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Section 8.1
Functions and Their
Representations
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Basic Concepts
• Representations of a Function
• Definition of a Function
• Identifying a Function
• Graphing Calculators (Optional)
The notation y = f(x) is called function notation. The input is x, the output is y, and the name of the function is f.
Name
y = f(x)
Output Input
FUNCTION NOTATION
The variable y is called the dependent variable and the variable x is called the independent variable. The expression f(4) = 28 is read “f of 4 equals 28” and indicates that f outputs 28 when the input is 4. A function computes exactly one output for each valid input. The letters f, g, and h, are often used to denote names of functions.
Representations of a Function
Verbal Representation (Words)
Numerical Representation (Table of Values)
Symbolic Representation (Formula)
Graphical Representation (Graph)
Diagrammatic Representation (Diagram)
Example
Evaluate each function f at the given value of x.a. f(x) = 2x – 4, x = –3 b.
Solutiona. b.
( ) ; 21
xf x x
x
( ) 2 4f x x
( 3) 2( 3) 4
( 3) 10
f
f
( ) ; 21
xf x x
x
2
(2)1 22
(2) 21
f
f
Example
Let a function f compute a sales tax of 6% on a purchase of x dollars. Use the given representation to evaluate f(3).Solutiona. Verbal Representation Multiply a purchase of x dollars by 0.06 to obtain a sales tax of y dollars.
b. Numerical Representation x f(x)
$1.00 $0.06
$2.00 $0.12
$3.00 $0.18
$4.00 $0.24
Example (cont)
c. Symbolic Representation f(x) = 0.06xd. Graphical Representation
e. Diagrammatic Representation
X
Y
1 2 3 4 5 6
0.1
0.2
0.3
0.4
0.5
0.6
0
1 ●
2 ●
3 ●
4 ●
● 0.06
● 0.12
● 0.18
● 0.24
f(3) = 0.18
Definition of a Function
A function receives an input x and produces exactly one output y, which can be expressed as an ordered pair:
(x, y).
Input Output
A relation is a set of ordered pairs, and a function is a special type of relation.
A function f is a set of ordered pairs (x, y), where each x-value corresponds to exactly one y-value.
Function
The domain of f is the set of all x-values, and the range of f is the set of all y-values.
Example
Use the graph of f to find the function’s domain and range. SolutionThe arrows at the ends of the graph indicate that the graph extends indefinitely. Thus the domain includes all real numbers.The smallest y-value on the graph is y = −4. Thus the range is y ≥ −4.
X
Y
-5 -4 -3 -2 -1 1 2 3 4 5
-5
-4
-3
-2
-1
1
2
3
4
5
0
Example
Use f(x) to find the domain of f.
a. f(x) = 3x b.
Solutiona. Because we can multiply a real number x by 3, f(x) = 3x is defined for all real numbers. Thus the domain of f includes all real numbers.
b. Because we cannot divide by 0, the input x = 4 is not valid. The domain of f includes all real numbers except 4, or x ≠ 4.
1
4f x
x
Example
Determine whether the table of values represents a function.
x f(x)
2 −6
3 4
4 2
3 −1
1 0
Solution
The table does not represent a function because the input x = 3 produces two outputs; 4 and −1.
If every vertical line intersects a graph at no more than one point, then the graph represents a function.
Vertical Line Test
Example
Determine whether the graphs represent functions.a. b.
Any vertical line will cross the graph at most once. Therefore the graph does represent a function.
The graph does not represent a function because there exist vertical lines that can intersect the graph twice.
Section 8.2
Linear Functions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Basic Concepts
• Representations of Linear Functions
• Modeling Data with Linear Functions
• The Midpoint Formula (Optional)
, 0 and 1,xf x a a a
A function f defined by f(x) = mx + b, where m and b are constants, is a linear function.
LINEAR FUNCTION
Example
Determine whether f is a linear function. If f is a linear function, find values for m and b so that f(x) = mx + b.a. f(x) = 6 – 2x b. f(x) = 3x2 – 5Solutiona. Let m = –2 and b = 6. Then f(x) = −2x + 6, and f is a linear function.
b. Function f is not linear because its formula contains x2. The formula for a linear function cannot contain an x with an exponent other than 1.
Example
Use the table of values to determine whether f(x) could represent a linear function. If f could be linear, write the formula for f in the form f(x) = mx + b.
SolutionFor each unit increase in x, f(x) increases by 7 units so f(x) could be linear with a = 7. Because f(0) = 4, b = 4. thus f(x) = 7x + 4.
x 0 1 2 3
f(x) 4 11 18 25
Example
Sketch the graph of f(x) = x – 3 . Use the graph to evaluate f(4).SolutionBegin by creating a table.
x y
−1 −4
0 −3
1 −2
2 −1
Plot the points and sketch a line through the points.
Example (cont)
Sketch the graph of f(x) = x – 3 . Use the graph to evaluate f(4).
To evaluate f(4), first find x = 4 on the x-axis. Then find the corresponding y-value. Thus f(4) = 1.
The formula f(x) = ax + b may be interpreted as follows.
f(x) = mx + b
(New amount) = (Change) + (Fixed amount)When x represents time, change equals (rate of change) × (time).
f(x) = m × x + b(Future amount) = (Rate of change) × (Time) + (Initial amount)
MODELING DATA WITH A LINEAR FUNCTION
Example
Suppose that a moving truck costs $0.25 per mile and a fixed rental fee of $20. Find a formula for a linear function that models the rental fees.
SolutionTotal cost is found by multiplying $0.25 (rate per mile) by the number of miles driven x and then adding the fixed rental fee (fixed amount) of $20. Thus f(x) = 0.25x + 20.
Example
The temperature of a hot tub is recorded at regular intervals.
a. Discuss the temperature of the water during this time interval.
The temperature appears to be a constant 102°F.
b. Find a formula for a function f that models these data.Because the temperature is constant, the rate of
change is 0. Thus f(x) = 0x + 102 or f(x) = 102.
Elapsed Time (hours) 0 1 2 3
Temperature 102°F 102°F 102°F 102°F
Example (cont)
The temperature of a hot tub is recorded at regular intervals.
c. Sketch a graph of f together with the data.
Elapsed Time (hours) 0 1 2 3
Temperature 102°F 102°F 102°F 102°F
0
20
40
60
80
100
120
0 1 2 3
Time (hours)
Te
mp
era
ture
(d
eg
ree
s)
The midpoint of the line segment with endpoints(x1, y1) and (x2, y2) in the xy-plane is
Midpoint Formula in the xy-Plane (Optional)
1 2 1 2,2 2
x x y yM
Example
Find the midpoint of the line segment connecting the points (3, 4) and (5, 3).
Solution1 2 1 2,
2 2
x x y yM
3 5 4 3,
2 2
2 1,
2 2
11,
2
Section 8.3
Compound Inequalities
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Basic Concepts
• Symbolic Solutions and Number Lines
• Numerical and Graphical Solutions
• Interval Notation
Basic Concepts
A compound inequality consists of two inequalities joined by the words and or or.
2x > –5 and 2x ≤ 8
x + 3 ≥ 4 or x – 2 < –6
Example
Determine whether the given x-values are solutions to the compound inequalities. x + 2 < 7 and 2x – 3 > 3 x = 4, –4
Solutionx + 2 < 7 and 2x – 3 > 3
Substitute 4 into the given compound inequality.4 + 2 < 7 and 2(4) – 3 > 3 6 < 7 and 5 > 3 True and True
Both inequalities are true, so 4 is a solution.
Example (cont)
Determine whether the given x-values are solutions to the compound inequalities. x + 2 < 7 and 2x – 3 > 3 x = 4, –4
Solutionx + 2 < 7 and 2x – 3 > 3
Substitute –4 into the given compound inequality. –4 + 2 < 7 and 2(–4) – 3 > 3 – 2 < 7 and –11 > 3 True and False
To be a solution both inequalities must be true, so –4 is not a solution.
Symbolic Solutions and Number Lines
We can use a number line to graph solutions to compound inequalities, such as x < 7 and x > –3.
x < 7
x > –3
x < 7 and x > –3
Note: A bracket, either [ or ] or a closed circle is used when an inequality contains ≤ or ≥. A parenthesis, either ( or ), or an open circle is used when an inequality contains < or >.
Example
Solve 3x + 6 > 12 and 5 – x < 11 . Graph the solution.
Solution3x + 6 > 12 and 5 – x < 11
3 6x 6x
2x 6x
Example
Solve each inequality. Graph each solution set. Write the solution in set-builder notation. a. b. c.Solutiona. b.
6 2 10w 4 4 8y 4 2
53 3
w
6 2 2 22 10w 4 8w
6 2 10w
| 4 8w w
1 2y
4 4 8y
| 1 2 y y
Example (cont)
c.
4 25
3 3
w
4 2 5
3 3
w
4 23 5
33
33
w
4 2 15w
24 2 22 15w
6 13w
( 6) ( 11 ) 31 1w
6 13w 13 6w
| 13 6 w w
Example
Solve x + 3 < –2 or x + 3 > 2
Solutionx + 3 < –2 or x + 3 > 2 x < –5 or x > –1
Example
Write each expression in interval notation. a. –3 ≤ x < 7
b. x ≥ 4
c. x < –3 or x ≥ 5
d. {x|x > 0 and x ≤ 5}
e. {x|x ≤ 2 or x ≥ 5}
3,7
4,
, 3 5,
0,5
, 2 5,
Example
Solve 2x + 3 ≤ –3 or 2x + 3 ≥ 5
Solution
2x + 3 ≤ –3 or 2x + 3 ≥ 5
2x ≤ –6 or 2x ≥ 2
x ≤ –3 or x ≥ 1
The solution set may be written as (, 3] [1, )
Section 8.4
Other Functions and Their Properties
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Expressing Domain and Range in Interval Notation
• Absolute Value Function
• Polynomial Functions
• Rational Functions (Optional)
• Operations on Functions
Expressing Domain and Range in Interval Notation
The set of all valid inputs for a function is called the domain, and the set of all outputs from a function is called the range.
Rather than writing “the set of all real numbers” for the domain of f, we can use interval notation to express the domain as (−∞, ∞).
Example
Write the domain for each function in interval notation.
a. f(x) = 3x b.Solutiona. The expression 3x is defined for all real numbers x. Thus the domain of f is
b. The expression is defined except when x – 4 = 0 or x = 4. Thus the domain of f includes all real numbers except 4 and can be written
1
4f x
x
, .
, 4 4, .
Absolute Value Function
We can define the absolute value function by f(x) = |x|.
To graph y = |x|, we begin by making a table of values.
x |x|
–2 2
–1 1
0 0
1 1
2 2
Example
Sketch the graph of f(x) = |x – 3|. Write its domain and range in interval notation. SolutionStart by making a table of values.
x y
0 3
2 1
3 0
4 1
6 3
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8
-5
-4
-3
-2
-1
1
2
3
4
5
0
The domain of f is , .
The range of f is [0, ).
Polynomial Functions
The following expressions are examples of polynomials of one variable.
As a result, we say that the following are symbolic representations of polynomial functions of one variable.
2 3, , and 51 515 3x xx x
32( ) 3, , and ( 5 1 ( )) 1 5 5f g x x xx xx hx
Example
Determine whether f(x) represents a polynomial function. If possible, identify the type of polynomial function and its degree. a.
b.
c.
3( ) 6 2 7f x x x
3.5( ) 4f x x
4( )
5f x
x
cubic polynomial, of degree 3
not a polynomial function because the exponent on the variable is negative
not a polynomial
Example
A graph of is shown. Evaluate f(1) graphically and check your result symbolically.Solution
3( ) 5f x x x
To calculate f(–1) graphically find –1 on the x-axis and move down until the graph of f is reached. Then move horizontally to the y-axis.
f(1) = –4 3( 1) 5( ( )1 1)f 5 1
4
Example
Evaluate f(x) at the given value of x.
Solution
3 2( ) 4 3 7, 2f x x x x
3 2( ) 4 3 7, 2f x x x x
3 2( 2) 4( 2) 3( 2) 7f
( 2) 4( 8) 3(4) 7f
( 2) 32 12 7f
( 2) 27f
Example
Use and to evaluate each of the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
2a. ( ) 3 11 (1)
4
f
2( ) 61 ( )
5
1g
( )(1) (1) (1)
4 5
9
f g f g
Example (cont)
Use and to evaluate each of the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
2b. ( ) 3( ) 1
3(4
2
13
2
) 1
f 2( ) 6 (
6 4
2
2 )2g ( )( 2) ( 2) ( 2)
13 2
11
f g f g
Example (cont)
Use and to evaluate each of the following.
Solution
2( ) 3 1f x x 2( ) 6g x x
a. ( )(1) b. ( )( 2) c. 0f
f g f gg
c. 0f
g
0
00
ff
g g
2
2
3( ) 1 =
6 )
0
(0
1 =
6
Section 8.5
Absolute Value Equations and
Inequalities
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
• Absolute Value Equations
• Absolute Value Inequalities
Absolute Value Equations
An equation that contains an absolute value is an absolute value equation.Examples: |x| = 2, |2x – 5| = 7, |6 – 2x| – 5 = 2
Consider the equation |x| = 2.This equation has two solutions: 2 and –2 because |2| = 2 and also |2| = 2.
Absolute Value Equations
Example
Solve each equation. a. |x| = 38 b. |x| = –4
Solutiona. |x| = 38
b. |x| = –4
38 and 38
no solutions
Example
Solve |2x – 7| = 5 symbolically.Solution2x – 7 = 5 or 2x – 7 = –5
The solutions are 1 and 6.
2x = 122x – 7 = 5 or 2x – 7 = –5
x = 6or 2x = 2 or x = 1
Numerical Solutionx 0 1 2 3 4 5 6
|2x –7| 7 5 3 1 1 3 5
Graphical Solution
Example
Solve. a. |6 – x| – 3 = 0 b.Solutiona. Start by adding 3 to each side.
1 7( 8)
2 8x
6 3x
or 6 63 3x x
3 or 9x x
3 or 9x x
The solutions are 3 and 9.
Example (cont)
b.
1 7( 8)
2 8x
1 7( 8)
2 8x
or 1 1
( 8) ( 8)2 2
7 7
8 8x x
4( 8) 7 or 4( 8) 7x x
Multiply by 8 to clear fractions.
4 32 7 or 4 32 7x x
4 39 or 4 25x x 39 25
or 4 4
x x
The solutions are 25/4 and 39/4.
Example
Solve.a. |3x – 2| = –7 b. |6 – 3x| = 0
Solutiona. |3x – 2| = –7 The absolute value is never negative, there are no solutions.
b. |6 – 3x| = 0
–3x = –6
x = 2
Example
Solve |3x| = |2x – 5|.
Solution3x = 2x – 5 or 3x = –(2x – 5) x = –5 or 3x = –2x + 5
The solutions are −5, and 1.
5x = 5
x = 1
Example
Solve each absolute value equation and inequality.a. |3 – 4x| = 5 b. |3 – 4x| < 5 c. |3 – 4x| > 5Solutiona. |3 – 4x| = 5
3 – 4x = 5 or 3 – 4x = –5 –4x = 2 or –4x = –8 x = –1/2 or x = 2
b. |3 – 4x| < 5 The solution includes x-values between, but not including –1/2 and 2. {x| –1/2 < x < 2} or (–1/2, 2)
Example (cont)
Solve each absolute value equation and inequality.a. |3 – 4x| = 5 b. |3 – 4x| < 5 c. |3 – 4x| > 5Solutionc. |3 – 4x| > 5
The solution includes x-values to the left of x = –1/2 or to the right of x = 2.
The solution set is: {x|x < –1/2 or x > 2}or (, –1/2) (2, )
Example
Solve Write the solution set in interval notation.
Solution
3 24.
4
x
3 24
4
x 3 2 3 2
4 or 44 4
x x
3 2 16 or 3 2 16x x
3 18 or 3 14x x
146 or
3x x
14 or 6
3x x 14
, 6,3
Example
An engineer is designing a circular cover for a container. The diameter d of the cover is to be 4.75 inches and must be accurate within 0.05 inch. Write an absolute value inequality that gives acceptable values for d.SolutionThe diameter d must satisfy 4.7 ≤ d ≤ 4.8.Subtracting 4.75 from each part gives –0.05 ≤ d – 4.75 ≤ 0.05, which is equivalent to |d – 4.75| ≤ 0.05.
Example
Solve if possible.a. b.
Solutiona.
Because the absolute value of an expression cannot be negative, |4 – 3x| is greater than 0 – 1 for every x-value. The solution set is all real numbers.
4 3 2 1x 6 5 0x
4 3 2 1x
4 3 1x
Example
Solve if possible.a. b.
Solutiona.
Because the absolute value is always greater than or equal to 0, no x-values satisfy this inequality. There are no solutions.
4 3 2 1x 6 5 0x
6 5 0x