    | We can add, subtract, multiply and divide functions! The result is a new function. |
Let us try doing those operations on f(x) and g(x):
| Addition |
We can add two functions:
(f+g)(x) = f(x) + g(x)
Note: we put the f+g inside () to show 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 we will talk about that later!
| 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)
| 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
| 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 easy so far, but now we must consider the Domains of the functions.
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:
Using set-builder notation it is written:
{ x
| x ≥ 0}
"the set of all x's that are a member of the Real Numbers,
such that x is greater than or equal to zero"
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.
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.
Example: f(x)=√x and g(x)=√(3−x)
The domain for f(x)=√x is from 0 onwards:
The domain for g(x)=√(3−x) is up to and including 3:
So the new domain (after adding or whatever) is from 0 to 3:
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) = { x | xDom(f) and xDom(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" (means "member of")
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:
2. The domain for g(x)=√(3−x) is up to and including 3:
3. AND √(3−x) cannot be zero, so x cannot be 3:
(Notice the open circle at 3, which means not including 3)
So all together we end up with:
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.
1193, 1194, 2459, 2460, 8461, 559, 560, 2461, 2462, 8462
Function Composition Algebra Index
