# Operations with Functions

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):

## 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) = x^{2}

(f+g)(x) = (2x+3) + (x^{2}) = x^{2}+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!

## Subtraction |

We can also subtract two functions:

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

### Example: f(x) = 2x+3 and g(x) = x^{2}

(f-g)(x) = (2x+3) - (x^{2})

## Multiplication |

We can multiply two functions:

(f·g)(x) = f(x) · g(x)

### Example: f(x) = 2x+3 and g(x) = x^{2}

(f·g)(x) = (2x+3)(x^{2}) = 2x^{3} + 3x^{2}

## Division |

And we can divide two functions:

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

### Example: f(x) = 2x+3 and g(x) = x^{2}

(f/g)(x) = (2x+3)/x^{2}

## 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.

The domain is The function must work for all values we give it, so it is |

### 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 |

Here is an example:

### 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"

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.