)) = f(5√3) = 25√3
- g(g(√3)) = g(1/√3) = √3
- If we start with address x = sin60°:
- f(g(sin60°)) = f(1/sin60°) = 5/(1/sin60°) = 5sin60° = 5√3/2
- g(f(sin60°)) = g(5sin60°) = 1/(5sin60°)
- f(f(sin60°)) = f(5sin60°) = 25sin60° = 25√3/2
- g(g(sin60°)) = g(1/sin60°) = sin60°
The possible addresses we can move to include 5/x, 1/(5x), 25x, and x. However, since Infinite Drive spans infinitely, there are infinitely many possible addresses. The examples above provide at least three possible addresses: 5/π, 1/(5π), and 25π.