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
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[(x1)] and g(x) = Ö[(x2)]. Find (f °g)(x).
The domain of g is [2,¥). The domain of f is
[1,¥). If g(x) is contained in the domain of f then
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[(x1)] and g(x) = x^{2}. Find (f °g)(x).
The domain of g is \reals. The domain of f is
[1,¥). If g(x) is contained in the domain of f then
Note that x^{2} 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/(x1)] and g(x) = [(x1)/(x+1)]. Find (f °g)(x).
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
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
T_{E}X
by
T_{T}H,
version 2.61.
On 4 Jan 2001, 11:35.