# Composition of Functions

Suppose that `y=f(u)=ln(u)` and `u=g(x)=sin(x)` . Since `y` is a function of `u` and `u` is afunction of `x` the we obtain that `y` is a function of `x`: `y=f(u)=f(g(x))=f(sin(x))=ln(sin(x))` .

The procedure is called **composition** because the new function is composed of the two given functions `f` and `g`.

If we are given two functions `f` and `g`, we start with a number `x` in the domain of `g` and find value `g(x)`. If this number is in the domain of `f`, then we can calculate the value of `f(g(x))`. The result is a new function `h(x)=f(g(x))` obtained by substituting `g` into `f`.

**Definition**. Given two functions `f` and `g`, the **composite function** `f@g` is defined by `(f@g)(x)=f(g(x))`.

Domain of composite functions is set of all `x` such that `x` is in domain of `g` and `g(x)` is in domain of `f`.

**Example 1**. Find `f@g`, `g@f`, `f@f`, `g@g` if `f(x)=x^3` and `g(x)=sqrt(x)/(x-1)` .

`f@g=f(g(x))=f(sqrt(x)/(x-1))=(sqrt(x)/(x-1))^3`

`g@f=g(f(x))=g(x^3)=sqrt(x^3)/(x^3-1)` .

Notice, that in general `f@g!=g@f` .

`f@f=f(f(x))=f(x^3)=(x^3)^3=x^9`

`g@g=g(g(x))=g(sqrt(x)/(x-1))=sqrt(sqrt(x)/(x-1))/(sqrt(x)/(x-1)-1)`.

Note, that we can take composition of more than two functions: `(f@g@h)(x)=f(g(h(x)))`.

**Example 2**. Find `f@g@h` if `f(x)=x^2`, `g(x)=x/(x-1)` and `h(x)=sin(x)`.

`f@g@h=f(g(h(x)))=f(g(sin(x)))=f(sin(x)/(sin(x)-1))=(sin(x)/(sin(x)-1))^2`.

Sometimes we need to perform inverse task, in other words given composite function, we need to find functions it is formed from.

**Example 3**. Find f, g, and h if `f@g@h=(cos(x+9))^2`.

The formula for `f@g@h` says: first add 9, then take the cosine of the result, and finally square.

So, we take `h(x)=x+9`, `g(x)=cos(x)` and `f(x)=x^2`.

Then

`f@g@h=f(g(h(x)))=f(g(x+9))=f(cos(x+9))=(cos(x+9))^2`.

It is worth noting, that characteristic of function as composite is not connected with natural functional dependence of `f` and `x`, it is just a way to represent this dependence.

For example, let `f=sqrt(1-u^2)` for `u in[-1,1]` and `y=sin(x)` for `x in [-pi/2,pi/2]` then `(f@g)(x)=f(g(x))=f(sin(x))=sqrt(1-(sin(x))^2)=cos(x)`. Here `cos(x)` is represented as composite function.