James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 1.3 New Functions from Old Functions 41
x (input)
g
©
f • g
We compute this by substitution:
y − f sud − f stsxdd − f sx 2 1 1d − sx 2 1 1
The procedure is called composition because the new function is composed of the two
given functions f and t.
In general, given any two functions f and t, we start with a number x in the domain
of t and calculate tsxd. If this number tsxd is in the domain of f, then we can calculate
the value of f stsxdd. Notice that the output of one function is used as the input to the next
function. The result is a new function hsxd − f stsxdd obtained by substituting t into f. It is
called the composition (or composite) of f and t and is denoted by f 8 t (“ f circle t”).
f
f{©}
(output)
Definition Given two functions f and t, the composite function f 8 t (also called
the composition of f and t) is defined by
s f 8 tdsxd − f stsxdd
FIGURE 11
The f 8 t machine is composed of
the t machine (first) and then the
f machine.
The domain of f 8 t is the set of all x in the domain of t such that tsxd is in the domain
of f. In other words, s f 8 tdsxd is defined whenever both tsxd and f stsxdd are defined.
Figure 11 shows how to picture f 8 t in terms of machines.
Example 6 If f sxd − x 2 and tsxd − x 2 3, find the composite functions f 8 t and t 8 f.
SOLUtion We have
s f 8 tdsxd − f stsxdd − f sx 2 3d − sx 2 3d 2
st 8 f dsxd − ts f sxdd − tsx 2 d − x 2 2 3
n
NOTE You can see from Example 6 that, in general, f 8 t ± t 8 f. Remember, the
notation f 8 t means that the function t is applied first and then f is applied second. In
Example 6, f 8 t is the function that first subtracts 3 and then squares; t 8 f is the function
that first squares and then subtracts 3.
ExamplE 7 If f sxd − sx and tsxd − s2 2 x , find each of the following functions
and their domains.
(a) f 8 t (b) t 8 f (c) f 8 f (d) t 8 t
SOLUTION
(a)
s f 8 tdsxd − f stsxdd − f (s2 2 x) − ss2 2 x − s 4 2 2 x
The domain of f 8 t is hx | 2 2 x > 0j − hx |
x < 2j − s2`, 2g.
(b)
st 8 f dsxd − ts f sxdd − t( sx ) − s2 2 sx
If 0 < a < b, then a 2 < b 2 .
For sx to be defined we must have x > 0. For s2 2 sx to be defined we must have
2 2 sx > 0, that is, sx < 2, or x < 4. Thus we have 0 < x < 4, so the domain of
t 8 f is the closed interval f0, 4g.
(c)
s f 8 f dsxd − f s f sxdd − f (sx ) − ssx − s 4 x
The domain of f 8 f is f0, `d.
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