Operations with Functions

add, subtract, multiply and divide

We can add, subtract, multiply and divide functions!

The result will be a new function.

Let us try doing those operations on f(x) and g(x):

add

Addition

We can add two functions:

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

Note: I put the f+g inside () so you know they both work on x.

Example: f(x) = 2x+3 and g(x) = x2

(f+g)(x) = (2x+3) + (x2) = x2+2x+3

Sometimes we may need to combine like terms:

Example: v(x) = 5x+1, w(x) = 3x-2

(v+w)(x) = (5x+1) + (3x-2) = 8x-1

The only other thing to worry about is the Domain (the set of numbers that go into the function), but I will talk about that later!

 

Subtract

Subtraction

We can also subtract two functions:

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

Example: f(x) = 2x+3 and g(x) = x2

(f-g)(x) = (2x+3) - (x2)

 

Multiply

Multiplication

We can multiply two functions:

(f·g)(x) = f(x) · g(x)

Example: f(x) = 2x+3 and g(x) = x2

(f·g)(x) = (2x+3)(x2) = 2x3 + 3x2

 

Divide

Division

And we can divide two functions:

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

Example: f(x) = 2x+3 and g(x) = x2

(f/g)(x) = (2x+3)/x2

Function Composition

There is another special operation called Function Composition,
read that page to find out more!
  (g º f)(x)

Domains

It has been very easy so far, but now we must consider the Domains of the functions.

doman and range

The domain is the set of all the values that go into a function.

The function must work for all values we give it, so it is up to us to make sure we get the domain correct!

 

Example: the domain for √x (the square root of x)

We can't have the square root of a negative number (unless we use imaginary numbers, but we aren't doing that here), so we must exclude negative numbers:

The Domain of √x is all non-negative Real Numbers

On the Number Line it looks like:

zero onwards

Using set-builder notation it is written:

{ xmember ofreals | x ≥ 0}

Or using interval notation it is:

[0,+∞)

It is important to get the Domain right, or we will get bad results!

So how do we work out the new domain after doing an operation?

How to Work Out the New Domain

When we do operations on functions, we end up with the restrictions of both.

chicken dish

It is like cooking for friends:

  • one can't eat peanuts,
  • the other can't eat dairy food.

So what we cook can't have peanuts and also can't have dairy products.

Here is an example:

Example: f(x)=√x and g(x)=√(3-x)

The domain for f(x)=√x is from 0 onwards:

zero onwards

The domain for g(x)=√(3-x) is up to and including 3:

zero onwards

So the new domain (after adding or whatever) is from 0 to 3:

zero onwards

If we choose any other value, then one or the other part of the new function won't work.

In other words we want to find where the two domains intersect.

Note: we can put this whole idea into one line using Set Builder Notation:

Dom(f+g) = { xmember ofreals | xmember ofDom(f) and xmember ofDom(g) }

Which says "the domain of f plus g is the set of all Real Numbers that are in the domain of f AND in the domain of g"

The same rule applies when we add, subtract, multiply or divide, except divide has one extra rule.

An Extra Rule for Division

There is an extra rule for division:

As well as restricting the domain as above, when we divide:

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

we must also make sure that g(x) is not equal to zero (so we don't divide by zero).

Here is an example:

Example: f(x)=√x and g(x)=√(3-x)

(f/g)(x) = √x / √(3-x)

1. The domain for f(x)=√x is from 0 onwards:

zero onwards

2. The domain for g(x)=√(3-x) is up to and including 3:

zero onwards

3. AND √(3-x) cannot be zero, so x cannot be 3:

zero onwards
(Notice the open circle at 3, which means not including 3)

So all together we end up with:

zero onwards

 

Summary

  • To add, subtract, multiply or divide functions just do as the operation says.
  • The domain of the new function will have the restrictions of both functions that made it.
  • Divide has the extra rule that the function we are dividing by cannot be zero.