# Operations on Functions - addition, substraction, functions multiplication, functions division

Operations on FunctionsFunctions can be added

Functions can be subtracted

Functions can be multiplied

Functions can be divided

Functions can be composed with each other

Let us take two function

f(x) = x

^{2}and g(x) = x

The sum of these functions are

f(x) + g(x) = x

^{2}+ x

Sum of two functions

*f*and

*g*is denoted as

*f + g*

Definition for Operations on Functions

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

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

(f.g)(x) = f(x).g(x) Multiplication

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

For the function

*f + g, f - g, f.g,*the domains are defined as the inrersection of the domains of

*f*and

*g*

For

*f/g*, the domains is the intersection of the domains of

*f*and

*g*except for the points where

*g(x) = 0*

Example

f(x) = 1 + √x - 2 and g(x) = x - 1

Then their sum is defined by

(f + g)(x) = f(x) + g(x) = (1 + √x - 2) + (x - 1) = x + √x - 2

Now we compare the domains of original functions f and g, and their sum:

Function |
Domain |

f(x) = 1 + √x - 2 | [2; +∞) |

g(x) = x - 1 | (-∞ +∞) |

(f + g)(x) = x + √x - 2 | [2; ∞)∩(-∞ +∞) = [2; ∞) |

Example:

Now consider the two functions

f(x) = 3√x and g(x) = √x

Then their product is defined by

(f.g)(x) = f(x).g(x) = (3√x)(√x) = 3x

Note that

**Natural domain of 3x is (-∞; +∞)**

Now we compare the domains of original functions

*f*and

*g*, and their product:

Function |
Domain |

f(x) = 3√x | [0; +∞) |

g(x) = √x | [0; +∞) |

(f.g)(x) = 3x, x ≥ 0 | [0; +∞) ∩ [0; +∞) = [0; +∞) |

Some time we write the product of two same functions as

f

^{2}(x) = f(x).f(x)

In general, if

*n*is positive integer, then

f

^{n}(x) = f(x).f(x)...f(x)

For example,

sin(x).sin(x) = (sin(x))

^{2}= sin

^{2}x

Let us suppose two functions

f(x) = x

^{3}and g(x) = x + 4

Now if we substitute

*g(x)*for

*x*in the formula for

*f*we obtain a new function denoted by

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

^{3}= (x + 4)

^{3}

In order to compute

*f(g(x))*one needs to first compute

*g(x)*for an

*x*from the domain of

*g*, then one needs

*g(x)*in the domain of

*f*to compute

*f(g(x))*

Example:

Here

f(x) = x

^{2}+ 3 g(x) = √x

Then composition of these functions are

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

^{2}+ 3 = (√x)

^{2}+ 3 = x + 3

Now we compare the domains of original functions

*f*and

*g*, and their composition

Function |
Domains |

f(x) = x^{2} + 3 |
(-∞; +∞) |

g(x) = √x | [0; +∞) |

(f o g)(x) = x + 3 | All x in [0; +∞) such that g(x)lies in (-∞; +∞) so domain is (-∞; +∞) |

Consider the function

h(x) = (x + 1)

^{2}

we can break up function

*h*as

f(x) = x + 1

g(x) = x

^{2}

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

Remark :

Note that we can express the function as

(x

^{2}+ 1)

^{10}= [(x

^{2}+ 1)

^{2}]

^{5}= f(g(x))

g(x) = (x

^{2}+ 1)

^{2}, f(x) = x

^{5}

Also we can write (x

^{2}+ 1) = [(x

^{2}+ 1)

^{3}]

^{10/3}= f(g(x))

g(x) = (x

^{2}+ 1)

^{3}, f(x) = x

^{10/3}

Note that in general we cannot write

(f o g) ≠ (g o f)

Domain of

*(f o g)*consists of all

*x*in the domain of

*g*for which

*g(x)*is in the domain of

*f*