Sketch the graph of an invertible function

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To sketch the graph of an invertible function that intersects its inverse at exactly three points, one must select an injective function, which is either strictly increasing or decreasing. Functions like y=x^3 or y=x^5 serve as examples since they maintain a positive slope and can be designed to intersect the line y=x at three distinct points. The key is to ensure that the function f(x) - x equals zero at exactly three values of x, indicating the points of intersection. This problem highlights the unique nature of functions and their inverses, as they typically intersect only on the identity line y=x. Ultimately, constructing a cubic function that meets these criteria is a viable approach.
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Homework Statement


Sketch the graph of an invertible function f(x) that intersects its inverse in exactly three points.



The Attempt at a Solution


I don't know where to start... Can I pick any function to start with?
 
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You can't choose any function. If, for example, you choose the function y=x2 then x=2 -> y=4 and x=-2 -> y=4. Clearly this function doesn't have an inverse because given a value for y (other than 0) there are two corresponding values for x.

You need to choose a function y=f(x) with the property that any two different values of x always yield two different values for y. Such functions are called injective functions and have graphs that are monotonic (i.e. either increasing or decreasing but not both).

Lastly, to find the inverse of such a graph, you need to reflect it in the line x=y (the 45 degree line). Try this with some examples and look at what all of the points of intersection between the function and its inverse have in common--that should help you to figure out how to get three points of intersection.
 
This is an interesting problem because from what I remember functions and their inverses only intersect on the identity function y=x
 
armolinasf said:
This is an interesting problem because from what I remember functions and their inverses only intersect on the identity function y=x
Yes, that's right. So in order to satisfy "intersects its inverse in exactly three points" there must be exactly three values of x such that f(x)= x. That is the same as saying f(x)- x= 0 for exactly three values of x. However the problem is making sure f is invertible then. You need to go from each such x to the next without having negative slope.

My first thought was to make f(x)- x a cubic. It's easy to construct a cubic that is 0 at three given values but a little harder to make sure it is always increasing. (Actually, there is a very simple example!)
 

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