Function composition

In the set of the real functions of real variable we can define an operation completely different called function composition.

Let's consider the functions $f (x) = x + 3$ and $g (x)=x^2-1$, and any real number, for example $x = 2$.

We can calculate the image of $2$ under $f$ and we obtain $f (2) = 5$.

Next we can calculate the image of $5$ under $g$ and obtain $g (5) = g (f (2)) = 24$

In general, given two functions $f$ and $g$, the function that assigns to every $x$ the value of $g (f (x))$ is called a function composition of f and g and is denoted by $g\circ f$ (it is read $f$ composed with $g$).

Therefore:

$$(g \circ f) (x) = g (f (x))$$

The function $g \circ f$ is well defined when x belongs to the domain of f and f (x) belongs to the domain of g. Namely $$Dom( g\circ f)=\{x \in Dom(f) \mid f(x) \in Dom(g) \}=$$

$$=Dom(f)-\{x \in \mathbb{R} \mid f(x) \notin Dom(g)\}$$

Consider the functions $f (x) = x + 3$ and $g (x) =x^2-1$, compute the functions $(g \circ f)$ and $(f \circ g)$, and determine its domain.

$$(g \circ f) (x) = g (f (x)) = g (x + 3) =(x+3)^2-1=$$

$$= x^2+6x+9-1=x^2+6x+8$$

and since $Dom (f) = Dom (g) =\mathbb{R}$, we have:

$$Dom (g \circ f) = \mathbb{R}$$

$$(f \circ g) (x) = f (g (x)) = f(x^2-1) = x^2-1+3= x^2+2$$As in the previous case, $$Dom (f \circ g) =\mathbb{R}$$

We observe that we obtain two different functions, that is, the function composition is not commutative.