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Main Question or Discussion Point
If we have a relation, ##R##, and it's inverse, ##R^{1}## they behave such that a point on ##R##, say (a,b), corresponds to the point (b,a) on ##R^{1}## This is a reflections across the line y=x.
This relation does not mean that ##R^{1}## is a function. For example,
Let ##R## be ##y=x^25##. Then ##R^{1}## is ##x=y^25## These are both parabolas reflected across the line y=x. However, since ##R## is not onetoone, then ##R^{1}## is not a function.
Now, consider the two relations as functions, rather than just relations.
Then we have, ##f\left(x\right)=x^25##
To find the inverse function of ##f##, we swap the variables x and y, then solve for y.
This results in ##f^{1}\left(x\right)=\sqrt{x5}##
Now I have a attached a graph of these functions with respect to one another as reference to my question:
Take a look at the domains and ranges of ##R## and ##R^{1}##.
They swap, as they are reflections of each other across the line y=x.
Domain ##R##: ##\left(∞,∞\right)##, Range of ##R##: ##\left(∞,5\right]##
Domain ##R^{1}##: ##\left(∞,5\right]##, Range of ##R^{1}##: ##\left(∞,∞\right)##
Now in looking at the graph of ##f^{1}\left(x\right)##, the range is restricted to the interval ##\left[0,∞\right)##, which is what makes it a function by having one input to exactly one output.
So my question: When we find the inverse relation of a function, there are no restrictions. The domains and ranges swap. When we find the inverse function of a function, restrictions apply which allow the inverse to be a function.
Where and how does this "restriction take place"? Does it happen because when we swap x and y then solve for y, we execute an operation that is the opposite of the main operation of ##f\left(x\right)##? And in this case, the main operation for ##f^{1}\left(x\right)## is the square root, which in its nature has restrictions, so the inverse graph itself is then restricted by the main inverse function operator? Or does the restriction take place before we even find the inverse, by restricting the domain of ##f\left(x\right)##?
This relation does not mean that ##R^{1}## is a function. For example,
Let ##R## be ##y=x^25##. Then ##R^{1}## is ##x=y^25## These are both parabolas reflected across the line y=x. However, since ##R## is not onetoone, then ##R^{1}## is not a function.
Now, consider the two relations as functions, rather than just relations.
Then we have, ##f\left(x\right)=x^25##
To find the inverse function of ##f##, we swap the variables x and y, then solve for y.
This results in ##f^{1}\left(x\right)=\sqrt{x5}##
Now I have a attached a graph of these functions with respect to one another as reference to my question:
Take a look at the domains and ranges of ##R## and ##R^{1}##.
They swap, as they are reflections of each other across the line y=x.
Domain ##R##: ##\left(∞,∞\right)##, Range of ##R##: ##\left(∞,5\right]##
Domain ##R^{1}##: ##\left(∞,5\right]##, Range of ##R^{1}##: ##\left(∞,∞\right)##
Now in looking at the graph of ##f^{1}\left(x\right)##, the range is restricted to the interval ##\left[0,∞\right)##, which is what makes it a function by having one input to exactly one output.
So my question: When we find the inverse relation of a function, there are no restrictions. The domains and ranges swap. When we find the inverse function of a function, restrictions apply which allow the inverse to be a function.
Where and how does this "restriction take place"? Does it happen because when we swap x and y then solve for y, we execute an operation that is the opposite of the main operation of ##f\left(x\right)##? And in this case, the main operation for ##f^{1}\left(x\right)## is the square root, which in its nature has restrictions, so the inverse graph itself is then restricted by the main inverse function operator? Or does the restriction take place before we even find the inverse, by restricting the domain of ##f\left(x\right)##?
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