# 1.7 - Inverse Functions

## Notation

The inverse of the function f is denoted by f -1 (if your browser doesn't support superscripts, that is looks like f with an exponent of -1) and is pronounced "f inverse". Although the inverse of a function looks like you're raising the function to the -1 power, it isn't. The inverse of a function does not mean the reciprocal of a function.

## Inverses

A function normally tells you what y is if you know what x is. The inverse of a function will tell you what x had to be to get that value of y.

A function f -1 is the inverse of f if

• for every x in the domain of f, f -1[f(x)] = x, and
• for every x in the domain of f -1, f[f -1(x)] = x

The domain of f is the range of f -1 and the range of f is the domain of f -1.

## Graph of the Inverse Function

The inverse of a function differs from the function in that all the x-coordinates and y-coordinates have been switched. That is, if (4,6) is a point on the graph of the function, then (6,4) is a point on the graph of the inverse function.

Points on the identity function (y=x) will remain on the identity function when switched. All other points will have their coordinates switched and move locations.

The graph of a function and its inverse are mirror images of each other. They are reflected about the identity function y=x.

## Existence of an Inverse Function

A function says that for every x, there is exactly one y. That is, y values can be duplicated but x values can not be repeated.

If the function has an inverse that is also a function, then there can only be one y for every x.

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.

If the inverse of a function is also a function, then the inverse relation must pass a vertical line test. Since all the x-coordinates and y-coordinates are switched when finding the inverse, saying that the inverse must pass a vertical line test is the same as saying the original function must pass a horizontal line test.

If a function passes both the vertical line test (so that it is a function in the first place) and the horizontal line test (so that its inverse is a function), then the function is one-to-one and has an inverse function.

## Finding Inverses Informally

The inverse of some functions, especially those where there is only one occurrence of the independent variable, can be solved by undoing the operations. To undo the operations, you must not only reverse the order, but use the inverse operation.

### Example 1

#### The function f(x) = 5x-2

2. Multiply by 5: 5x
3. Subtract 2: 5x-2

#### The inverse f -1(x) = (x+2)/5

3. Divide by 5: (x+2) / 5

### Example 2

#### The function f(x) = 2(x-3)2-5, x≥3

Note that there is a restriction on x.

2. Subtract 3: x-3
3. Square: (x-3)2
4. Multiply by 2: 2(x-3)2
5. Subtract 5: 2(x-3)2-5

#### The inverse f -1(x) = 3 + sqrt[(x+5)/2]

3. Divide by 2: (x+5)/2
4. Take the square root: ± sqrt[(x+5)/2]
5. Add 3: 3 ± sqrt[(x+5)/2]
6. Wait! That inverse isn't a function because there are two values of y for every x. That's because of the ± that appeared when we took the square root of both sides. Now we go back to the original domain of x≥3. That means that for the inverse, the range is y≥3. Since y must be at least 3, we need the positive square root and not the negative. Without the restriction on x in the original function, it wouldn't have had an inverse function: 3 + sqrt[(x+5)/2]

### Example 3

#### The function f(x) = x2 - 4x + 6, x≤2

Uhhh????

What happens when there is more than one occurrence of the independent variable in the function? You don't know what you did to x because you did it to two different x's and you didn't do the same thing to both of them.

## Finding Inverses Formally

Now, that last example is not to be said it can't be done, but it involves completing the square to obtain f(x) = (x-2)2+2, then inversing it so that you get f -1(x) = 2-sqrt(x-2).

However, there is another way that doesn't rely so much on informality and will work whether or not you can figure out exactly what you did with exactly one x.

2. Replace f(x) by y if necessary
3. Switch the x's and y's. At this point you are dealing with the inverse
4. Solve for y
5. Replace y with f -1(x) if the inverse is also a function, otherwise leave it as y

### Example 4

#### The function f(x) = x2 / (x2+1), x≥0

The restriction is important to make it 1-1.

2. Replace f(x) by y: y = x2 / (x2+1), x≥0
3. Switch the x's and y's: x = y2 / (y2+1), y ≥0
4. Solve for y:
1. Multiply by denominator: x(y2+1) = y2
2. Distribute: xy2 + x = y2
3. Move y's to one side and everything else to other: xy2-y2 = -x
4. Factor: y2(x-1)=-x
5. Divide by coefficient on y2: y2 = -x/(x-1)
6. Simplify right hand side: y2 = x/(1-x)
7. Take square root: y = ± sqrt[ x/(1-x) ]
8. Since y≥0, we need the positive square root: y = sqrt[ x/(1-x) ]
5. Call this f -1(x): f -1(x) = sqrt[ x/(1-x) ]

On this last function, the implied domain of the inverse is [0,1). That means that the range of the original function must have been [0,1), also. Check it on your calculator, and you'll see it is.

Sometimes the instructions say if the function is not one-to-one, then don't find the inverse function (because there's not one). So, always check before wasting time trying to find the inverse function. Now, if you're supposed to find the inverse, regardless of whether it is a function or not, then go ahead.

### Good Stuff!

One-to-One Functions are wonderful things.

When solving equations, you can add the same thing to both sides, subtract the same thing from both sides, multiply both sides by the same non-zero thing, and divide both sides by the same non-zero thing and still get the same solution without worrying about having to check your answer.

You can also apply a one-to-one function to both sides of an equation without worrying about introducing extraneous solutions (solutions that work after doing something that didn't work before). This is not necessarily true with functions that aren't one-to-one like the squaring function where you should always check answers after you square both sides of an equation. For example, the equation sqrt(x) = -2 has no solution, but if you square both sides, you get x = 4, but it doesn't check in the original problem. With one-to-one functions, you won't be introducing any extraneous solutions.

Wow! Talk about powerful. You don't appreciate it now, and the book doesn't deal with it properly until you get to chapter 4 and deal with logarithmic and exponential functions, and even then they don't make as big of deal out of it as it is.

Okay, let's try one now. Take my word for it that exp(x) is a one-to-one function and is the inverse of ln(x).

ln(x) = 3
Solve for x.
exp[ ln(x) ] = exp [ 3 ]
"Wait a minute, Mr. Jones" is your response. You've never seen such a beast. That's okay. Take the inverse of the function, and apply it to both sides.
x = exp ( 3 )
Go back to the definition of an inverse at the top of this document. When inverses are applied to each other, they inverse each other out, and you're just left with the argument (input) to the function.
x = e3
On the calculator, the exp(x) function is written e^x, and is found on the [2nd] [ln] key.

Wow - more cohesiveness. The inverse of a function is found by taking the [2nd] function. Look at it for other things on the calculator.

The square root is the inverse of the square. If you look at the three trigonometric keys [sin], [cos], and [tan], their inverses are all found by using the [2nd] key.

Soapbox mode on.

I'm telling you - it all fits together. For those who remember the line Hannibal Smith used in the A-Team, "I love it when a plan comes together".

Mathematics is one of the most together subjects there is. Everything complements everything else. What I'm hoping you get out of this course is much more than just the mechanics of mathematics, but a comprehension, understanding, and appreciation of the way the system works. With that solid foundation, mathematics can be less stressful, and yes, even enjoyable. You have to stop dealing with concepts as individual things that don't relate to each other and stand alone. They are all related to each other and intertwined. You can not separate them and understand.

Soapbox mode off.