College Algebra 9th txtbk

SECTION 3–6 Inverse Functions 243
Step 1. Find the domain of f and verify that f is one-to-one. Price and demand are never
negative, so p 0 and
x 20,000 1,000p
1,000(20 p) 0
20 p 0
20 p
Factor.
Divide both sides by 1,000.
Add p to both sides.
or p 20
Since p must satisfy both p 0 and p 20, the domain of f is [0, 20]. The
graph of f (Fig. 5) shows that f is one-to-one.
x
20,000
x 20,000 1,000p
0
Z Figure 5
20
p
Step 2. Since x and p have specific meaning in the context of this problem, interchanging
them does not apply here.
Step 3. Solve the equation x 20,000 1,000p for p.
0.001x 20 p
x 20,000 1,000p
x 20,000 1,000p
Subtract 20,000 from both sides.
Divide both sides by 1,000.
The inverse of the demand function is
p f 1 (x) 20 0.001x
Step 4. From Figure 5, we see that the range of f is [0, 20,000], so this must also be the
domain of f 1 .
p f 1 (x) 20 0.001x
0 x 20,000
We should check that f ( f 1 (x)) x and f 1 ( f( p)) p, but we will leave that to
the reader.
The revenue R is given by
R xp
R(x) x(20 0.001x)
20x 0.001x 2
and the domain of R is [0, 20,000].
MATCHED PROBLEM 4
Repeat Example 3 for the demand function
x f ( p) 10,000 1,000p
0 p 10
The demand function in Example 4 was defined with independent variable p and
dependent variable x. When we found the inverse function, we did not rewrite it with independent
variable p. Because p represents price and x represents number of players, to
interchange these variables would be confusing. In most applications, the variables have
specific meaning and should not be interchanged as part of the inverse process.