 # Composite Functions

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.

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