Find Inverse of Rational Function

[mathdad]

Find the inverse of
f(x) = 2/(x - 3).

Let y = f(x)

y = 2/(x - 3)

Replace y for x.

x = 2/(y - 3)

x(y - 3) = 2

Solve for y.

xy - 3x = 2

xy = 2 + 3x

y = (2 + 3x)/x

Replace y with f^-1 (x).

f^-1(x) = (2 + 3x)/x

1. Is f^-1(x) the inverse of f(x)?

2. What does f(x) and f^-1(x) look like together on the same xy-plane?

 

[MarkFL]

Re: Find Inverse of Rational Number

1. Is f^-1(x) the inverse of f(x)?

If you did the work correctly, then you should find:

\(\displaystyle f\left(f^{-1}(x)\right)=f^{-1}\left(f(x)\right)=x\)

2. What does f(x) and f^-1(x) look like together on the same xy-plane?

A function and it inverse will appear to be reflected across the line $y=x$.

 

[MarkFL]

About the reflection part, if we have some point on a function $(x,y)$, then the point on the inverse corresponding to this point is naturally $(y,x)$. The mid-point of these points is:

\(\displaystyle \left(\frac{x+y}{2},\frac{x+y}{2}\right)\)

This point is of course on the line $y=x$.

And the slope of the line segment between them is:

\(\displaystyle m=\frac{x-y}{y-x}=-1\)

The two points are the same perpendicular distance from the line $y=x$.

Thus, we see that a point reflected across the line $y=x$ will simply require swapping $xy$ coordinates, which is what the inverse is.

 

[mathdad]

So, the inverse of (x, y) is simply (y, x). What if the function is too complicated to find its inverse by hand? Must we then rely on a computer system or graphing calculator?

 

[DrWahoo]

So, the inverse of (x, y) is simply (y, x). What if the function is too complicated to find its inverse by hand? Must we then rely on a computer system or graphing calculator?

If the function has an inverse that is also a function, then there can only be one y for every x. In two dimensions. Higher dimensions, yes I would use numerical compilers.

A one-to-one function, is a function in which for every x there is exactly one y and for every y, there is exactly one x.

A one-to-one function has an inverse that is also a function.

There are functions which have inverses that are not functions. There are also inverses for relations. For the most part, we disregard these, and deal only with functions whose inverses are also functions.

Later on in abstract algebra this is one of the key ingredients for an isomorphism. Its the basis for a lot of topology classes, analysis classes, combinatorics. Its really a fundamental necessity to understand what it means for a function to be an inverse. I.E. the identitiy is one key.

 

[mathdad]

If the function has an inverse that is also a function, then there can only be one y for every x. In two dimensions. Higher dimensions, yes I would use numerical compilers.

A one-to-one function, is a function in which for every x there is exactly one y and for every y, there is exactly one x.

A one-to-one function has an inverse that is also a function.

There are functions which have inverses that are not functions. There are also inverses for relations. For the most part, we disregard these, and deal only with functions whose inverses are also functions.

Later on in abstract algebra this is one of the key ingredients for an isomorphism. Its the basis for a lot of topology classes, analysis classes, combinatorics. Its really a fundamental necessity to understand what it means for a function to be an inverse. I.E. the identitiy is one key.

1. Thank you for the data.

2. I am doing a self-study of math.

3. I am not a student.

4. I am a middle-aged man.

5. I do not need to go beyond Calculus 3, which is my ultimate goal that will be accomplished in time. Linear algebra and beyond is not my concern. Abstract algebra is so not possible given my age and interest.

 

[DrWahoo]

So, the inverse of (x, y) is simply (y, x). What if the function is too complicated to find its inverse by hand? Must we then rely on a computer system or graphing calculator?

Not exactly. The function y could possibly not have an inverse at all. If you are in two dimensions and the function is one- to- one then yes, your logic holds.

The typical vertical line test tells you by graphing a function is in fact a function (vertical line test). Use the horizontal line test to check if the graph of the inverse is indeed true. This follows from the result of the transformation MikeFL discussed about the reflection.

- - - Updated - - -

1. Thank you for the data.

2. I am doing a self-study of math.

3. I am not a student.

4. I am a middle-aged man.

5. I do not need to go beyond Calculus 3, which is my ultimate goal that will be accomplished in time. Linear algebra and beyond is not my concern. Abstract algebra is so not possible given my age and interest.

Very good questions! It is always good to keep old topics fresh. I am a professor and I tend to forgot little details at times. Just remember practice, practice, practice. Good luck to you in those courses!

 

[mathdad]

Not exactly. The function y could possibly not have an inverse at all. If you are in two dimensions and the function is one- to- one then yes, your logic holds.

The typical vertical line test tells you by graphing a function is in fact a function (vertical line test). Use the horizontal line test to check if the graph of the inverse is indeed true. This follows from the result of the transformation MikeFL discussed about the reflection.

- - - Updated - - -
Very good questions! It is always good to keep old topics fresh. I am a professor and I tend to forgot little details at times. Just remember practice, practice, practice. Good luck to you in those courses!

Thanks. Not courses but self-study.