Lesson INVERSE FUNCTIONS

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This lesson provides a brief overview of INVERSE FUNCTIONS.

If you're not familiar with functions, please see the lesson on FUNCTIONS.

SUMMARY OF FUNCTIONS AND FUNCTIONAL NOTATION

My main source of reference for this lesson comes from the following web address: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut32b_inverfun.htm. If you want a full explanation with plenty of examples and exercises, then go there. There are other websites that are very good as well.  To find them, all you need to do is go to google or yahoo or any other search engine and look for the subject of interest.

Check out the <A HREF=Plot-Inverse-Function.solver>solver that graphs functions and their inverses</A>.

REVIEW OF FUNCTIONS, RELATIONS, AND EQUATIONS

A very brief review of Functions, Relations, and Equations follows.

Equations are a set of rules that map input values to output values.

Examples of Equations would be:
y = {{{x^2}}}
y = {{{sqrt(x)}}}
y = x
etc.
In these examples, the output value is represented by the letter y and the input value is represented by the letter x.

Relations map input values to output values where there can be more than one output value for one input value.

An example of a relation would be:
y = {{{sqrt(x)}}}

This a relation and not a function because each input value can result in more than one output value.
Example:
x = 25, y = +/- 5

Functions map input values to output values where there can only be one output value for each input value.

An example of a function would be:
y = {{{x^2}}}

This is a function because each x value can result in one and only one y value.

Example:
x = 5, y = 25
x = -5, y = 25

The value of 25 can be the output value for more than one input value, but each input value will only have one output value associated with it.  When x = 5, y = 25 is the only y value for it.  When x = -5, y = 25 is the only y value for it as well.

FUNCTIONAL NOTATION

y = f(x) means y equals a function of x with the name of f.
y = g(x) means y equals a function of x with the name of g.
y = h(t) means y equals a function of t with the name of h.
functions f and g are working with the variable x.
function h is working with the variable t.
x and t would be called the arguments of their respective functions.

COMPOSITE FUNCTIONS

let g(x) = {{{x^3}}}
let h(x) = {{{x^4}}}
g(h(x)) means the function g of the function h of x.
the argument of the function h is x.  that stays the same.
the argument of the function g is normally x, but you are replacing it with the function h(x).  since the function h(x) = {{{x^4}}}, then you are replacing x with {{{x^4}}}.
since g(x) = {{{x^3}}}, then g(h(x)) = g({{{x^4}}})) = {{{(x^4)^3}}} = {{{x^(12)}}}

you can chain composite functions as many times as you need to.
the equation f(g(h(x))) means the function f of the function g of the function h of x.
here's how that works.
let f(x) = {{{sqrt(x)}}}
let g(x) = {{{x^3}}}
let h(x) = {{{x^4}}}
g(h(x)) = {{{(x^4)^3}}}
f(g(h(x)) = {{{sqrt((x^4)^3)}}} = {{{sqrt(x^12)}}} = {{{x^6}}}

INVERSE FUNCTIONS

Inverse functions are essentially the reverse of functions.
They undo what the functions do.

An example of this would be:
y = {{{x^3}}}
The inverse function is:
y = {{{root(3,x)}}}

let x = 3 in the original function.
{{{x^3}}} = {{{(3)^3}}} = 27
your ordered pair would be (3,27)
the x value is 3 and the y value is 27

let x = 27 in the inverse function.
{{{root(3,x)}}} = {{{root(3,27)}}} = 3
your ordered pair would be (27,3)

The inverse function has undone what the original function did.
Where the x value of 3 in the original equation went to a y value of 27, the x value of 27 in the inverse equation went to a y value of 3.

This is a property of inverse functions.
The ordered pair (x,y) in the original function has a corresponding (x,y) in the inverse equation where the x value in the original equation becomes the y value in the inverse equation and vice versa.
(a,b) = (b,a)
(3,27) = (27,3)
you take the value of 3 in the original function and cube it to get 27.  you take the value of 27 in the inverse equation and take the cube root of it to get 3.

(a,b) is said to be a reflection of (b,a) about the line y = x as you will see later on.

To be called Inverse Functions they have to follow the rules of functions meaning there can only be one output value for each input value.
If this rule is violated, the inverse equation exists, but it would be a relation rather than a function and you would have to say that an inverse function does not exist for the original function.

FORMAL DEFINITION OF INVERSE FUNCTIONS

You are given f(x) and g(x)

If f(g(x)) = x and g(f(x)) = x, then g(x) = {{{f^(-1)}}}(x)

g(x) = {{{f^(-1)}}}(x) means that g(x) is the inverse function of x.
you have a function of x which is called f(x).
you have an inverse function of x which is called {{{f^(-1)}}}(x)

The domain of f is equal to the range of {{{f^(-1)}}}.
The range of f is equal to the domain of {{{f^(-1)}}}.

EXAMPLE OF APPLYING THE DEFINITION OF AN INVERSE FUNCTION TO DETERMINE IF THE INVERSE FUNCTION EXISTS.

Let y = f(x) = {{{x^3}}}
let y = g(x) = {{{root(3,x)}}}
The domain of f(x) is all real values of x.
The range of f(x) is all real values of y.

First we test to see if f(g(x)) = x
f(g(x)) = f({{{root(3,x)}}}) = {{{(root(3,x))^3}}} = x
First part is good.
Next we test to see if g(f(x)) = x
g(f(x)) = g({{{x^3}}}) = {{{root(3,x^3)}}} = x
Second part is good.

What was the domain of f(x) is now the range of {{{f^(-1)}}}(x)
What was the range of f(x) is now the domain of {{{f^(-1)}}}(x)
Bear in mind that g(x) can now be called {{{f^(-1)}}}(x) because it passes the test which included the following:
1.  It has to be a function (only one value of y for each x)
2.  f(g(x)) = g(f(x)) = x

THE INVERSE FUNCTION IS A REFLECTION OF THE ORIGINAL FUNCTION ABOUT THE LINE Y = X

What does this mean?
If you draw a line perpendicular to the line y = x, you will see that if the function intersects that perpendicular line at the point (a,b), then the inverse function will intersect that same perpendicular line at the point (b,a).
You will also see further that the distance from the point (a,b) to the line y = x will be the same as the distance from the point (b,a) to the same line y = x.

The function and the inverse function are mirror images with the line y = x acting as the mirror.

A simple example will show this to be true.
let the point (3,27) be one of the points on the graph of the equation y = {{{x^3}}}.
let the point (27,3) be one of the points on the graph of the equation y = {{{root(3,x)}}}.
We want to show that (3,27) and (27,3) are reflections about the line y = x.

A line perpendicular to the line y = x will have a slope which is a negative reciprocal of the slope of the line y = x.
Since the slope of the line y = x is 1, then the slope of the perpendicular line would be -1.
An equation of the line perpendicular to the line y = x and passing through the points (3,27) and (27,3) would be:
y = -x + 30

(3,27) is on this line because 27 = -3 + 30 = 27.
(27,3) is on this line because 3 = -27 + 30 = 3.

To be reflections they have to be on the line y = -x + 30 and be equidistant in opposite directions from the line y = x.

The line y = -x + 30 intersects with the line y = x at the point (15,15).

The distance from (3,27) to (15,15) = {{{sqrt((15-3)^2 + (15-27)^2)}}} = 16.9705.
The distance from (27,3) to (15,15) = {{{sqrt((15-27)^2 + (15-3)^2)}}} = 16.9705.

They are on opposite sides of the line and they are equidistant to the line making them reflections of each other.

The domain of {{{f^(-1)}}}(x) is the same as the range of f(x) which is all real values of x.
The range of {{{f^(-1)}}}(x) is the same as the domain of f(x) which is all real values of x.

I can show you this on a graph as follows:

{{{graph(600,600,-30,30,-30,30,x^3,x^(1/3),x,-x+30)}}}

You can see the line y = x going through the origin and ascending from left to right.
You can see the line y = -x + 30 perpendicular to the line y = x and going through the points (3,27) and (27,3).
If you look from x = 0 to x = 30, you will see that the graph of the equation y = {{{x^3}}} is a mirror image of the graph of the equation y = {{{root(3,x)}}} about the line y = x, with the points (3,27) and (27,3) just being one of the points on the graph of their respective equations.

You can see the graph of y = {{{x^3}}} when x is smaller than 0.  
If you do not see the graph of y = {{{root(3,x)}}} when x is smaller than 0, it's because there is a problem with the graph rendering routines.  It has been reported and will hopefully soon be fixed.
If you see the graph of y = {{{root(3,x)}}} when x is smaller than 0, then the problem has been fixed.

CREATING AN INVERSE FUNCTION

The rules for creating an inverse function are:
1.  Take the function and solve for the input value (x)
2.  Reverse the Input and Output Values (x = y and y = x)
3.  Check to make sure you have a function and not a relation.

Some will tell you to do the following:
1.  Reverse the Input and Output Values (x = y and y = x)
2.  Solve for the output Value (y)
3.  Check to make sure you have a function and not a relation.

I prefer to solve for x and then reverse x and y .
Others prefer to reverse x and y and then solve for y.
Either way will get you the correct answer if you do it properly.

The last step (checking) can be done ahead of time as will be shown later in this lesson.

An important part of this is to also keep track of the domain and the range because they will be flipped, i.e. the domain of the function will become the range of the inverse function and the range of the function will become the domain of the inverse function.

EXAMPLE OF CREATING AN INVERSE FUNCTION FROM A FUNCTION

Your function is:  y = {{{x^2}}}

1.  Take the function and solve for the input value.
{{{x^2}}} = y
take the square root of both sides of this equation to get:
x = {{{sqrt(y)}}}

2.  Reverse the input and output values
This means you take the x and make it a y, and you take the y and make it an x.
Your formula becomes:
y = {{{sqrt(x)}}}

3.  Check to make sure you have a function and not a relation.
In this case you have a relation instead of a function because you can have multiple y values for each x value.
x = 25, y can be + 5 or y can be -5.

This is a poor example because of the special way that the equation {{{y = sqrt(x)}}} is defined in order to make it a function rather than a relation, but let that slide for now and assume that this interpretation is correct and that we have a relation instead of a function.   The error in the  interpretation will be explained further down after the main reason for allowing it to continue  has been completed.  That reason is to show how the vertical and horizontal line tests tell you whether you have a relation or a function.

Is the inverse relation useful?
Absolutely.
But, .....
It's not a function.

You would have to conclude that the function y = {{{x^2}}} does not have an inverse function if you allow the inverse equation to have a plus or minus value associated with it.  If you allow it to have only a plus value associated with it, then it would be a function as described further down in this lesson.

GRAPHICAL TESTS TO DETERMINE WHETHER AN INVERSE FUNCTION IS A FUNCTION OR A RELATION

AFTER you have created the inverse equation that could either be a relation or a function, you can give your inverse function the VERTICAL LINE TEST to see if it is a function.

BEFORE you create the inverse equation that could either be a relation or a function, you can give your original function the HORIZONTAL LINE TEST to determine if an Inverse Function exists for that function.

USE THE GRAPHICAL TEST TO SEE IF THE INVERSE EQUATION IS A FUNCTION AFTER YOU HAVE CREATED IT.

You have created the inverse equation of y = {{{sqrt(x)}}} and you give it the VERTICAL LINE TEST to see if it is a function.

The vertical line test is where you check each x value on the graph of the equation to see if more than one y value exists for the same x value.

The vertical line test would show that the inverse equation of y = {{{sqrt(x)}}} for the original function of y = {{{x^2}}} is a relation and not a function because the vertical line passes through the graph of it more than once on at least one occasion.

Please disregard the fact that my vertical line is not exactly vertical.  Assume that what you are seeing is a vertical line.  I was not able to exactly represent the vertical line.  This was the closest I could get.

{{{graph(600,300,-5,5,-3,3,sqrt(x),-sqrt(x),(1000/(x-3)))}}}

USE THE GRAPHICAL TEST TO SEE IF AN INVERSE FUNCTION EXISTS BEFORE YOU CREATE THE EQUATION FOR IT.

You can check your original function BEFORE you create the inverse equation to see if an Inverse Function exists for it by applying the HORIZONTAL LINE TEST to it.

The horizontal line test is where you check each y value on the graph of the equation to see if more than one x value exists for the same y value.

Your original function is y = {{{x^2}}}
The horizontal line test would show that an inverse function for the original function would not exist because the horizontal line passes through the graph of it more than once on at least one occasion.

{{{graph(600,300,-5,5,-5,5,x^2,3))}}}

The HORIZONTAL LINE TEST is a quick way to see if an inverse function exists before even creating the inverse equation.

DOMAIN AND RANGE OF A FUNCTION AND OF THE INVERSE EQUATION

With no restriction on the domain of the function y = {{{x^2}}}, the inverse function does NOT exist.
We can restrict the domain of the function y = {{{x^2}}} in such a way that the inverse function DOES exist.

CREATING THE INVERSE EQUATION FOR f(x) = {{{x^2}}} WITH A RESTRICTION ON THE DOMAIN OF THE ORIGINAL FUNCTION SO THAT THE DOMAIN IS POSITIVE REAL VALUES OF X RATHER THAN ALL REAL VALUES OF X

suppose we state that the domain of x is all positive real values of x.  Negative values of x would not allowed.

The original function is f(x) = {{{x^2}}}
Domain (x) = {{{x >= 0}}} = all positive real values of x because we specified the domain as such.  Negative numbers are not allowed.
Range (f(x)) = {{{y >= 0}}} = all positive real values of y because the function does not have the capability for generating negative values of y.

When we create the inverse equation y = {{{sqrt(x)}}}, we make x in the inverse equation equal to what was y in the original equation, and we make y in the inverse equation equal to what was x in the original equation (they are interchanged).  
This reverses the domain and range of the inverse equation.
The domain of the original function becomes the range of the inverse equation not yet proven to be a function.
The range of the original function becomes the domain of the inverse equation not yet proven to be a function.

The inverse equation states that y = {{{sqrt(x)}}}

The domain of the inverse equation is all positive real values of x.  This is because the range of the original function was all positive real values of y.

The range of the inverse equation is all positive real values of y.  This is because the domain of the original function was all positive real values of x.

The inverse equation is now a function because there does not exist at least one value of x for which there is more than one value of y.  This is because the range of y can now only be positive. The square root of a number can still be positive or negative (sqrt(25) = +/- 5), but only the positive number is allowed.  The negative is not.

A graph of the original function would be:

{{{graph(600,300,-1,5,-5,5,x^2,3))}}}

Only the values of x >= 0 are valid in this graph because the domain is restricted to positive values of x.

You can see from the graph of the original equation that an Inverse Function does exist because the HORIZONTAL LINE TEST will show that the horizontal line will only cross the x value in the graph of the equation once for each value of y as long as we restrict the domain of the original function to x >= 0.

A graph of the inverse equation would be:

{{{graph(600,300,-5,5,-1,5,sqrt(x),1000/(x-3))}}}

You can see from the graph of the inverse equation that an Inverse Function does exist because the VERTICAL LINE TEST will show that the vertical line will only cross the y value in the graph of the equation once for each value of x as long as we restrict the range of the inverse equation to y >= 0.

Remember that the domain of the original function equals the range of the inverse equation.
Remember also that the range of the original function equals the domain of the inverse equation.
The inverse equation is a function if the rules of functions are obeyed.
Otherwise the inverse equation is a relation.

The problem of interpretation stated above is that I called the inverse equation of {{{y = x^2}}} a relation rather than a function.
The inverse equation was {{{y = sqrt(x)}}}
By convention, {{{y = sqrt(x)}}} is defined for all positive values of y only.
By calling it a relation, I was violating that convention because I was allowing negative values of y as well.

The violation was allowed to continue because it enabled me to show you the graph of a relation rather than a function.

In truth, by the strict rules of inverse equations, my interpretation was correct because the original function of {{{x^2}}} allowed negative and positive values of x in the domain which resulted in negative and positive values in the range of the inverse equation.

For practical purposes, however, you should assume the following:

If you are given a function of {{{y = x^2}}}, and you are solving for x, then x = +/- {{{sqrt(x)}}} is a valid answer.  negative and positive roots apply.  Note that you are solving for x here, and not y.  This allows the function to continue to be a function because you may have multiple values of x for each y but you still only have one value y for each x.

If you are given a function of {{{y = sqrt(x)}}}, and you are solving for y, then y = + {{{sqrt(x)}}} is a valid answer.  positive root applies.  negative root doesn't.  Note that you are solving for y here, and not x.  In order for this function to continue to be a function, only the positive roots are allowed.  In order to get the negative roots, your equation would need to be {{{y = -sqrt(x)}}}