**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

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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, }, 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:

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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)).

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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

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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

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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).

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