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7
Functions and Graphs
7.1
Introduction to Functions
7.2
Domain and Range
7.3
Graphs of Functions
7.4
The Algebra of Functions
7.5
Formulas, Applications, and Variation
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
7.1
Introduction to Functions

Correspondences and Functions

Functions and Graphs

Function Notations and Equations

Applications
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Function
A function is a special kind of correspondence
between two sets. The first set is called the
domain. The second set is called the range.
For any member of the domain, there is
exactly one member of the range to which it
corresponds. This kind of correspondence is
called a function.
Domain
Correspondence
Range
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Slide 7- 4
Example
Determine if the correspondence is a
function.
8
2
0
17
–5
Solution
The correspondence is a function because
each member of the domain corresponds to
exactly one member of the range.
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Slide 7- 5
Example
Determine if the correspondence is a
function.
Football
Jackson
Wrestling
Max
Cade
Soccer
Solution
The correspondence is not a function because a
member of the domain (Jackson) corresponds to
more than one member of the range.
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Slide 7- 6
Function
A function is a correspondence between a
first set called the domain, and a second set,
called the range, such that each member of
the domain corresponds to exactly one
member of the range.
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Slide 7- 7
Example
Determine whether the correspondence is a function.
Domain
A set of
rectangles
Correspondence
Range
Each
rectangle’s
area
A set of
numbers
Solution
The correspondence is a function, because
each rectangle has only one area.
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Slide 7- 8
Example
Determine if the correspondence is a function.
Domain
Correspondence
Famous
singers
A song that
the singer
has recorded
Range
A set of
song titles
Solution
The correspondence is not a function,
because some singers have recorded more
than one song.
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Slide 7- 9
Relation
A relation is a correspondence between a
first set called the domain, and a second set,
called the range, such that each member of
the domain corresponds to at least one
member of the range.
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Slide 7- 10
Functions and Graphs
When a function is given as a set of
ordered pairs, the domain is the set of
all first coordinates and the range is the
set of all second coordinates. Function
names are generally represented by
lower- or upper-case letters.
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Slide 7- 11
Example
Find the domain and range of the function f
below.
6
f
5
4
3
2
1
-5 -4 -3 -2 -1-1
1
2
3 4
5
-2
-3
-4
-5
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Slide 7- 12
Solution
Here f can be written
{(–5, 1), (1, 0), (3, –5), (4, 3)}.
The domain is the set
of all first coordinates,
f
{–5, 1, 3, 4}, and the
range is the set of all
second coordinates,
-5 -4 -3
{1, 0, –5, 3}.
6
5
4
3
2
1
-2 -1-1
1
2
3 4
5
-2
-3
-4
-5
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Slide 7- 13
Example
For the function f represented below, determine each
y
of the following.
a) What member of the
range is paired with -2
7
6
5
4
c) What member of the
domain is paired with 6
f
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5
x
-2
-3
-4
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Slide 7- 14
a) What member of the range is paired with -2
Solution
y
Locate -2 on the horizontal
axis (this is where the
domain is located). Next,
find the point directly
above -2 on the graph of f.
From that point, look to
the corresponding ycoordinate, 3. The “input”
-2 has the “output” 3.
7
6
5
f
4
3
2
Output
1
-5 -4 -3 -2 -1
-1
Input -2
1
2 3 4 5
x
-3
-4
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Slide 7- 15
b) What member of the domain is paired with 6
Solution
Locate 6 on the vertical
axis (this is where the range
is located). Next, find the
point to the right of 6 on the
graph of f. From that point,
look to the corresponding
x-coordinate, 2.5. The
“output” 6 has the “input”
2.5.
y
7
Output
6
5
f
4
3
2
1
-5 -4 -3 -2 -1
-1
-2
1
2 3 4 5
Input
x
-3
-4
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Slide 7- 16
The Vertical-Line Test
If it is possible for a vertical line to cross a
graph more than once, then the graph is not
the graph of a function.
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Slide 2- 17
Not a function. Three
y-values correspond
to one x-value
A function
Not a function. Two
y-values correspond to
one x-value
Graphs that do not represent functions still do
represent relations.
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Slide 2- 18
Function Notation and Equations
To understand function notation, it helps to
imagine a “function machine.” Think of putting a
member of the domain (an input) into the
machine. The machine is programmed to produce
the appropriate member of the range (the output)
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Slide 7- 19
Function Notation and Equations
The function pictured has been named f. Here x
represents an arbitrary input, and f (x) – read “f of
x,” “f at x,” or “the value of f at x” represents the
corresponding output.
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Slide 7- 20
Function Notation and Equations
Most functions are described by equations. For
example, f (x) = 5x +2 describes the function that
takes an input x, multiplies it by 5 and then adds 2.
Input
f (x) = 5x + 2
To calculate the output f (3), take the input 3,
multiply it by 5, and add 2 to get 17. That is,
substitute 3 into the formula for f (x).
f (3) = 5(3) + 2 = 17
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Slide 7- 21
Function Notation and Equations
Sometimes in place of f (x) = 5x +2, we
write y = 5x +2, where it is understood that
the value of y, the dependent variable,
depends on our choice of x, the independent
variable.
a) If f (x) = 5x +2, then f(3) = 17.
b) If y = 5x +2, then the value of y is 17
when x is 3.
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Slide 7- 22
Example
Find each indicated function value.
a) f(–2), for f(x) = 3x 2 + 2x
b) g(4), for g(t) = 6t + 9
c) h(m +2), for h(x) = 8x + 1
Solution
a) f (–2) = 3(–2)2 + 2(–2) = 12 – 4 = 8
b) g(4) = 6(4) + 9 = 24 + 9 = 33
c) h(m +2) = 8(m+ 2) + 1 = 8m + 16 + 1
= 8m + 17
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Slide 7- 23
Note that whether we write f(x) = 5x +2 or
f(m) = 5m +2, we still have f(3) = 17. Thus
the independent variable can be thought of
as a dummy variable.
When a function is described by an equation,
the domain is often unspecified. In such
cases, the domain is the set of all numbers for
which function values can be calculated.
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Slide 7- 24
Example
A small business started out in the year
1996 with 10 employees. By the start of
2000 there were 28 employees, and by the
beginning of 2004 the business had grown
to 34 employees. Estimate the number of
employees in 1998 and also predict the
number of employees in 2007.
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Slide 7- 25
Solution
1. and 2. Familiarize. and Translate.
Year
2004
2000
50
40
30
20
10
1996
Number of
Employees
Plot the points and connect the three points. Let the
horizontal axis represent the year and the vertical axis the
number of employees. Label the function itself E.
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Slide 7- 26
3. Carry out.
Year
2004
2000
1998
50
40
30
20
19 10
1996
Number of
Employees
To estimate the number of employees in 1998, locate the point
directly above the year 1998. Then estimate its second
coordinate by moving horizontally from that point to the
y-axis. We see that E (1998)  19.
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Slide 7- 27
3. Carry out (continued).
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2007
Year
2004
2000
50
40
30
20
10
1996
Number of
Employees
To predict the number of employees in 2007, extend the graph
and extrapolate. We see that E (2007)  40.
Slide 7- 28
4. Check.
A precise check would involve knowing more
information. Since 19 is between 10 and 28 and 40
is greater than 34, the estimate seems plausible.
5. State.
In 1997, there were about 19 employees at the small
business. By 2007, the number of employees
should grow to 40.
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Slide 7- 29
7.2
Domain and Range

Determining the Domain and Range

Restrictions on Domain

Functions Defined Piecewise
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When a function is given as a set of ordered pairs, the
domain is the set of all first coordinates and the range
is the set of all second coordinates.
Example
Find the domain and range for the function f given by
f = {(–5, 1), (1, 0), (3, –5), (4, 3)}.
Solution
The domain is the set of all first coordinates:
{–5, 1, 3, 4}.
The range is the set of all second coordinates:
{1, 0, –5, 3}.
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Slide 7- 31
Example
Find the domain and range of the function f shown
y
here.
7
6
5
4
f
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5
x
-2
-3
-4
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Slide 7- 32
The domain of f
Solution
The domain of f is the
set of all x-values that
are used in the points on
the curve. These extend
continuously from -5 to
3 and can be viewed as
the curve’s shadow, or
projection, on the x-axis.
Thus the domain is
y
7
6
5
f
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5
The domain
of f
-2
x
-3
-4
{x | 5  x  3}.
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Slide 7- 33
The range of f
Solution
The range of f is the set
of all y-values that are
used in the points on the
curve. These extend
continuously from -1 to
7 and can be viewed as
the curve’s shadow, or
projection, on the y-axis.
Thus the range is
{ y | 1  y  7}.
y
7
The
range of
f
6
5
f
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2 3 4 5
x
-2
-3
-4
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Slide 7- 34
Example
Determine the domain of f ( x)  3x  4.
2
Solution
We ask, “Is there any number x for which
we cannot compute 3x2 – 4?” Since the
answer is no, the domain of f is , the set
of all real numbers.
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Slide 7- 35
Example
2
.
Determine the domain of f ( x) 
x 8
Solution
We ask, “Is there any number x for which
2
cannot be computed?” Since 2 cannot
x 8
x 8
be computed when x – 8 = 0 the answer is yes.
To determine what x-value would cause x – 8 to
be 0, we solve an equation: x – 8 = 0,
x=8
Thus 8 is not in the domain of f, whereas all other
real numbers are. The domain of f is
{x | x is a real number and x  8}.
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Slide 7- 36
Functions Defined Piecewise
Some functions are defined by different equations
for various parts of their domains. Such functions
are said to be piecewise defined.
To evaluate a piecewise-defined function for an
input a, we first determine what part of the domain
a belongs to. The we use the appropriate formula
for that part of the domain.
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Slide 7- 37
Example
Find each function value for the function f given by
 x  3, if x  3
 2
f ( x)   x ,
if  3  x  4
 4 x,
if x  4

a) f(3)
b) f(2)
c) f(7)
Solution
a) f(x) = x + 3: f(3) = 3 + 3 = 0
b)f(x) = x2; f(2) = 22 = 4
c)f(7) = 4x = 4(7) = 28
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Slide 7- 38
7.3
Graphs of Functions

Linear Functions

Nonlinear Functions
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Equations of Lines
Standard form:
Ax + By = C
Slope-intercept form: y = mx + b
Point-slope form:
y  y1 = m(x  x1)
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Slide 7- 40
Example
Graph 5x + 2y = 10.
Solution
Let x = 0 to find the y-intercept:
5 • 0 + 2y = 10
2y = 10
y=5
The y-intercept is (0, 5).
Let y = 0 to find the x-intercept:
5x + 2• 0 = 10
5x = 10
x=2
The x-intercept is (2, 0).
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y-intercept (0, 5)
x-intercept (2, 0)
5x + 2y = 10
Slide 7- 41
Example
Graph: 3x + 4y = 12
Solution
Rewrite the equation in slope-intercept form.
3 x  4 y  12
4 y  3x  12
1
y   3 x  12 
4
3
y   x3
4
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Slide 7- 42
Solution
The slope is 3/4 and the
y-intercept is (0, 3).
We plot (0, 3), then move
down 3 units and to the
right 4 units.
An alternate approach
would be to move up 3
units and to the left 4 units.
left 4
(4, 6)
up 3
(0, 3)
down 3
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right 4 (4, 0)
Slide 7- 43
Linear Function
A function described by an equation of the
form f(x) = mx + b is a linear function. Its
graph is a straight line with slope m and
y-intercept at (0, b).
When m = 0, the function is described by
f(x) = b is called a constant function. Its graph
is a horizontal line through (0, b).
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Slide 7- 44
Example
4
Graph: f ( x)  x  2
3
Solution
The slope is 4/3 and the
y-intercept is (0, 2).
We plot (0, 2), then move up
4 units and to the right 3 units.
We could also move down
4 units and to the left 3 units.
Then draw the line.
right 3
(3, 2)
up 4 units
(0, 2)
y
4
x2
3
down 4
(3, 6)
left 3
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Slide 7- 45
Example
Graph y = 2
Solution
This is a constant function. For every input x, the output is 2.
The graph is a horizontal line.
y=2
(0, 2)
(4, 2)
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(4, 2)
Slide 7- 46
Domain and Range of a Linear
Function
The domain of any linear function f(x) = mx + b is
{x|x is a real number}.
The range of any linear function f(x) = mx + b,
m  0 is {y|y is a real number}.
The range of a constant function f(x) = b is b.
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Slide 7- 47
Example
Determine the domain and range of each of the
following functions.
a) f, where f(x) = 3x + 2
b) h, where h(x) = 5
Solution
a) f(x) = 3x + 2 describes a linear function, but not a
constant function,
Domain = All R and
Range = All R
b) h(x) = 5 is a constant function.
Domain = All R and Range = {5}
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Slide 7- 48
Nonlinear Functions

A function for which the graph is not a straight line
is a nonlinear function.
Type of function
Example
Absolute-value
f(x) = |x + 2|
Polynomial
p(x) = x4 + 3x2 – 2
Quadratic
h(x) = x2 – 4x + 2
Rational
r ( x) 
x3
x4
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Slide 7- 49
Example
Graph the function given by g(x) = |x + 2|.
Solution
Calculate function values for several choices of x and
list the results in a table.
x
g(x) = |x + 2| (x, f(x))
1
3
(1, 3)
2
4
(2, 4)
–1
1
(–1, 1)
–2
0
(–2, 0)
–3
1
(–3, 1)
0
2
(0, 2)
Domain:
Range: {y|y > 0}
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Slide 7- 50
7.4
The Algebra of Functions

The Sum, Difference, Product, or Quotient of
Two Functions

Domains and Graphs
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The Sum, Difference, Product,
or Quotient of Two Functions
Suppose that a is in the domain of two
functions, f and g. The input a is paired
with f (a) by f and with g(a) by g. The
outputs can then be added to get
f (a) + g(a).
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Slide 7- 52
The Algebra of Functions
If f and g are functions and x is in the
domain of both functions, then:
1. ( f
2. ( f
3. ( f
4. ( f
 g )( x)  f ( x)  g ( x);
 g )( x)  f ( x)  g ( x);
 g )( x)  f ( x)  g ( x);
g )( x)  f ( x) g ( x), provided g ( x)  0.
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Slide 7- 53
Example
For f ( x)  2 x  x and g ( x)  3x  1,
find the following.
a) ( f + g)(4)
b) ( f – g)(x)
2
c) ( f /g)(x)
d) ( f  g )(1)
Solution
a) Since f (4) = –8 and g(4) = 13, we have
( f + g)(4) = f (4) + g(4) = –8 + 13 = 5.
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Slide 7- 54
Solution
b) We have, ( f  g )( x)  f ( x)  g ( x)
 2 x  x  (3x  1)
2
  x  x  1.
2
c) We have, ( f / g )( x)  f ( x) / g ( x)
2x  x

.
3x  1
2
We assume
1
x
3
d) Since f (–1) = –3 and g(–1) = –2, we have
( f  g )(1)  f (1)  g (1)  (3)(2)  6.
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Slide 7- 55
Domains and Graphs
To find ( f  g )(a), ( f  g )(a), ( f  g )(a),
or ( f g )(a), we must first be able to find
f (a) and g(a). This means a must be in the
domain of both f and g.
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Slide 7- 56
Example
3
Given f ( x) 
and g ( x)  x  2,
x 1
find the domains of
( f  g )( x), ( f  g )( x),( f  g )( x) and ( f / g )( x).
Solution
The domain of f is {x | x  1}.
The domain of g is .
Thus the domain of f + g, f – g, and
f  g is {x | x  1}.
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Slide 7- 57
Solution continued
To find the domain of f /g, note that
f ( x) 1/( x  1)
( f / g )( x) 

g ( x)
x2
can not be evaluated if x + 1 = 0 or x – 2 = 0.
Thus the domain of f /g is
{x | x is a real number and x  1 and x  2}.
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Slide 7- 58
7.5
Formulas, Applications and
Variation

Formulas

Direct Variation

Inverse Variation

Joint and Combined Variation
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Formulas
Formulas occur frequently as mathematical
models. Many formulas contain rational
expressions, and to solve such formulas for a
specified letter, we proceed as when solving
rational equations.
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Slide 7- 60
Example
In a hydraulic system, a fluid is confined to two
connecting chambers. The pressure in each
chamber is the same and is given by finding the
force exerted (F) divided by the surface area (A).
Therefore, we know
F1 F2
 .
A1 A2
Solve for A2.
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Slide 7- 61
Solution
F1
F2
A1 A2   A1 A2 
A1
A2
A2 F1  A1F2
A1F2
A2 
F1
Multiplying both sides by the LCD
Simplifying by removing factors
Dividing both sides by F1
This formula can be used to calculate A2 whenever
A1, F2, and F1 are known.
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Slide 7- 62
To Solve a Rational Equation for a
Specified Variable
1. If necessary, multiply both sides by the LCD to
clear fractions.
2. Multiply, as needed, to remove parentheses.
3. Get all terms with the specified variable alone on
one side.
4. Factor out the specified variable if it is in no more
than one term.
5. Multiply or divide on both sides to isolate the
specified variable.
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Slide 7- 63
Variation
To extend our study of formulas and
functions, we now examine three real-world
situations: direct variation, inverse variation,
and combined variation.
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Slide 7- 64
Direct Variation
A mass transit driver earns $17 per hour. In 1 hr, $17
is earned. In 2 hr, $34 is earned. In 3 hr, $51 is
earned, and so on. This gives rise to a set of ordered
pairs: (1, 22), (2, 34), (3, 51), (4, 68), and so on.
Note that the ratio of earnings E to time t is 17/1 in
every case.
If a situation gives rise to pairs of numbers in which
the ratio is constant, we say that there is direct
variation. Here earnings vary directly as the time:
We have E/t = 17, so E = 17t, or using function
notation, E(t) = 17t.
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Direct Variation
When a situation gives rise to a linear
function of the form f(x) = kx, or y = kx,
where k is a nonzero constant, we say that
there is direct variation, that y varies directly
as x, or that y is proportional to x. The
number k is called the variation constant, or
constant of proportionality.
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Example
Find the variation constant and an equation of
variation if y varies directly as x, and y = 15
when x = 3.
Solution
We know that (3, 15) is a solution of y = kx.
Therefore,
Substituting
15  k  3
15
 k ; or k  5
3
Solving for k
The variation constant is 5. The equation of variation
is y = 5x. The notation y(x) = 5x or f(x) = 5x is also
used.
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Inverse Variation
To see what we mean by inverse variation,
suppose it takes one person 8 hours to paint
the baseball fields for the local park district. If
two people do the job, it will take only 4
hours. If three people paint the fields, it will
take only 2 and 2/3 hours, and so on. This
gives rise to pairs of numbers, all have the
same product: (1, 8), (2, 4), (3, 8/3), (4, 2),
and so on.
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Inverse Variation
Note that the product of each pair of numbers
is 8. Whenever a situation gives rise to pairs
of numbers for which the product is constant,
we say that there is inverse variation. Since
pt = 8, the time t, in hours, required for the
fields to be painted by p people is given by
t = 8/p or, using function notation, t(p) = 8/p.
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Inverse Variation
When a situation gives rise to a rational
function of the form f(x) = k/x, or y = k/x,
where k is a nonzero constant, we say that
there is inverse variation, that is y varies
inversely as x, or that y is inversely
proportional to x. The number k is called
the variation constant, or constant of
proportionality.
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Example
The time, t, required to empty a tank varies
inversely as the rate, r, of pumping. If a pump can
empty a tank in 90 minutes at the rate of 1080
kL/min, how long will it take the pump to empty
the same tank at the rate of 1500 kL/min?
Solution
1. Familiarize. Because of the phrase
“ . . . varies inversely as the rate, r, of pumping,”
we express the amount of time needed to empty
the tank as a function of the rate: t(r) = k/r
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Solution (continued)
2. Translate. We use the given information to
solve for k. Then we use that result to write the
equation of variation.
k
t (r ) 
r
Using function notation
k
Replacing r with 1080
1080
k
Replacing t(90) with 1080
90 
1080
Solving for k, the variation constant
97, 200  k
t (1080) 
The equation of variation is t(r) = 97,200/r. This
is the translation.
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Solution (continued)
3. Carry out. To find out how long it would
take to pump out the tank at a rate of 1500
mL/min, we calculate t(1500).
97, 200
t (1500) 
 64.8.
1500
t = 64.8 when r = 1500
4. Check. We could now recheck each step.
Note that, as expected, as the rate goes up, the
time it takes goes down.
5. State. If the pump is emptying the tank at a
rate of 1500 mL/min, then it will take 64.8
minutes to empty the entire tank.
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Joint and Combined Variation
When a variable varies directly with more
than one other variable, we say that there is
joint variation. For example, in the formula
for the volume of a right circular cylinder,
V = πr2h, we say that V varies jointly as h
and the square of r.
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Joint Variation
y varies jointly as x and z if, for some
nonzero constant k, y = kxz.
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Example
Find an equation of variation if a varies
jointly as b and c, and a = 48 when b = 4 and
c = 2.
Solution
We have a = kbc, so
48  k  4  2
6k
The variation constant is 6.
The equation of variation is a = 6bc.
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Joint variation is one form of combined
variation. In general, when a variable varies
directly and/or inversely, at the same time,
with more than one other variable, there is
combined variation.
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Example
Find an equation of variation if y varies jointly as x
and z and inversely as the product of w and p, and
y = 60 when x = 24, z = 5, w = 2, and p = 3.
Solution The equation if variation is of the form
xz
y k
wp
so, substituting, we have:
24  5
60  k 
23
60  k  20
3k
xz
Thus, y  3 
.
wp
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