Handout 4: Composition of Functions Handout 4: Composition of Functions

Given two functions f and g, the composition of the f and g denoted by f °g is defined by
(f °g)(x) = f(g(x)).
The domain of f°g is slightly more complicated and is given by
{x: x Î Domain of g,  g(x) Î Domain of f}.
In other words the domain of f°g consists of the values x such that x is in the domain of g and g(x) is in the domain of f. Note that this is not the same as the intersections of the domains of f and g. To see why we make this definition consider what we have to do to form f(g(x)). We must first compute g(x) and in order to be able to do this we must have x in the domain of g. Secondly we apply f to g(x). To do this we must have g(x) in the domain of f.

Let f(x) = Ö[(x-1)] and g(x) = Ö[(x-2)]. Find (f °g)(x).
(f °g)(x)
=
f(g(x))
=
  æ
Ö
  ___
Öx-2
 
 -1
 
The domain of g is [2,¥). The domain of f is [1,¥). If g(x) is contained in the domain of f then
  ___
Öx-2
 
³
1
Þ x-2
³
1
Þ x ³ 3.
So for x to be in the domain of f °g we must have x Î Domain of g (i.e. x ³ 2) and g(x) Î Domain of f (i.e. x ³ 3). Hence the domain of f °g is [3,¥).

Let f(x) = Ö[(x-1)] and g(x) = x2. Find (f °g)(x).
(f °g)(x)
=
f(g(x))
=
  ____
Öx2-1
 
The domain of g is \reals. The domain of f is [1,¥). If g(x) is contained in the domain of f then
x2
³
1
Þ
x
Î (-¥,-1] È[1, ¥)
Note that x2 ³ 1 does not imply that x ³ 1 because x could be negative. So for x to be in the domain of f °g we must have x Î Domain of g (i.e. x Î \reals) and g(x) Î Domain of f (i.e. x Î (-¥,-1] È[1,¥)). Hence the domain of f °g is (-¥,-1] È[1,¥). Notice that there are negative values in the domain. Try computing (f°g)(x) (by first computing g(x)) for a negative value x £ -1.

Let f(x) = [1/(x-1)] and g(x) = [(x-1)/(x+1)]. Find (f °g)(x).
(f °g)(x)
=
f(g(x))
=
1
x-1
x+1
 -1
=
x+1
(x-1)-(x+1)
=
- x+1
2
To get from the second to the third line above we multiplied the top and the bottom of the fraction by x+1. This is the simplest way to manipulate fractions, just multiply the top and bottom by the same thing, something that will simplify either top or bottom. The domain of g is (-¥,-1)È(-1,¥). The domain of f is (-¥,1)È(1,¥). If g(x) is not contained in the domain of f then
x-1
x+1
=
1
x-1
=
x+1
but this is clearly impossible so g(x) is always contained in the domain of f. So for x to be in the domain of f °g we must have x Î Domain of g (i.e. x \not = -1) and g(x) Î Domain of f. Hence the domain of f °g is (-¥,-1)È(-1,¥).

File translated from TEX by TTH, version 2.61.
On 4 Jan 2001, 11:35.