Another useful combination of two functions f and g is the composition of these two functions. Let f : X → Y and g : Y → Z be two functions.

We define a function h : X → Z by setting h(x) = g(f(x). To obtain h(x), we first take the f-image f(x), of an element x in X so that f(x) ε Y, which is the domain of g(x) and then take the g-image of f(x), that is, g(f(x)), which is an element of Z. The scheme is shown in the figure.

The function h, defined above, is called the composition of f and g and is written gof. Thus (gof)(x) = g(f(x)). Domain of gof = {x : x in domain f, f(x) in domain g}.

e.g. Let f : R → R be a function defined by f(x) = x² + 4 and g[0, ∞) → R be a function defined by g(x) = √x. Then gof(x) = g(f(x)) = √(x² + 4). Domain of gof = R. Thus we have gof : R → R defined by (gof)(x) = √(x² + 4). Similarly, we shall have fog : [0, ∞) → R defined by (fog)(x) = x + 4. Note that (gof)(x) ≠ (fog)(x).

Illustration:      Two functions are defined as under :

             Find fog and gof.

Solution:   (fog)(x) = f(g(x)) 

                Let us consider, g(x) < 1 :

                (i) x² < 1, -1 < x < 2  =>  -1 < x < 1, -1 < x < 2 => -1 < x < 1

                (ii) x² + 2 < 1, 2 < x < 3  =>  x < -1, 2 < x < 3 => x = φ

                Let us consider, 1 <  g(x)  <  2,

                (iii) 1 < x² < 2,  -1 < x < 2

                        => x ε [-√2, -1) υ (1,√2] ,  -1 < x < 2  =>  1 < x < √2

                (iv) 1 < x+2 < 2, 2 < x < 3 => -1 < x < 0, 2 < x < 3, x = φ

                Let us consider -1 < f(x) < 2 :

                (i)     -1 < x+1 < 2, x < 1 => -2 < x < 1, x < 1 => -2 < x < 1

                (ii)    -1 < 2x+1 < 2, 1 < x < 2 => -1 , x < ½, 1 < x < 2 => x= φ

                Let us consider 2 < f(x) < 3:

                (iii)    2 < x+1 < 3 ,  x < 1 => x < 2 , x < 1 => x = 1

                (iv)   2 < 2x+1 < 3, 1 < x < 2 => 1 < 2x < 2, 1 < x < 2

                        => ½  <  x  <  1 , 1 < x < 2 => x = φ

                If we like we can also write g(f(x)) = (x+1)2, -2 < x < 1.

Problem of finding out fog and gof can also be handled using graphical methods

                                    f(g(x))  

Here g(x) becomes the variable that means we would draw the graph of g(x). It is clear that g(x) < 1 ∀ x ε [-1, 1] and 1 < g(x) < 2 ∀ x ε (1, √2]

In this case f(x) becomes the variable and we will draw the graph of f(x). From the graph we observe that -1 < f(x) < 2 ∀ x ε [-2, 1) and f(x) = x + 1.

2 < f(x)) < 3 => x = 1 and f(x) = x + 1.

i.e. g(f(x)) = (x + 1)2, -2 < x < 1