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Lecture 1.1
College Algebra
Instruction: Definition of Functions
The BIG IDEA in College Algebra is the idea of a "function." A function is a rule that produces
a correspondence between two sets of elements such that to each element in the first set there
corresponds one and only one element in the second set.
Consequently, a relation between a set of first components (called the domain) and a set of
second components (called the range) can be readily identified as a functional relation if each
element in the first set of components does not correspond to more than one element in the
second set.
Alternately defined, a function is a rule that pairs each element x from a set called the domain
with one and only one element y from a set called the range.
In some situations, it is more convenient to think of functions as sets of ordered pairs. In other
situations, functions are best viewed as rules, which are stated in the form of equations.
Functions relate two quantities, the elements of two sets, domain and range. These quantities are
called variables because they can vary or change. The variable from the domain is the
independent or input variable while the variable from the range is the dependent or output
variable. Commonly, the independent variable is represented by x, and the dependent variable is
represented by y.
The following table of ordered pairs represents a set that defines a function since no two ordered
pairs have the same first component and different second components.
x
–2
–1
0
1
2
y
4
1
0
1
4
The next table of ordered pairs, however, does not represent a functional set since it contains
ordered pairs with the same first component and different second components.
x
4
1
0
1
4
y
–2
–1
0
1
2
Lecture 1.1
In terms of equations, a function is an equation that generates a set of solutions or ordered pairs
with the property that no two ordered pairs have the same first component and different second
components.
The equation y + x2 = 5 represents a function because every x-value has only one corresponding
y-value.
The equation y2 + x2 = 5 does not represent a function because some x-values have two
corresponding y-values. For instance, if x = 1, y = 2 or y = –2.
Relations graphed on a Cartesian plane (an x-y plane) can be easily identified as functions using
the vertical line test. If all vertical lines on the Cartesian plane will intersect the relation no more
than once, then the relation is a function. Consequently, the following relations are functions.
The following relations are not functions since it is possible to intersect each more than once
with a vertical line.
Lecture 1.1
An Informal Discussion . . .
Let's think of a paperboy who gets paid per subscriber. Let's assume that the paperboy delivers
papers to x number of subscribers. If the newspaper pays the paperboy $4.50 per month per
subscriber, we can write a monthly income function for the paperboy:
y = $4.50x
The "rule" says multiply the input (or independent) variable by 45 to get the output (or
dependent) variable. The input (independent) variable is the number of subscribers. The monthly
income is the output of the function (or dependent variable). For example, if the paperboy has
100 subscribers, he earns $450.00 per month, i.e., if x = 100, y = $450.00.
The input/independent (x) variable represents a set of values called the domain.
The output/dependent (y) variable represents a set of values called the range.
For the relation to be a function, there can be only one range value for any given domain value,
and we see that this is the case with the paperboy function. If he has 100 subscribers, he gets paid
$450.00 not any other amount.
Example Exercises 1.1
Instruction: Definition of Functions
Example 1
Identifying the Graph of a Function
Which of the graphs below are graphs of functions?
I.
II.
III.
IV.
Graphs I., II., and IV. are graphs of functions because any vertical line imposed over the graphs
will only intersect the graph in at most one point. Graph III., however, is not the graph of a
function since a vertical line will intersect the graph more than once. Consider the vertical line
x = 4, which will intersect the graph at (4,0) and (4,4).
Example Exercises 1.1
Example 2
Determining Whether an Equation is a Function
Does the equation x + y 2 = 4 represent a function?
Since the question does not state which variable represents the domain (input) values and which
represents the range (output) values, the question is ambiguous. This course assumes, however,
that y will always represent the range, which is traditional. Solving the equation for y, will help
determine if the equation represents a function.
x + y2 = 4
y2 = 4 − x
y2 = ± 4 − x
y = ± 4− x
Remember to use the ± symbol
whenever taking the square root of
both sides of an equation.
For particular values of x, e.g., x-values between –4 and 4, the equation is true for more than one
y-value. For instance, substituting 3 for x, the equation yields the following result.
y = ± 4−3
y=± 1
y = ±1
The equation represents a rule that assigns two numbers, 1 and –1, to correspond with the x-value
3; therefore, the equation does not represent a function since a function will only assign at most
one and only one y-value to any given x-value.
Practice Set 1.1
Instructions: Decide if each relation is a function. If the relation is a function, state why. If it is
not, state why not.
#1
#6
x
1
2
3
4
5
10
5
0
-4
-2
0
2
4
-5
y
7
9
11
13
15
-10
#2
#7
10
x
-5
-1
0
1
5
8
6
4
2
-4
0
-2 0
-2
#3
2
4
#8
10
8
x
-1
0
1
6
4
2
-4
#4
0
-2 0
-2
y
5
1
0
1
5
2
y
2, -2
0
2, -2
4
#9 5x + 7y = 32
3
2
1
-2
0
-1 0
2
4
6
-2
-3
#5
#10 x2 + y2 = 25
3
2
1
-2
0
-1 0
2
4
-2
-3
#1 function/passes vertical line test, #2 function/passes vertical line test, #3 not a function/fails vertical line test, #4 not a function/fails vertical
line test, #5 function/passes vertical line test, #6 function because every x-value has only one corresponding y-value, #7 function because every xvalue has only one corresponding y-value, #8 not a function because some of the x-values have more than one corresponding y-value, #9 function
(students should recognize this as a linear function written in standard form) because every x-value has only one corresponding y-value, #10 not a
function (students might recognize this as an equation of a circle) because x-values have more than one corresponding y-value, for instance both
ordered pairs (4,3) and (4,-3) are solutions to the equation
Study Exercise 1.1
Problems
Use the vertical line test to determine whether the following relations are functions. State the
answer with a sentence on the line provided.
#1
R1
______________________________________________________________________________
#2
R2
______________________________________________________________________________
Lecture 1.2
College Algebra
Instruction: Functional Notation
f(x) = x –4
Functional notation replaces the dependent variable with the symbol f(x) such that the
ordered pair (x, f(x)) belongs to the function f. The symbol f(x), therefore, represents the
real number in the range of f (where range is the set of values of the dependent variable)
corresponding to the domain value x (where domain is the set of values of the
independent variable). Consequently, the notation f(4) = 0 represents the ordered pair
(4,0). Evaluating f(2) requires the substitution of 2 in for the independent variable:
f ( x) = x − 4
f (2) = 2 − 4
f (2) = −2
(2,−2)
If x is a real number that is not in the domain of f, then f is not defined at x and f(x) does
not exist.
Any symbol can be similarly substituted for f so that one can discuss multiple functions
such as f and g.
An Informal Discussion continued. . .
In Section 1.1, we discussed the monthly income of a paperboy who earns $4.50 for every
subscriber to whom he delivers the paper. We used the equation below to describe the paperboy's
monthly income.
y = $4.50x
For functions like this, we use a type of notation called function notation where f(x) represents y
(and the "f" is arbitrary and can be any letter or symbol).
For example, we might write p(x) = $4.50x instead of y = $4.50x. This notation gives us a
shorthand where p(100) stands for the function's value (that is, the function's y-value) when
x = 100 so that p(100) = $450.00.
Evaluating p(50) equates to finding the function's value (the range value) when x = 50 as shown
below.
p(50) = $4.5(50)
p(50) = $225.00
So, the paperboy earns $225.00 per month when he has 50 subscribers. :)
Example Exercises 1.2
Instruction: Functional Notation
Example 1
Evaluating Functions
Given f ( x ) = 3 x and g ( x ) = 2 x , which is greater f (125 ) or g ( 50 ) ?
To evaluate f (125 ) , substitute 125 for x in the equation for f.
f ( x) = 3 x
f (125 ) = 3 125 = 5
To evaluate g ( 50 ) , substitute 50 for x in the equation for g.
g ( x) = 2x
g ( 50 ) = 2 ⋅ 50 = 100 = 10
Clearly, 10 is greater than 5; thus, g ( 50 ) > f (125 ) .
Example Exercises 1.2
Example 2
Evaluating Functions
Given q ( x ) = x 2 − 2 , which of the following is negative?
⎛2⎞
q ( 8 ) , q ( −5 ) , q ⎜ ⎟ , q ( 0 ) , q ( −2 )
⎝3⎠
To evaluate q ( 8 ) , substitute 8 for x in the equation for q.
q ( x ) = x2 − 2
q ( 8 ) = ( 8 ) − 2 = 64 − 2 = 62
2
To evaluate q ( −5 ) , substitute -5 for x in the equation for q.
q ( x ) = x2 − 2
q ( −5 ) = ( −5 ) − 2 = 25 − 2 = 23
2
⎛2⎞
To evaluate q ⎜ ⎟ , substitute 2/3 for x in the equation for q.
⎝3⎠
q ( x ) = x2 − 2
2
4
14
⎛2⎞ ⎛2⎞
q⎜ ⎟ = ⎜ ⎟ − 2 = − 2 = −
9
9
⎝3⎠ ⎝3⎠
To evaluate q ( 0 ) , substitute 0 for x in the equation for q.
q ( x ) = x2 − 2
q ( 0 ) = ( 0 ) − 2 = −2
2
To evaluate q ( −2 ) , substitute -2 for x in the equation for q.
q ( x ) = x2 − 2
q ( −2 ) = ( −2 ) − 2 = 4 − 2 = 2
2
⎛2⎞
Only q ⎜ ⎟ and q ( 0 ) are negative.
⎝3⎠
Example Exercises 1.2
Example 3
Using Function Notation
Given f ( x ) = 5 − x 2 , simplify the expression
f ( x + h ) − f ( x)
h
.
Note that the numerator of the expression is a difference of f-values. Evaluate f ( x + h ) .
f ( x ) = 5 − x2
f ( x + h) = 5 − ( x + h)
2
f ( x + h ) = 5 − ⎡⎣( x + h )( x + h ) ⎤⎦
f ( x + h ) = 5 − ⎡⎣ x 2 + xh + hx + h 2 ⎤⎦
f ( x + h ) = 5 − ⎡⎣ x 2 + 2 xh + h 2 ⎤⎦
f ( x + h ) = 5 − x 2 − 2 xh − h 2
Find the difference of f ( x + h ) and f ( x ) .
f ( x + h ) − f ( x) = 5 − x 2 − 2 xh − h 2 − ( 5 − x 2 )
= 5 − x 2 − 2 xh − h 2 − 5 + x 2
= −2 xh − h 2
Simplify the ratio
f ( x + h) − f ( x)
h
f ( x + h ) − f ( x)
h
−2 xh − h 2
h
−2 x h h 2
=
−
h
h
=
= −2 x − h
.
Practice Set 1.2
Given the functions: f ( x ) = 22 − x , g ( x ) = −5 x + 1 , j ( x ) = x 2 + 1 , evaluate the
following expressions.
#1 f (6)
#2 g (0)
#3 f (−3)
#4 j (−1)
#5 f (−27)
#6 g (6)
#7 j (0)
#8 f ( 22)
#9 j (12)
#10 g (−2)
#11 f ( 26)
#12 f ( 20)
#13 j (θ )
#14 g (λ )
#15 g (−1)
#16 g ( a + 2)
#17 f ( 2 )
#18 j (a )
#19 g ( j ( x ))
#20 j ( g ( x ))
#21 g ( a − 1)
#22 g ( a + h )
#23 g ( a ) + g ( 2)
#24 j ( a + h )
#25 j ( a + h ) − j ( a )
#26
j(a + h) − j(a )
h
#1 4, #2 1, #3 5, #4 2, #5 7, #6 –29, #7 1, #8 0, #9 145, #10 11, #11 not a real number, #12
2 , #13 θ2 + 1, #14 –5λ + 1, #15 6,
#16 –5a – 9, #17 22 − 2 , #18 a2 + 1, #19 –5x2 – 4, #20 25x2 – 10x + 2, #21 -5a + 6, #22 –5a – 5h + 1, #23 –5a – 8,
#24 a2 + 2ah + h2 + 1, #25 h2 + 2ah, #26 2a + h
Study Exercise 1.2
Problems
Consider functions f and g given below.
f ( x ) = 5x2 + 1
Evaluate the following expressions.
#1
g ( −1)
#2
g (1)
#3
f ( 3)
#4
f (a + h)
g ( x) = 7 − x
Lecture 1.4
College Algebra
Instruction: Six Elementary Functions
This lecture discusses six elementary functions. The serious student would memorize these six
functions.
I. f ( x) = x
As defined in Box I, f(x) is a linear function. The function increases throughout its domain,
which includes all real numbers. The range also includes all real numbers.
x
–2
–1
0
1
2
f(x)
–2
–1
0
1
2
II. g ( x) = x
As defined in Box II, g(x) is an absolute value function. It decreases from negative infinity to
zero (– ∞ ,0) and increases thereafter (0, ∞ ). The function is symmetrical about the y-axis.
Consequently, the function is an even function where g(x) = g(–x). The minimum value of g(x)
is zero, so the range is (0, ∞ ).
x
–2
–1
0
1
2
g(x)
2
1
0
1
2
Lecture 1.4
III. h( x) = x
2
As defined in Box III, h(x) is a quadratic function with a vertex at the origin (0,0). The function
decreases from negative infinity to zero (– ∞ ,0) and increases thereafter (0, ∞ ). Thus, the yvalue of the vertex represents the absolute minimum value of the function, and the range is
(0, ∞ ). The function is symmetrical about the y-axis. Consequently, the function is an even
function where h(x) =
h(–x).
x
–2
–1
0
1
2
h(x)
4
1
0
1
4
IV. j ( x) = x
As defined in Box IV, j(x) is a square root function. The domain of j(x) is limited to where the
radicand is positive; thus, the domain is where x > 0. The function increases throughout its
domain. Its minimum value is zero, so the range is (0, ∞ ).
x
–1
0
1
2
4
j(x)
non-real number
0
1
2
2
Lecture 1.4
V. k ( x) = x
3
As defined in Box V, k(x) is a cubic function. This function is symmetrical about the origin, so
it is an odd function where –k(x) = k(–x). The domain and range both include all real numbers.
x
–2
–1
0
1
2
k(x)
–8
–1
0
1
8
VI. q ( x) = 3 x
As defined in Box VI, q(x) is a cube-root function. This function is symmetrical about the
origin, so it is an odd function where –q(x) = q(–x). The domain and range both include all real
numbers.
x
–8
–2
–1
0
1
2
8
q(x)
–2
−3 2
–1
0
1
3
2
2
Example Exercises 1.4
Instruction: Elementary Functions
Example 1
Elementary Functions
Name the function graphed below.
The function is a quadratic function, y ( x) = x 2 .
Example 2
Elementary Functions
Name the function graphed below?
The function is the linear function y ( x) = x .
Example 3
Elementary Functions
State the domain of y ( x) = x .
The radicand of a square root must be a non-negative value. The domain of the square
root function includes non-negative numbers: [ 0, ∞ ) .
Example Exercises 1.4
Example 4
Elementary Functions
State the range of y ( x) = x .
The absolute value of a non-zero number is positive. The absolute value of zero is zero.
The range of the absolute value function is [ 0, ∞ ) .
Example 5
Elementary Functions
Consider the cubic function y ( x) = x3 . Is the function an increasing or decreasing
function?
The cubes of successively larger numbers are themselves successively larger. As x-values
increase, the y-values increase. The function increases.
Example 6
Elementary Functions
Consider the cube-root function y ( x) = 3 x . Is the function an increasing or
decreasing function?
The cubed-roots of successively larger numbers are themselves successively larger. As xvalues increase, the y-values increase. The function increases.
Practice Set 1.4
Problems
Label each of the functions below as linear, absolute value, quadratic, square-root, cubic,
or cube-root functions. Describe the transformation applied to the elementary function
(described in Section 1.7) of the same name.
1
3
#2
s (t ) = −t 2
T ( x) = − x
#4
V ( x) =
#5
p( x) = x 2 − 10 x + 25
Hint: The trinomial factors.
#6
A(t ) = −17 + x
L( x) = x 3 + π
#8
F ( x) = 5 x
#1
C ( x) = x + 1.5
s
Hint: a r = r a s .
#3
#7
.
x+7
2
#1 cube-root function, shifted up 1.5 units
#2 quadratic function, reflected over the x-axis
#3 square-root function, reflected over the y-axis
1
#4 absolute value function, V ( x) = x + 7 , shifted left seven units and dilated by a factor of ½
2
#5 quadratic function, p( x) = ( x + 5 ) , shifted right 5 units
2
#6 linear function, shifted down 17 units
#7 cubic function shifted up pi units
#8 linear function, dilated by a factor of 5
Study Exercise 1.4
Problems
Label each of the functions below as linear, absolute value, quadratic, square-root, cubic, or cuberoot functions.
#1 y ( x) = 2 − x
#2 z (t ) = t + 3
#3 A(t ) = t + 3
#4 f ( x ) = x 2 − 9