1- Operations on Functions

First you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.

• Given f(x) = 3x + 2 and g(x) = 4 – 5x,

find (f + g)(x), (f – g)(x), (f×g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f(x) + g(x) = [3x + 2] + [4 – 5x] = 3x – 5x + 2 + 4 = –2x + 6

(f – g)(x) = f(x) – g(x) = [3x + 2] – [4 – 5x] = 3x + 5x + 2 – 4 = 8x – 2

(f×g)(x) = [f(x)][g(x)] = (3x + 2)(4 – 5x) = 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

Given f(x) = 2x, g(x) = x + 4, and h(x) = 5 – x3,

find (f + g)(2), (h – g)(2), (f × h)(2), and (h / g)(2).

To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:

f(2) = 2(2) = 4

g(2) = (2) + 4 = 6

h(2) = 5 – (2)3 = 5 – 8 = –3

Now I can evaluate the listed expressions:

(f + g)(2) = f(2) + g(2) = 4 + 6 = 10

(h – g)(2) = h(2) – g(2) = –3 – 6 = –9

(f × h)(2) = f(2) × h(2) = (4)(–3) = –12

(h / g)(2) = h(2) ÷ g(2) = –3 ÷ 6 = –0.5

2- Operations on Functions

________________________________________

Introduction

In this tutorial we will be working with functions. Note as you are going through this lesson that a lot of the things we are doing we have done before with expressions. Like adding, subtracting, multiplying and dividing. What is new here is we are specifically looking at these same operations with functions this time. I think we are ready to forge ahead.

Tutorial

________________________________________

The following show us how to perform the different operations on functions.

Use the functions and to illustrate the operations:

Sum of f + g

(f + g)(x) = f(x) + g(x)

This is a very straight forward process. When you want the sum of your functions you simply add the two functions together.

Example 1: If and then find (f + g)(x)

*Add the 2 functions

*Combine like terms

Difference of f - g

(f - g)(x) = f(x) - g(x)

Another straight forward idea, when you want the difference of your functions you simply take the first function minus the second function.

Example 2: If and then find (f - g)(x) and (f - g)(5)

*Take the difference of the 2 functions

*Subtract EVERY term of the 2nd ( )

*Plug 5 in for x in the diff. of the 2 functions found above

Since the difference function had already been found, we didn't have to take the difference of the two functions again. We could just merely plug in 5 into the already found difference function.

Product of f g

(f g)(x) = f(x)g(x)

Along the same idea as adding and subtracting, when you want to find the product of your functions you multiply the functions together.

Example 3: If and then find (fg)(x)

*Take the product of the 2 functions

*FOIL method to multiply

Quotient of f/ g

(f /g)(x) = f(x)/g(x)

Well, we don't want to leave division of functions out of the loop. It stands to reason that when you want to find the quotient of your functions you divide the functions.

Example 4: If and then find (f/g)(x) and (f/g)(1)

*Write as a quotient of the 2 functions

*Use the quotient found above to plug 1 in for x

Composite Function

Be careful, when you have a composite function, one function is inside of the other. It is not the same as taking the product of those functions.

Example 5: If and then find

*g is inside of f

*Substitute in x + 2 for g

*Plug x + 2 in for x in function f

Example 6: Let , , and . Find an equation defining each function and state the domain.

a) b) c) d) e) .

a)

*Add the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

b)

*Subtract the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

c)

*Multiply the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

d)

*Divide the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator h(x) CANNOT equal zero:

*The den. h(x) = x + 3 CANNOT equal zero

Putting these two sets together we get the domain:

e)

*h is inside of f is inside of g

*Substitute in x + 3 for h

*Substitute in sqroot(x + 3 + 1) for f(x + 3)

*Plug sqroot(x + 4) in for x in function g

Domain:

The radicand CANNOT be negative AND the denominator CANNOT equal zero. In other words it has to be greater than zero:

*Set radicand of x + 4 greater than 0 and solve

The domain would be:

Example 7: Let f = {(-3, 2), (-2, 4), (-1, 6), (0,8)}, and g = {(-2, 5), (0, 7), (2, 9)} and h = {(-3, 0), (-2, 1)}. Find the following functions and state the domain.

a) b) c) d) e)

a)

When you are looking for the domain of the sum of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 9)

*Add together the corresponding y values to x = -2

x = 0:

(0, 15)

*Add together the corresponding y values to x = 0

Putting it together in ordered pairs we get:

b)

When you are looking for the domain of the difference of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, -1)

*Subtract the corresponding y values to x = -2

x = 0:

(0, 1)

*Subtract the corresponding y values to x = 0

Putting it together in ordered pairs we get:

c)

When you are looking for the domain of the product of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 20)

*Multiply the corresponding y values to x = -2

x = 0:

(0, 56)

*Multiply the corresponding y values to x = 0

Putting it together in ordered pairs we get:

d)

When you are looking for the domain of the quotient of two functions that are given as sets, you are looking for the intersection of their domains AND values of x that do NOT cause the denominator to equal 0.

The x values that f and h have in common are -3 and -2. However, h(-3) = 0, which would cause the denominator of the quotient to be 0.

So, the domain would be {-2}.

x = -2:

(-2, 4)

*Find the quotient of the corresponding y values to x = -2

Putting it together in an ordered pair we get:

e)

When you are looking for the domain of the composition of two functions that are given as sets, you are looking for values that come from the domain of the inside function AND when you plug those values of xinto the inside function, the output is in the domain of the outside function.

The x values of h are -3 and -2. However, g(h(-2)) = g(1), which is undefined. In other words, there are no ordered pairs in g that have a 1 for their x value.

So, the domain would be {-3}.

x = -3:

(-3, 7) *h(-3) = 0

*Find the y value that corresponds to x = 0

Putting it together in an ordered pair we get:

Example 8: Let and . Write as a composition function using f and g.

When you are writing a composition function keep in mind that one function is inside of the other. You just have to figure out which function is the inside function and which is the outside function.

Note that if you put g inside of f you would get:

*Put g inside of f

Note that if you put f inside of g you would get:

*Put f inside of g

Hey this looks familiar.

Our answer is .

This is the video:

http://www.youtube.com/watch?v=fieyNGo8Tbw

http://www.youtube.com/watch?v=qbf_VDtu7ww

Part 2: the Inverse of a Function:

________________________________________

Definition of Inverse Function

Before defining the inverse of a function we need to have the right mental image of function.

Consider the function f(x) = 2x + 1. We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. In this section it helps to think of f as transforming a 3 into a 7, and f transforms a 5 into an 11, etc.

Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 7 back to 3, and take -3 back to -2, etc.

Let g(x) = (x - 1)/2. Then g(7) = 3, g(-3) = -2, and g(11) = 5, so g seems to be undoing what f did, at least for these three values. To prove that g is the inverse of f we must show that this is true for any value of x in the domain of f. In other words, g must take f(x) back to x for all values of x in the domain of f. So, g(f(x)) = x must hold for all x in the domain of f. The way to check this condition is to see that the formula for g(f(x)) simplifies to x.

g(f(x)) = g(2x + 1) = (2x + 1 -1)/2 = 2x/2 = x.

This simplification shows that if we choose any number and let f act it, then applying g to the result recovers our original number. We also need to see that this process works in reverse, or that f also undoes what g does.

f(g(x)) = f((x - 1)/2) = 2(x - 1)/2 + 1 = x - 1 + 1 = x.

Letting f-1 denote the inverse of f, we have just shown that g = f-1.

Definition:

Let f and g be two functions. If

f(g(x)) = x and g(f(x)) = x,

then g is the inverse of f and f is the inverse of g.

Exercise 1:

(a) Open the Java Calculator and enter the formulas for f and g. Note that you take a cube root by raising to the (1/3), and you do need to enter the exponent as (1/3), and not a decimal approximation. So the text for the g box will be

(x - 2)^(1/3)

Use the calculator to evaluate f(g(4)) and g(f(-3)). g is the inverse of f, but due to round off error, the calculator may not return the exact value that you start with. Try f(g(-2)). The answers will vary for different computers. However, on our test machine f(g(4)) returned 4; g(f(-3)) returned 3; but, f(g(-2)) returned -1.9999999999999991, which is pretty close to -2.

The calculator can give us a good indication that g is the inverse of f, but we cannot check all possible values of x.

(b) Prove that g is the inverse of f by simplifying the formulas for f(g(x) and g(f(x)).

Return to Contents

Graphs of Inverse Functions

We have seen examples of reflections in the plane. The reflection of a point (a,b) about the x-axis is (a,-b), and the reflection of (a,b) about the y-axis is (-a,b). Now we want to reflect about the line y = x.

The reflection of the point (a,b) about the line y = x is the point (b,a).

Let f(x) = x3 + 2. Then f(2) = 10 and the point (2,10) is on the graph of f. The inverse of f must take 10 back to 2, i.e. f-1(10)=2, so the point (10,2) is on the graph of f-1. The point (10,2) is the reflection in the line y = x of the point (2,10). The same argument can be made for all points on the graphs of f and f-1.

The graph of f-1 is the reflection about the line y = x of the graph of f.

• Videos: x3 + c Animated Gif, MS Avi File, or Real Video File

• Videos: x2 + c Animated Gif, MS Avi File, or Real Video File

Return to Contents

Existence of an Inverse

Some functions do not have inverse functions. For example, consider f(x) = x2. There are two numbers that f takes to 4, f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f.

Look at the same problem in terms of graphs. If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below.

Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.

This generalizes as follows: A function f has an inverse if and only if when its graph is reflected about the line y = x, the result is the graph of a function (passes the vertical line test). But this can be simplified. We can tell before we reflect the graph whether or not any vertical line will intersect more than once by looking at how horizontal lines intersect the original graph!

Horizontal Line Test

Let f be a function.

If any horizontal line intersects the graph of f more than once, then f does not have an inverse.

If no horizontal line intersects the graph of f more than once, then f does have an inverse.

The property of having an inverse is very important in mathematics, and it has a name.

Definition: A function f is one-to-one if and only if f has an inverse.

The following definition is equivalent, and it is the one most commonly given for one-to-one.

Alternate Definition: A function f is one-to-one if, for every a and b in its domain, f(a) = f(b) implies a = b.

Exercise 2:

Graph the following functions and determine whether or not they have inverses.

(a) f(x) = (x - 3) x2. Answer

(b) f(x) = x3 + 3x2 +3x. Answer

(c) f(x) = x ^(1/3) ( the cube root of x). Answer

Return to Contents

Finding Inverses

Example 1. First consider a simple example f(x) = 3x + 2.

The graph of f is a line with slope 3, so it passes the horizontal line test and does have an inverse.

There are two steps required to evaluate f at a number x. First we multiply x by 3, then we add 2.

Thinking of the inverse function as undoing what f did, we must undo these steps in reverse order.

The steps required to evaluate f-1 are to first undo the adding of 2 by subtracting 2. Then we undo multiplication by 3 by dividing by 3.

Therefore, f-1(x) = (x - 2)/3.

Steps for finding the inverse of a function f.

1. Replace f(x) by y in the equation describing the function.

2. Interchange x and y. In other words, replace every x by a y and vice versa.

3. Solve for y.

4. Replace y by f-1(x).

Example 2. f(x) = 6 - x/2

Step 1 y = 6 - x/2.

Step 2 x = 6 - y/2.

Step 3 x = 6 - y/2.

y/2 = 6 - x.

y = 12 - 2x.

Step 4 f-1(x) = 12 - 2x.

Step 2 often confuses students. We could omit step 2, and solve for x instead of y, but then we would end up with a formula in y instead of x. The formula would be the same, but the variable would be different. To avoid this we simply interchange the roles of x and y before we solve.

Example 3. f(x) = x3 + 2

This is the function we worked with in Exercise 1. From its graph (shown above) we see that it does have an inverse. (In fact, its inverse was given in Exercise 1.)

Step 1 y = x3 + 2.

Step 2 x = y3 + 2.

Step 3 x - 2 = y3.

(x - 2)^(1/3) = y.

Step 4 f-1(x) = (x - 2)^(1/3).

How to find the Inverse of a Function

The inverse of a function is the function that undoes its operation. The notation for the inverse of the function f is f-1. In function notation this can be written like this.

f-1(f(x))=x

f(f-1(x))=x

We can also write it in terms of composition of functions. (See How to find Compositions of Functions .)

f-1°f(x)=x

f°f-1(x)=x

Not all functions have inverses. A function must be one to one in order to have one. A one to one function passes the horizontal line test as well as the vertical line test. In the graph of a function, no vertical line can pass through more than one point. In order to be a one to one function the same also has to be true for horizontal lines, no horizontal line can pass through more than one point of the graph. In terms of the function as an operation, this means that there can be no collapsing, two different inputs can't give you the same output, so for example f(x)=x2 isn't one to one, because when you put in negative numbers you get out the same thing as when you put in positive numbers.

Now suppose we have a function that is one to one. How can we find its inverse? Even this is not always possible, but for a good number of simple functions it is not too difficult and here is how to do it.

1. Replace f(x) with y.

2. Reverse the roles of x and y, that is replace every occurrence of x with y and every occurrence of y with x.

3. Solve for y in terms of x.

4. Replace y with f-1(x).

Examples

In the following problems the instruction is to find f-1.

Example 1:

f(x)=3x+2

Solution:

1.

y=3x+2

2.

x=3y+2

3.

3y=x-2

y = x-2

________________________________________

3

4.

f-1(x)= x-2

________________________________________

3

Example 2:

f(x)=x3+1

Solution:

1.

y=x3+1

2.

x=y3+1

3.

y3=x-1

You can do this with cube root because there is no plus or minus needed, because every number has only one real cube root. If this had been a square instead of a cube, though, the function would not have been one to one, so it wouldn't have had an inverse.

4.

Example 3:

f(x)= 1

________________________________________

x+1

Solution:

(y+1)x=1

xy+x=1

xy=1-x

y= 1-x

________________________________________

x

1.

y= 1

________________________________________

x+1

2.

x= 1

________________________________________

y+1

3.

Here I am multiplying both sides by y+1 at the beginning to clear the denominator. Then we have to get y alone, so we get everything without a y on the other side of the equation and then divide by the coefficient on y.

4.

f-1(x)= 1-x

________________________________________

x

Example 4:

ã

f(x)= 2x

________________________________________

x-3

Solution:

1.

y= 2x

________________________________________

x-3

2.

x= 2y

________________________________________

y-3

3.

(y-3)x=2y

yx-3x=2y

-3x=2y-xy

2y-xy=-3x

(2-x)y=-3x

y=

-3x

________________________________________

2-x

= 3x

________________________________________

x-2

This one is a little bit harder because there is more than one occurrence of y. What we have to do is think of y as the only variable and pretend that x is a number and then do what we would normally do in such a situation, get all of the variables on one side of the equation and all of the numbers on the other side. The factoring out of the y in the 5th step is like combining like terms, just like you would do if it had been 7y-3y. The only difference is that you can do the subtraction in 2-x, because you don't know what x is, so you just have to write it. At the very end I multiplied top and bottom by -1 to make it look prettier. This isn't really required, but it is better form, so nice to do if you think of it.

4.

f-1(x)= 3x

________________________________________

x-2

Finding the Inverse of a Function:

• Find the inverse of y = 3x – 2.

Here's how the process works:

Here's my original function:

Now I'll try to solve for "x =":

Once I have "x =", I'll switch x and y; the "y =" is the inverse.

Then the inverse is y = (x + 2) / 3

If you need to find the domain and range, look at the original function and its graph. The domain of the original function is the set of all allowable x-values; in this case, the function was a simple polynomial, so the domain was "all real numbers". The range of the original function is all the y-values you'll pass on the graph; in this case, the straight line goes on for ever in either direction, so the range is also "all real numbers". To find the domain and range of the inverse, just swap the domain and range from the original function.

• Find the inverse function of y = x2 + 1, if it exists.

There will be times when they give you functions that don't have inverses.

From the graph, it's easy to see that this function can't possibly have an inverse, since it violates the Horizontal Line Test:

It is usually considered acceptable to draw the above graph, draw a horizontal line across it that crosses the graph twice, and then say something like "The inverse of this function is not itself a function, because of the Horizontal Line Test". But some teachers want to see the algebra anyway. Be sure to check with your teacher and verify what will be an acceptable answer -- and do this before the test! Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

What will this look like when I try to find the inverse algebraically? The Vertical Line Test says that I can't have two y's that share an x-value. That is, each x has to have a UNIQUE corresponding y value. But look at what happens when I try to solve for "x =":

My original function:

Solving for "x =":

Well, I solved for "x =", but I didn't get a UNIQUE "x =". Instead, I've shown that any given x-value will actually correspond to two different y-values, one from the "plus" on the square root and the other from the "minus".

The inverse is not a function.

Any time you come up with a "±" sign, you can be pretty sure that the inverse isn't a function.

• Find the inverse function of y = x2 + 1, x < 0.

The only difference between this function and the previous one is that the domain has been restricted to only the negative half of the x-axis. This restriction makes the graph look like this:

This function will have an inverse that is also a function. Just about any time they give you a problem where they've taken the trouble to restrict the domain, you should take care with the algebra and draw a nice picture, because the inverse probably is a function, but it will probably take some extra effort to show this. In this case, since the domain is x < 0 and the range (from the graph) is 1 < y, then the inverse will have a domain of 1 < x and a range of y < 0. Here's how the algebra looks:

The original function:

Solve for "x =":

By figuring out the domain and range of the inverse, I know that I should choose the negative sign for the square root:

Now I'll switch the x and y;

the new "y =" is the inverse:

(The "x > 1" restriction comes from the fact that x is inside a square root.)

So the inverse is y = –sqrt(x – 1), x > 1, and this inverse is also a function.

1- Operations on Functions

First you learned (back in grammar school) that you can add, subtract, multiply, and divide numbers. Then you learned that you can add, subtract, multiply, and divide polynomials. Now you will learn that you can also add, subtract, multiply, and divide functions. Performing these operations on functions is no more complicated than the notation itself. For instance, when they give you the formulas for two functions and tell you to find the sum, all they're telling you to do is add the two formulas. There's nothing more to this topic than that, other than perhaps some simplification of the expressions involved.

• Given f(x) = 3x + 2 and g(x) = 4 – 5x,

find (f + g)(x), (f – g)(x), (f×g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f(x) + g(x) = [3x + 2] + [4 – 5x] = 3x – 5x + 2 + 4 = –2x + 6

(f – g)(x) = f(x) – g(x) = [3x + 2] – [4 – 5x] = 3x + 5x + 2 – 4 = 8x – 2

(f×g)(x) = [f(x)][g(x)] = (3x + 2)(4 – 5x) = 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

Given f(x) = 2x, g(x) = x + 4, and h(x) = 5 – x3,

find (f + g)(2), (h – g)(2), (f × h)(2), and (h / g)(2).

To find the answers, I can either work symbolically (like in the previous example) and then evaluate, or I can find the values of the functions at x = 2 and then work from there. It's probably simpler in this case to evaluate first, so:

f(2) = 2(2) = 4

g(2) = (2) + 4 = 6

h(2) = 5 – (2)3 = 5 – 8 = –3

Now I can evaluate the listed expressions:

(f + g)(2) = f(2) + g(2) = 4 + 6 = 10

(h – g)(2) = h(2) – g(2) = –3 – 6 = –9

(f × h)(2) = f(2) × h(2) = (4)(–3) = –12

(h / g)(2) = h(2) ÷ g(2) = –3 ÷ 6 = –0.5

2- Operations on Functions

________________________________________

Introduction

In this tutorial we will be working with functions. Note as you are going through this lesson that a lot of the things we are doing we have done before with expressions. Like adding, subtracting, multiplying and dividing. What is new here is we are specifically looking at these same operations with functions this time. I think we are ready to forge ahead.

Tutorial

________________________________________

The following show us how to perform the different operations on functions.

Use the functions and to illustrate the operations:

Sum of f + g

(f + g)(x) = f(x) + g(x)

This is a very straight forward process. When you want the sum of your functions you simply add the two functions together.

Example 1: If and then find (f + g)(x)

*Add the 2 functions

*Combine like terms

Difference of f - g

(f - g)(x) = f(x) - g(x)

Another straight forward idea, when you want the difference of your functions you simply take the first function minus the second function.

Example 2: If and then find (f - g)(x) and (f - g)(5)

*Take the difference of the 2 functions

*Subtract EVERY term of the 2nd ( )

*Plug 5 in for x in the diff. of the 2 functions found above

Since the difference function had already been found, we didn't have to take the difference of the two functions again. We could just merely plug in 5 into the already found difference function.

Product of f g

(f g)(x) = f(x)g(x)

Along the same idea as adding and subtracting, when you want to find the product of your functions you multiply the functions together.

Example 3: If and then find (fg)(x)

*Take the product of the 2 functions

*FOIL method to multiply

Quotient of f/ g

(f /g)(x) = f(x)/g(x)

Well, we don't want to leave division of functions out of the loop. It stands to reason that when you want to find the quotient of your functions you divide the functions.

Example 4: If and then find (f/g)(x) and (f/g)(1)

*Write as a quotient of the 2 functions

*Use the quotient found above to plug 1 in for x

Composite Function

Be careful, when you have a composite function, one function is inside of the other. It is not the same as taking the product of those functions.

Example 5: If and then find

*g is inside of f

*Substitute in x + 2 for g

*Plug x + 2 in for x in function f

Example 6: Let , , and . Find an equation defining each function and state the domain.

a) b) c) d) e) .

a)

*Add the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

b)

*Subtract the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

c)

*Multiply the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

d)

*Divide the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator h(x) CANNOT equal zero:

*The den. h(x) = x + 3 CANNOT equal zero

Putting these two sets together we get the domain:

e)

*h is inside of f is inside of g

*Substitute in x + 3 for h

*Substitute in sqroot(x + 3 + 1) for f(x + 3)

*Plug sqroot(x + 4) in for x in function g

Domain:

The radicand CANNOT be negative AND the denominator CANNOT equal zero. In other words it has to be greater than zero:

*Set radicand of x + 4 greater than 0 and solve

The domain would be:

Example 7: Let f = {(-3, 2), (-2, 4), (-1, 6), (0,8)}, and g = {(-2, 5), (0, 7), (2, 9)} and h = {(-3, 0), (-2, 1)}. Find the following functions and state the domain.

a) b) c) d) e)

a)

When you are looking for the domain of the sum of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 9)

*Add together the corresponding y values to x = -2

x = 0:

(0, 15)

*Add together the corresponding y values to x = 0

Putting it together in ordered pairs we get:

b)

When you are looking for the domain of the difference of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, -1)

*Subtract the corresponding y values to x = -2

x = 0:

(0, 1)

*Subtract the corresponding y values to x = 0

Putting it together in ordered pairs we get:

c)

When you are looking for the domain of the product of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 20)

*Multiply the corresponding y values to x = -2

x = 0:

(0, 56)

*Multiply the corresponding y values to x = 0

Putting it together in ordered pairs we get:

d)

When you are looking for the domain of the quotient of two functions that are given as sets, you are looking for the intersection of their domains AND values of x that do NOT cause the denominator to equal 0.

The x values that f and h have in common are -3 and -2. However, h(-3) = 0, which would cause the denominator of the quotient to be 0.

So, the domain would be {-2}.

x = -2:

(-2, 4)

*Find the quotient of the corresponding y values to x = -2

Putting it together in an ordered pair we get:

e)

When you are looking for the domain of the composition of two functions that are given as sets, you are looking for values that come from the domain of the inside function AND when you plug those values of xinto the inside function, the output is in the domain of the outside function.

The x values of h are -3 and -2. However, g(h(-2)) = g(1), which is undefined. In other words, there are no ordered pairs in g that have a 1 for their x value.

So, the domain would be {-3}.

x = -3:

(-3, 7) *h(-3) = 0

*Find the y value that corresponds to x = 0

Putting it together in an ordered pair we get:

Example 8: Let and . Write as a composition function using f and g.

When you are writing a composition function keep in mind that one function is inside of the other. You just have to figure out which function is the inside function and which is the outside function.

Note that if you put g inside of f you would get:

*Put g inside of f

Note that if you put f inside of g you would get:

*Put f inside of g

Hey this looks familiar.

Our answer is .

This is the video:

http://www.youtube.com/watch?v=fieyNGo8Tbw

http://www.youtube.com/watch?v=qbf_VDtu7ww

Part 2: the Inverse of a Function:

________________________________________

Definition of Inverse Function

Before defining the inverse of a function we need to have the right mental image of function.

Consider the function f(x) = 2x + 1. We know how to evaluate f at 3, f(3) = 2*3 + 1 = 7. In this section it helps to think of f as transforming a 3 into a 7, and f transforms a 5 into an 11, etc.

Now that we think of f as "acting on" numbers and transforming them, we can define the inverse of f as the function that "undoes" what f did. In other words, the inverse of f needs to take 7 back to 3, and take -3 back to -2, etc.

Let g(x) = (x - 1)/2. Then g(7) = 3, g(-3) = -2, and g(11) = 5, so g seems to be undoing what f did, at least for these three values. To prove that g is the inverse of f we must show that this is true for any value of x in the domain of f. In other words, g must take f(x) back to x for all values of x in the domain of f. So, g(f(x)) = x must hold for all x in the domain of f. The way to check this condition is to see that the formula for g(f(x)) simplifies to x.

g(f(x)) = g(2x + 1) = (2x + 1 -1)/2 = 2x/2 = x.

This simplification shows that if we choose any number and let f act it, then applying g to the result recovers our original number. We also need to see that this process works in reverse, or that f also undoes what g does.

f(g(x)) = f((x - 1)/2) = 2(x - 1)/2 + 1 = x - 1 + 1 = x.

Letting f-1 denote the inverse of f, we have just shown that g = f-1.

Definition:

Let f and g be two functions. If

f(g(x)) = x and g(f(x)) = x,

then g is the inverse of f and f is the inverse of g.

Exercise 1:

(a) Open the Java Calculator and enter the formulas for f and g. Note that you take a cube root by raising to the (1/3), and you do need to enter the exponent as (1/3), and not a decimal approximation. So the text for the g box will be

(x - 2)^(1/3)

Use the calculator to evaluate f(g(4)) and g(f(-3)). g is the inverse of f, but due to round off error, the calculator may not return the exact value that you start with. Try f(g(-2)). The answers will vary for different computers. However, on our test machine f(g(4)) returned 4; g(f(-3)) returned 3; but, f(g(-2)) returned -1.9999999999999991, which is pretty close to -2.

The calculator can give us a good indication that g is the inverse of f, but we cannot check all possible values of x.

(b) Prove that g is the inverse of f by simplifying the formulas for f(g(x) and g(f(x)).

Return to Contents

Graphs of Inverse Functions

We have seen examples of reflections in the plane. The reflection of a point (a,b) about the x-axis is (a,-b), and the reflection of (a,b) about the y-axis is (-a,b). Now we want to reflect about the line y = x.

The reflection of the point (a,b) about the line y = x is the point (b,a).

Let f(x) = x3 + 2. Then f(2) = 10 and the point (2,10) is on the graph of f. The inverse of f must take 10 back to 2, i.e. f-1(10)=2, so the point (10,2) is on the graph of f-1. The point (10,2) is the reflection in the line y = x of the point (2,10). The same argument can be made for all points on the graphs of f and f-1.

The graph of f-1 is the reflection about the line y = x of the graph of f.

• Videos: x3 + c Animated Gif, MS Avi File, or Real Video File

• Videos: x2 + c Animated Gif, MS Avi File, or Real Video File

Return to Contents

Existence of an Inverse

Some functions do not have inverse functions. For example, consider f(x) = x2. There are two numbers that f takes to 4, f(2) = 4 and f(-2) = 4. If f had an inverse, then the fact that f(2) = 4 would imply that the inverse of f takes 4 back to 2. On the other hand, since f(-2) = 4, the inverse of f would have to take 4 to -2. Therefore, there is no function that is the inverse of f.

Look at the same problem in terms of graphs. If f had an inverse, then its graph would be the reflection of the graph of f about the line y = x. The graph of f and its reflection about y = x are drawn below.

Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function.

This generalizes as follows: A function f has an inverse if and only if when its graph is reflected about the line y = x, the result is the graph of a function (passes the vertical line test). But this can be simplified. We can tell before we reflect the graph whether or not any vertical line will intersect more than once by looking at how horizontal lines intersect the original graph!

Horizontal Line Test

Let f be a function.

If any horizontal line intersects the graph of f more than once, then f does not have an inverse.

If no horizontal line intersects the graph of f more than once, then f does have an inverse.

The property of having an inverse is very important in mathematics, and it has a name.

Definition: A function f is one-to-one if and only if f has an inverse.

The following definition is equivalent, and it is the one most commonly given for one-to-one.

Alternate Definition: A function f is one-to-one if, for every a and b in its domain, f(a) = f(b) implies a = b.

Exercise 2:

Graph the following functions and determine whether or not they have inverses.

(a) f(x) = (x - 3) x2. Answer

(b) f(x) = x3 + 3x2 +3x. Answer

(c) f(x) = x ^(1/3) ( the cube root of x). Answer

Return to Contents

Finding Inverses

Example 1. First consider a simple example f(x) = 3x + 2.

The graph of f is a line with slope 3, so it passes the horizontal line test and does have an inverse.

There are two steps required to evaluate f at a number x. First we multiply x by 3, then we add 2.

Thinking of the inverse function as undoing what f did, we must undo these steps in reverse order.

The steps required to evaluate f-1 are to first undo the adding of 2 by subtracting 2. Then we undo multiplication by 3 by dividing by 3.

Therefore, f-1(x) = (x - 2)/3.

Steps for finding the inverse of a function f.

1. Replace f(x) by y in the equation describing the function.

2. Interchange x and y. In other words, replace every x by a y and vice versa.

3. Solve for y.

4. Replace y by f-1(x).

Example 2. f(x) = 6 - x/2

Step 1 y = 6 - x/2.

Step 2 x = 6 - y/2.

Step 3 x = 6 - y/2.

y/2 = 6 - x.

y = 12 - 2x.

Step 4 f-1(x) = 12 - 2x.

Step 2 often confuses students. We could omit step 2, and solve for x instead of y, but then we would end up with a formula in y instead of x. The formula would be the same, but the variable would be different. To avoid this we simply interchange the roles of x and y before we solve.

Example 3. f(x) = x3 + 2

This is the function we worked with in Exercise 1. From its graph (shown above) we see that it does have an inverse. (In fact, its inverse was given in Exercise 1.)

Step 1 y = x3 + 2.

Step 2 x = y3 + 2.

Step 3 x - 2 = y3.

(x - 2)^(1/3) = y.

Step 4 f-1(x) = (x - 2)^(1/3).

How to find the Inverse of a Function

The inverse of a function is the function that undoes its operation. The notation for the inverse of the function f is f-1. In function notation this can be written like this.

f-1(f(x))=x

f(f-1(x))=x

We can also write it in terms of composition of functions. (See How to find Compositions of Functions .)

f-1°f(x)=x

f°f-1(x)=x

Not all functions have inverses. A function must be one to one in order to have one. A one to one function passes the horizontal line test as well as the vertical line test. In the graph of a function, no vertical line can pass through more than one point. In order to be a one to one function the same also has to be true for horizontal lines, no horizontal line can pass through more than one point of the graph. In terms of the function as an operation, this means that there can be no collapsing, two different inputs can't give you the same output, so for example f(x)=x2 isn't one to one, because when you put in negative numbers you get out the same thing as when you put in positive numbers.

Now suppose we have a function that is one to one. How can we find its inverse? Even this is not always possible, but for a good number of simple functions it is not too difficult and here is how to do it.

1. Replace f(x) with y.

2. Reverse the roles of x and y, that is replace every occurrence of x with y and every occurrence of y with x.

3. Solve for y in terms of x.

4. Replace y with f-1(x).

Examples

In the following problems the instruction is to find f-1.

Example 1:

f(x)=3x+2

Solution:

1.

y=3x+2

2.

x=3y+2

3.

3y=x-2

y = x-2

________________________________________

3

4.

f-1(x)= x-2

________________________________________

3

Example 2:

f(x)=x3+1

Solution:

1.

y=x3+1

2.

x=y3+1

3.

y3=x-1

You can do this with cube root because there is no plus or minus needed, because every number has only one real cube root. If this had been a square instead of a cube, though, the function would not have been one to one, so it wouldn't have had an inverse.

4.

Example 3:

f(x)= 1

________________________________________

x+1

Solution:

(y+1)x=1

xy+x=1

xy=1-x

y= 1-x

________________________________________

x

1.

y= 1

________________________________________

x+1

2.

x= 1

________________________________________

y+1

3.

Here I am multiplying both sides by y+1 at the beginning to clear the denominator. Then we have to get y alone, so we get everything without a y on the other side of the equation and then divide by the coefficient on y.

4.

f-1(x)= 1-x

________________________________________

x

Example 4:

ã

f(x)= 2x

________________________________________

x-3

Solution:

1.

y= 2x

________________________________________

x-3

2.

x= 2y

________________________________________

y-3

3.

(y-3)x=2y

yx-3x=2y

-3x=2y-xy

2y-xy=-3x

(2-x)y=-3x

y=

-3x

________________________________________

2-x

= 3x

________________________________________

x-2

This one is a little bit harder because there is more than one occurrence of y. What we have to do is think of y as the only variable and pretend that x is a number and then do what we would normally do in such a situation, get all of the variables on one side of the equation and all of the numbers on the other side. The factoring out of the y in the 5th step is like combining like terms, just like you would do if it had been 7y-3y. The only difference is that you can do the subtraction in 2-x, because you don't know what x is, so you just have to write it. At the very end I multiplied top and bottom by -1 to make it look prettier. This isn't really required, but it is better form, so nice to do if you think of it.

4.

f-1(x)= 3x

________________________________________

x-2

Finding the Inverse of a Function:

• Find the inverse of y = 3x – 2.

Here's how the process works:

Here's my original function:

Now I'll try to solve for "x =":

Once I have "x =", I'll switch x and y; the "y =" is the inverse.

Then the inverse is y = (x + 2) / 3

If you need to find the domain and range, look at the original function and its graph. The domain of the original function is the set of all allowable x-values; in this case, the function was a simple polynomial, so the domain was "all real numbers". The range of the original function is all the y-values you'll pass on the graph; in this case, the straight line goes on for ever in either direction, so the range is also "all real numbers". To find the domain and range of the inverse, just swap the domain and range from the original function.

• Find the inverse function of y = x2 + 1, if it exists.

There will be times when they give you functions that don't have inverses.

From the graph, it's easy to see that this function can't possibly have an inverse, since it violates the Horizontal Line Test:

It is usually considered acceptable to draw the above graph, draw a horizontal line across it that crosses the graph twice, and then say something like "The inverse of this function is not itself a function, because of the Horizontal Line Test". But some teachers want to see the algebra anyway. Be sure to check with your teacher and verify what will be an acceptable answer -- and do this before the test! Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

What will this look like when I try to find the inverse algebraically? The Vertical Line Test says that I can't have two y's that share an x-value. That is, each x has to have a UNIQUE corresponding y value. But look at what happens when I try to solve for "x =":

My original function:

Solving for "x =":

Well, I solved for "x =", but I didn't get a UNIQUE "x =". Instead, I've shown that any given x-value will actually correspond to two different y-values, one from the "plus" on the square root and the other from the "minus".

The inverse is not a function.

Any time you come up with a "±" sign, you can be pretty sure that the inverse isn't a function.

• Find the inverse function of y = x2 + 1, x < 0.

The only difference between this function and the previous one is that the domain has been restricted to only the negative half of the x-axis. This restriction makes the graph look like this:

This function will have an inverse that is also a function. Just about any time they give you a problem where they've taken the trouble to restrict the domain, you should take care with the algebra and draw a nice picture, because the inverse probably is a function, but it will probably take some extra effort to show this. In this case, since the domain is x < 0 and the range (from the graph) is 1 < y, then the inverse will have a domain of 1 < x and a range of y < 0. Here's how the algebra looks:

The original function:

Solve for "x =":

By figuring out the domain and range of the inverse, I know that I should choose the negative sign for the square root:

Now I'll switch the x and y;

the new "y =" is the inverse:

(The "x > 1" restriction comes from the fact that x is inside a square root.)

So the inverse is y = –sqrt(x – 1), x > 1, and this inverse is also a function.