Intelligence Capitalism : The Techonomist's Edge: What happens 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} ?

1/04/2023

What happens 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 $f^{-1}$ ?

Suppose $f:R \rightarrow R$ is a function from the set of real numbers to the same set with $f(x)=x+1$ . We write $f^{2}$ to represent $f \circ f$ and $f^{n+1}=f^n \circ f$ . Is it true that $f^2 \circ f = f \circ f^2$ ? Why? Is the set { $g:R \rightarrow R$ l $g \circ f=f \circ g$ } infinite? Why?

Inherently, the range of a function f (x) is the domain of the inverse function $f^{-1}$ . The domain of $f$ is the range of . So, what happens 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 $f^{-1}$ ? Let's try some examples.

If we type $y=x^3$ , $y=x^{1/3}$ , and on the graph, all of the three functions will be limited in the same input and their output will also affected by such a input boundary. The domain of $f$ become the domain of , not its range.

Let's try another example: C° to F°: Celsius to Fahrenheit Conversion Formula

Again, as we can see that these these three lines are more like projections of each other from different angles rather than reflection.

The difficulty of setting the same set of axes for $f$ and $f^{-1}$ is that $f^{-1}$ will become not fully or true inverse function of $f$ since the output of $f^{-1}$ is limited. Instead of setting the input {-100 < x < 100}, the input of $f^{-1}$ should be the range of $f$

Suppose $f:R \rightarrow R$ is a function from the set of real numbers to the same set with $f(x)=x+1$ . We write $f^{2}$ to represent $f \circ f$ and $f^{n+1}=f^n \circ f$ .

When $f^{n+1}=f^n \circ f$ . and