85. State a conclusion based on the results of Exercises 79–84
2.8 Function Operations and Composition
Graph each function.
6. ƒ1x2 = e1x 2x+3
if xÚ 0
if x60 7. ƒ1x2= -x3+1 8. ƒ1x2=20x-10 +3 9. Connecting Graphs with Equations The func-
tion g1x2 graphed here is obtained by stretching, shrinking, reflecting, and/or translating the graph of ƒ1x2 = 1x. Give the equation that defines this function.
10. Determine whether each function is even, odd, or neither.
(a) ƒ1x2 =x2-7 (b) ƒ1x2= x3-x- 1 (c) ƒ1x2=x101-x99 4. For each basic function graphed, give the name of the function, the domain, the
range, and open intervals over which it is decreasing, increasing, or constant.
(a)
2 –8 8
0 x
y (b)
1 2 2
0 x
y (c)
2 8
–8 –2
2 8
x y
5. (Modeling) Long-Distance Call Charges A certain long-distance carrier provides service between Podunk and Nowheresville. If x represents the number of minutes for the call, where x70, then the function
ƒ1x2= 0.40Œxœ +0.75
gives the total cost of the call in dollars. Find the cost of a 5.5-min call.
2.8 Function Operations and Composition
Arithmetic Operations on Functions Figure 92 shows the situation for a company that manufactures DVDs. The two lines are the graphs of the linear functions for
revenue R1x2 = 168x and cost C1x2= 118x + 800,
where x is the number of DVDs produced and sold, and x, R1x2, and C1x2 are given in thousands. When 30,000 (that is, 30 thousand) DVDs are produced and sold, profit is found as follows.
P1x2 = R1x2 - C1x2 Profit function P1302 = R1302 -C1302 Let x=30.
P1302 = 5040- 4340 R1302=1681302; C1302=1181302+800
P1302 = 700 Subtract.
Thus, the profit from the sale of 30,000 DVDs is $700,000.
The profit function is found by subtracting the cost function from the revenue function. New functions can be formed by using other operations as well.
■ Arithmetic Operations on Functions
■ The Difference Quotient
■ Composition of Functions and Domain
10 20 30 40
2000 4000 6000
x y
P(30)
C(30) C(x)
R(30) R(x)
Dollars (in thousands)
DVDs (in thousands) DVD Production
Figure 92
1 x
y
–2 0 –4
–4
Operations on Functions and Domains
Given two functions ƒ and g, then for all values of x for which both ƒ1x2 and g1x2 are defined, the functions ƒ +g, ƒ -g, ƒg, and ƒg are defined as follows.
1ƒ +g2 1x2 = ƒ1x2 + g1x2 Sum function 1ƒ −g2 1x2 = ƒ1x2 − g1x2 Difference function
1ƒg2 1x2 = ƒ1x2 # g1x2 Product function aƒ
gb1x2 = ƒ1x2
g1x2 , g1x2 30 Quotient function
The domains of ƒ+ g, ƒ −g, and ƒg include all real numbers in the intersection of the domains of ƒ and g, while the domain of ƒg includes those real numbers in the intersection of the domains of f and g for which g1x2≠0.
NOTE The condition g1x2≠0 in the definition of the quotient means that the domain of AƒgB1x2 is restricted to all values of x for which g1x2 is not 0. The condition does not mean that g1x2 is a function that is never 0.
EXAMPLE 1 Using Operations on Functions
Let ƒ1x2 =x2 + 1 and g1x2= 3x+ 5. Find each of the following.
(a) 1ƒ+ g2112 (b) 1ƒ - g21-32 (c) 1ƒg2152 (d) aƒ gb102 SOLUTION
(a) First determine ƒ112= 2 and g112= 8. Then use the definition.
1ƒ +g2112
= ƒ112+ g112 1ƒ+g21x2=ƒ1x2+g1x2
= 2+ 8 ƒ112=12+1; g112=3112+5
= 10 Add.
(b) 1ƒ - g21-32
= ƒ1-32 - g1-32 1ƒ-g21x2=ƒ1x2-g1x2
= 10- 1-42 ƒ1-32=1-322+1; g1-32=31-32+5
= 14 Subtract.
(c) 1ƒg2152
= ƒ152 # g152 1ƒg21x2=ƒ1x2# g1x2
= 152 + 1213 #5 + 52 ƒ1x2=x2+1; g1x2=3x+5
= 26 # 20 ƒ152=26; g152=20
= 520 Multiply.
305
2.8 Function Operations and Composition
EXAMPLE 2 Using Operations on Functions and Determining Domains Let ƒ1x2 = 8x- 9 and g1x2= 22x- 1. Find each function in (a)–(d).
(a) 1ƒ+ g21x2 (b) 1ƒ- g21x2 (c) 1ƒg21x2 (d) af gb1x2 (e) Give the domains of the functions in parts (a)–(d).
SOLUTION (a) 1ƒ+ g21x2
= ƒ1x2 + g1x2
= 8x- 9+ 22x- 1
(b) 1ƒ- g21x2
= ƒ1x2 - g1x2
= 8x- 9- 22x- 1 (c) 1ƒg21x2
= ƒ1x2 # g1x2
= 18x- 9222x- 1
(d) aƒ gb1x2
= ƒ1x2 g1x2
= 8x -9 22x -1
(e) To find the domains of the functions in parts (a)–(d), we first find the domains of ƒ and g.
The domain of ƒ is the set of all real numbers 1-∞, ∞2.
Because g is defined by a square root radical, the radicand must be non- negative (that is, greater than or equal to 0).
g1x2= 22x- 1 Rule for g1x2
2x- 1 Ú 0 2x-1 must be nonnegative.
x Ú 1
2 Add 1 and divide by 2.
Thus, the domain of g is C12 , ∞B.
The domains of ƒ+g, ƒ- g, and ƒg are the intersection of the domains of ƒ and g, which is
1-∞, ∞2¨ c1
2 , ∞b = c1
2 , ∞b. The intersection of two sets is the set of all elements common to both sets.
The domain of ƒg includes those real numbers in the intersection above for which g1x2= 12x- 1≠0—that is, the domain of ƒg is A12 , ∞B.
■✔ Now Try Exercises 19 and 23.
(d) af gb102
= ƒ102
g102 AƒgB1x2=ƒg11xx22
= 02 + 1 3102+ 5
ƒ1x2=x2+1 g1x2=3x+5
= 1
5 Simplify. ■✔ Now Try Exercises 11, 13, 15, and 17.
EXAMPLE 3 Evaluating Combinations of Functions
If possible, use the given representations of functions ƒ and g to evaluate 1ƒ + g2142, 1ƒ -g21-22, 1ƒg2112, and a f
gb102. (a)
–4 –2 1 2 4
–3 –1 1 5 9
x y
y = f(x)
y = g(x)
x ƒ1x2 g1x2 -2 -3 undefined
0 1 0
1 3 1
4 9 2
(b)
(c) ƒ1x2= 2x +1, g1x2 = 2x
SOLUTION
(a) From the figure, ƒ142 = 9 and g142 = 2.
1ƒ+ g2142
= ƒ142 +g142 1ƒ+g21x2=ƒ1x2+g1x2
= 9 + 2 Substitute.
= 11 Add.
For 1ƒ- g21-22, although ƒ1-22 = -3, g1-22 is undefined because -2 is not in the domain of g. Thus 1ƒ -g21-22 is undefined.
The domains of ƒ and g both include 1.
1ƒg2112
= ƒ112 # g112 1ƒg21x2=ƒ1x2 #g1x2
= 3 # 1 Substitute.
= 3 Multiply.
The graph of g includes the origin, so g102 =0. Thus AgfB102 is undefined.
(b) From the table, ƒ142= 9 and g142= 2.
1ƒ +g2142
= ƒ142 + g142 1ƒ+g21x2=ƒ1x2+g1x2
= 9+ 2 Substitute.
= 11 Add.
In the table, g1-22 is undefined, and thus 1ƒ - g21-22 is also undefined.
1ƒg2112
= ƒ112 # g112 1fg21x2=ƒ1x2 #g1x2
= 3 # 1 ƒ112=3 and g112=1
= 3 Multiply.
The quotient function value AƒgB102 is undefined because the denominator, g102, equals 0.
307
2.8 Function Operations and Composition
(c) Using ƒ1x2 = 2x+ 1 and g1x2= 2x, we can find 1ƒ+ g2142 and 1ƒg2112. Because -2 is not in the domain of g, 1ƒ- g21-22 is not defined.
1ƒ + g2142 1ƒg2112
= ƒ142+ g142 = ƒ112 # g112
= 12 # 4 +12 + 24 = 12 # 1+ 12 # 21
= 9 +2 = 3112
= 11 = 3
AƒgB102 is undefined since g102 = 0. ■✔ Now Try Exercises 33 and 37.
The Difference Quotient Suppose a point P lies on the graph of y= ƒ1x2 as in Figure 93, and suppose h is a positive number. If we let 1x, ƒ1x22 denote the coordinates of P and 1x +h, ƒ1x +h22 denote the coordinates of Q, then the line joining P and Q has slope as follows.
m = ƒ1x+ h2 -ƒ1x2
1x+ h2 -x Slope formula
m = ƒ1x+ h2 − ƒ1x2
h , h 30 Difference quotient This boldface expression is the difference quotient.
Figure 93 shows the graph of the line PQ (called a secant line). As h approaches 0, the slope of this secant line approaches the slope of the line tangent to the curve at P. Important applications of this idea are developed in calculus.
x y
0
P(x, f (x)) y = f(x)
Q(x + h, f (x + h))
h
Secant line
Figure 93
EXAMPLE 4 Finding the Difference Quotient
Let ƒ1x2= 2x2- 3x. Find and simplify the expression for the difference quotient, ƒ1x+ h2- ƒ1x2
h .
SOLUTION We use a three-step process.
Step 1 Find the first term in the numerator, ƒ1x+ h2. Replace x in ƒ1x2 with x +h.
ƒ1x+ h2
= 21x +h22 - 31x+ h2 ƒ1x2=2x2-3x Step 2 Find the entire numerator, ƒ1x + h2 - ƒ1x2.
ƒ1x+ h2- ƒ1x2
= 321x +h22 - 31x+ h24 - 12x2 -3x2 Substitute.
= 21x2 + 2xh + h22- 31x +h2 - 12x2 - 3x2 Square x+h.
= 2x2 +4xh +2h2 - 3x- 3h- 2x2 +3x Distributive property
= 4xh + 2h2 - 3h Combine like terms.
From Step 1
Remember this term when squaring x+h
Step 3 Find the difference quotient by dividing by h.
ƒ1x + h2- ƒ1x2 h
= 4xh + 2h2 - 3h h
Substitute 4xh+2h2-3h for ƒ1x+h2-ƒ1x2, from Step 2.
= h14x+ 2h- 32
h Factor out h.
= 4x+ 2h- 3 Divide out the common factor.
■✔ Now Try Exercises 45 and 55.
LOOKING AHEAD TO CALCULUS The difference quotient is essential in the definition of the derivative of a function in calculus. The derivative provides a formula, in function form, for finding the slope of the tangent line to the graph of the function at a given point.
To illustrate, it is shown in calcu- lus that the derivative of
ƒ1x2=x2+3 is given by the function
ƒ′1x2=2x.
Now, ƒ′102=2102=0, meaning that the slope of the tangent line to ƒ1x2= x2+3 at x=0 is 0, which implies that the tangent line is horizontal. If you draw this tangent line, you will see that it is the line y=3, which is indeed a horizontal line.
Composition of Functions and Domain The diagram in Figure 94
shows a function g that assigns to each x in its domain a value g1x2. Then another function ƒ assigns to each g1x2 in its domain a value ƒ1g1x22. This two-step proc- ess takes an element x and produces a corresponding element ƒ1g1x22.
Input x Output f(g(x))
g
g(x)
f
Function Function
Figure 94
The function with y-values ƒ1g1x22 is the composition of functions ƒ and g, which is written ƒ°g and read “ƒ of g” or “ƒ compose g”.
Composition of Functions and Domain
If ƒ and g are functions, then the composite function, or composition, of ƒ and g is defined by
1ƒ° g2 1x2 = ƒ1g1x2 2.
The domain of ƒ°g is the set of all numbers x in the domain of g such that g1x2 is in the domain of ƒ.
As a real-life example of how composite functions occur, consider the following retail situation.
A $40 pair of blue jeans is on sale for 25% off. If we purchase the jeans before noon, they are an additional 10% off. What is the final sale price of the jeans?
We might be tempted to say that the jeans are 35% off and calculate
$4010.352 = $14, giving a final sale price of
$40- $14= $26. Incorrect
CAUTION In Example 4, notice that the expression ƒ1x+ h2 is not equivalent to ƒ1x2 + ƒ1h2. These expressions differ by 4xh.
ƒ1x + h2= 21x+ h22 - 31x+ h2= 2x2 +4xh+ 2h2 - 3x- 3h ƒ1x2+ ƒ1h2= 12x2- 3x2 +12h2 - 3h2 = 2x2 - 3x+ 2h2 - 3h In general, for a function ƒ, ƒ1x +h2 is not equivalent to ƒ1x2+ ƒ1h2.
309
2.8 Function Operations and Composition
$40 Input original
price
$27 Output sale
price g(40)
= 40 – 0.25(40)
= 40 – 10
= 30
$30
f(30)
= 30 – 0.10(30)
= 30 – 3
= 27 Function g
takes 25%
off.
Function f takes an additional
10% off.
g(x) = x – 0.25x
( f ° g)(40) = f( g(40)) = f(30) = 27 f(x) = x – 0.10x
Figure 95
EXAMPLE 6 Determining Composite Functions and Their Domains Given that ƒ1x2= 2x and g1x2= 4x+ 2, find each of the following.
(a) 1ƒ∘g21x2 and its domain (b) 1g∘ƒ21x2 and its domain SOLUTION
(a) 1ƒ∘g21x2
= ƒ1g1x22 Definition of composition
= ƒ14x+ 22 g1x2=4x+2
= 24x +2 ƒ1x2= 2x
$26 is not correct. To find the final sale price, we must first find the price after taking 25% off and then take an additional 10% off that price. See Figure 95.
EXAMPLE 5 Evaluating Composite Functions Let ƒ1x2 = 2x- 1 and g1x2= x -41 .
(a) Find 1ƒ∘g2122. (b) Find 1g∘ƒ21-32. SOLUTION
(a) First find g122: g122= 4 2- 1 = 4
1 = 4.
Now find 1ƒ∘g2122. 1ƒ∘g2122
= ƒ1g1222 Definition of composition
= ƒ142 g122=4
= 2142- 1 Definition of ƒ
= 7 Simplify.
(b) 1g∘ƒ21-32
= g1ƒ1-322 Definition of composition
= g321-32 -14 ƒ1-32=21-32-1
= g1-72 Multiply, and then subtract.
= 4
-7 -1 g1x2=
4 x+1
= -1 2
Subtract in the denominator.
Write in lowest terms. ✔ Now Try Exercise 57.
The screens show how a graphing calculator evaluates the expressions in Example 5.
The domain and range of g are both the set of all real numbers, 1-∞, ∞2. The domain of ƒ is the set of all nonnegative real numbers, 30, ∞2. Thus, g1x2, which is defined as 4x+ 2, must be greater than or equal to zero.
4x+ 2 Ú 0 Solve the inequality.
x Ú -1
2 Subtract 2. Divide by 4.
Therefore, the domain of ƒ∘g is C-12 , ∞B.
(b) 1g∘ƒ21x2
= g1ƒ1x22 Definition of composition
= gA2x B ƒ1x2= 2x
= 42x +2 g1x2=4x+2
The domain and range of ƒ are both the set of all nonnegative real numbers, 30, ∞2. The domain of g is the set of all real numbers, 1-∞, ∞2. Therefore, the domain of g∘ƒ is 30, ∞2. ✔ Now Try Exercise 75.
EXAMPLE 7 Determining Composite Functions and Their Domains Given that ƒ1x2 =x -6 3 and g1x2= 1x , find each of the following.
(a) 1ƒ∘g21x2and its domain (b) 1g∘ƒ21x2 and its domain SOLUTION
(a) 1ƒ∘g21x2
= ƒ1g1x22 By definition
= ƒ a1
xb g1x2=1x
= 6
1
x - 3 ƒ1x2=
6 x -3
= 6x 1- 3x
Multiply the numerator and denominator by x.
The domain of g is the set of all real numbers except 0, which makes g1x2 undefined. The domain of ƒ is the set of all real numbers except 3. The expression for g1x2, therefore, cannot equal 3. We determine the value that makes g1x2= 3 and exclude it from the domain of ƒ∘g.
1
x =3 The solution must be excluded.
1= 3x Multiply by x.
x= 1
3 Divide by 3.
Therefore, the domain of ƒ∘g is the set of all real numbers except 0 and 13 , written in interval notation as
1-∞, 02 ´ a0, 1
3b ´ a1 3 , ∞b.
The radicand must be nonnegative.
311
2.8 Function Operations and Composition
(b) 1g∘ƒ21x2
= g1ƒ1x22 By definition
= ga 6
x - 3b ƒ1x2=x-63
= 1
6 x -3
Note that this is meaningless if x=3; g1x2=1x .
= x -3 6
1
a b
=1,ab=1# ba=ba
The domain of ƒ is the set of all real numbers except 3. The domain of g is the set of all real numbers except 0. The expression for ƒ1x2, which is x -63 , is never zero because the numerator is the nonzero number 6. The domain of g∘ƒ is the set of all real numbers except 3, written 1-∞, 32 ´ 13, ∞2.
✔ Now Try Exercise 87.
LOOKING AHEAD TO CALCULUS Finding the derivative of a function in calculus is called differentiation.
To differentiate a composite function such as
h1x2=13x+224, we interpret h1x2 as 1ƒ∘g21x2, where
g1x2=3x+2 and ƒ1x2=x4. The chain rule allows us to differenti- ate composite functions. Notice the use of the composition symbol and func- tion notation in the following, which comes from the chain rule.
If h1x2=1ƒ∘g21x2, then h′1x2=ƒ′1g1x22#g′1x2.
EXAMPLE 8 Showing That 1g°f 2 1x2 Is Not Equivalent to 1f°g2 1x2 Let ƒ1x2= 4x+ 1 and g1x2= 2x2+ 5x. Show that 1g∘ƒ21x2≠1ƒ∘g21x2. (This is sufficient to prove that this inequality is true in general.)
SOLUTION First, find 1g∘ƒ21x2. Then find 1ƒ∘g21x2. 1g∘ƒ21x2
= g1ƒ1x22 By definition
= g14x +12 ƒ1x2=4x+1
= 214x +122 + 514x+ 12 g1x2=2x2+5x
= 2116x2 + 8x+ 12 + 20x + 5 Square 4x+1 and apply the distributive property.
= 32x2 + 16x +2 + 20x+ 5 Distributive property
= 32x2 + 36x +7 Combine like terms.
1ƒ∘g21x2
= ƒ1g1x22 By definition
= ƒ12x2 + 5x2 g1x2=2x2+5x
= 412x2 + 5x2 + 1 ƒ1x2=4x+1
= 8x2 + 20x+ 1 Distributive property
Thus, 1g∘ƒ21x2≠1ƒ∘g21x2. ✔ Now Try Exercise 91.
As Example 8 shows, it is not always true that ƒ°g= g°ƒ. One impor- tant circumstance in which equality holds occurs when ƒ and g are inverses of each other, a concept discussed later in the text.
NOTE It often helps to consider the unsimplified form of a composition expression when determining the domain in a situation like Example 7(b).
In calculus it is sometimes necessary to treat a function as a composition of two functions. The next example shows how this can be done.
EXAMPLE 9 Finding Functions That Form a Given Composite Find functions ƒ and g such that
1ƒ∘g21x2= 1x2 -523 - 41x2 -52 + 3.
SOLUTION Note the repeated quantity x2 - 5. If we choose g1x2 =x2 - 5 and ƒ1x2= x3 - 4x+ 3, then we have the following.
1ƒ∘g21x2
= ƒ1g1x22 By definition
= ƒ1x2 - 52 g1x2=x2-5
= 1x2- 523 - 41x2- 52+ 3 Use the rule for ƒ.
There are other pairs of functions ƒ and g that also satisfy these conditions. For instance,
ƒ1x2= 1x -523 - 41x- 52+ 3 and g1x2= x2.
✔ Now Try Exercise 99.
CONCEPT PREVIEW Without using paper and pencil, evaluate each expression given the following functions.
ƒ1x2=x +1 and g1x2 =x2