College Algebra 9th txtbk

SECTION 3–5 Operations on Functions; Composition 231
SOLUTION
The basic economic principle we are using is that profit is revenue minus cost. So the profit
function P is the difference of the revenue function R and the cost function C.
P(x) (R C)(x)
R(x) C(x)
(20x 0.001x 2 ) (25,000 3x)
17x 0.001x 2 25,000
This is a function of the demand x. We were asked to find the profit P as a function of the
price p; we can accomplish this using composition, because x f( p).
(P f )( p) P( f( p))
P(20,000 1,000p)
17(20,000 1,000p) 0.001(20,000 1,000p) 2 25,000
340,000 17,000p 400,000 40,000p 1,000p 2 25,000
85,000 23,000p 1,000p 2
Technically, P f and P are different functions, because the first has independent variable
p and the second has independent variable x. However, because both functions represent
the same quantity (the profit), it is customary to use the same symbol to name
each function. So
P( p) 85,000 23,000p 1,000p 2
expresses the weekly profit P as a function of price p. Now we can use the vertex formula
to find the maximum.
p b
2a 23,000 2,000 11.5
P(11.5) 85,000 23,000(11.5) 1,000(11.5) 2 47,250
Since a 0, the parabola opens downward, and the maximum value of P occurs at the
vertex. So the largest profit is $47,250 and it will occur when the price of the headphones
is $11.50.
MATCHED PROBLEM 8
Repeat Example 8 for the functions
x f( p) 10,000 1,000p 0 p 10
C(x) 10,000 2x R(x) 10x 0.001x 2
ANSWERS TO MATCHED PROBLEMS
1. ( f g)(x) 1x 110 x, ( f g)(x) 1x 110 x, ( fg)(x) 210x x 2 ,
( fg)(x) 1x(10 x); the functions f g, f g, and fg have domain:[0, 10] , the domain
of fg is [0, 10)
2. a f domain: all real numbers x except 2, 0, and 5
g b(x) x
(x 2)(x 5) ;
3. (h k)(10) 12; (h k)(8) 40; (h k)(0) 18
4. (h k)(x) 16x 2 8x 12
5. ( f g)(x) 1 3 7x 5, domain: (, )
(g f )(x) 71 3 x 5, domain: (, )
6. ( f g)(x) 110 x; domain: x 1 and x 10 or [1, 10]
7. h(x) ( f g)(x) where f(x) x 4 and g(x) 4x 3 7
8. P(p) 30,000 12,000p 1,000p 2 . The largest profit is $6,000 and occurs when the price
is $6.