What happens if we graph both ƒ and ƒ^(-1) on the same set of axes, using the x-axis for the input to both ƒ and ƒ^(-1)?
[Suggestion: go to www.desmos.com/calculator and type y = x^3 {-2 < x < 2}, y = x^(1/3) {–2 < x < 2}, and y = x {–2 < x < 2}, and describe the relationship between the three curves.] Then post your own example discussing the difficulty of graph both ƒ and ƒ^(-1) on the same set of axes.
Suppose ƒ: R → R is a function from the set of real numbers to the same set with ƒ(x) = x + 1. We write ƒ^2 to represent ƒ ∘ ƒ and ƒ^(n+1) = ƒ^n ∘ ƒ. Is it true that ƒ^2 ∘ ƒ = ƒ ∘ ƒ^2? Why? Is the set {g : R → R l g ∘ ƒ = ƒ ∘ g} infinite? Why?
If we graph both f and f^{-1} on the same set of axes, using the x-axis for the input to both f and f^{-1}, we can observe some interesting relationships between the curves. Let’s consider the suggested functions:
1. f(x) = x^3, where -2 < x < 2
2. f^{-1}(x) = x^{1/3}, where -2 < x < 2
3. g(x) = x, where -2 < x < 2
When we graph these three curves, we can see that f(x) = x^3 forms a curve that starts from the origin, passes through (-1, -1), (0, 0), and (1, 1), and then extends towards infinity in both positive and negative directions. This curve is symmetric about the origin.
The inverse function f^{-1}(x) = x^{1/3} reflects the f(x) curve across the line y = x. It starts from the origin, passes through (-1, -1), (0, 0), and (1, 1), and then extends towards infinity in both positive and negative directions. However, it is important to note that the inverse curve is restricted to the positive values of x due to the nature of the cube root function.
Finally, the function g(x) = x represents a straight line passing through the origin with a slope of 1. This line is also symmetric about the line y = x.
Now, let’s address your second question regarding the difficulty of graphing both f and f^{-1} on the same set of axes. In general, the difficulty arises when the inverse function is not explicitly given or when it is difficult to find analytically. In such cases, it becomes challenging to graph both the function and its inverse simultaneously. It might require numerical methods or approximation techniques to determine the inverse function.
Moving on to the third part of your question, let’s analyze the statement f^2 ∘ f = f ∘ f^2. In this case, f^2 represents the composition of f with itself, and f^n+1 represents the composition of f^n with f. So, f^2 ∘ f means applying f twice and then applying f again, while f ∘ f^2 means applying f once and then applying f twice.
For the given function f(x) = x + 1, we can evaluate both sides of the equation and observe that:
f^2 ∘ f = f ∘ f^2
(f ∘ f) ∘ f = f ∘ (f ∘ f)
(f+1) ∘ (f+1) ∘ (f+1) = (f+1) ∘ (f+1) ∘ (f+1)
Expanding both sides, we get:
(f+1)^3 = (f+1)^3
Simplifying further, we have:
f^3 + 3f^2 + 3f + 1 = f^3 + 3f^2 + 3f + 1
As we can see, both sides of the equation are equal, confirming that f^2 ∘ f = f ∘ f^2 is true for the given function.
Lastly, let’s consider the set {g:R → R l g ∘ f = f ∘ g}, where g represents a function from the set of real numbers to itself, and g ∘ f = f ∘ g. In this case, we can observe that the set is indeed infinite.
To see why, let’s consider a function g(x) = x + c, where c is a constant. If we evaluate g ∘ f and f ∘ g, we get:
g ∘ f = (x + c) ∘ (x + 1) = x + c + 1
f ∘ g = (x + 1) ∘ (x + c) = x + 1 + c
As we can see, for any constant c, the functions g(x) = x + c satisfy the condition g ∘ f = f ∘ g. Since there are infinitely many possible values for c, the set {g:R → R l g ∘ f = f ∘ g} is indeed infinite.