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Lecture 5.3
Contemporary Mathematics
Instruction: Types of Functions
Lecture 5.2 introduced the concept of a function. A function maps (or transforms)
elements from a set called the domain to a set called the range, which is a subset of the codomain. For most of this chapter, we will assume that the co-domain is always the set of real
numbers. We will further assume that the domain is the set of real numbers unless the operations
1
.
involved in an equation imply some restriction. For instance, consider the equation y =
x+5
Since division is not defined for a divisor of zero, the domain does not include negative five
1
, then
since negative five would make the divisor (the denominator) equal zero. If f : x
x+5
the domain is understood to be all real numbers except −5 , denoted ( −∞, −5 ) ∪ ( −5, ∞ ) using set
interval notation.
As long as we assume that the co-domain includes the set of real numbers, i.e., Y = ,
then functions involving even roots also have restricted domains. Recall that if a < 0, then a ,
4
a,
6
a ,… , n a where n is even are all non-real because any real number multiplied by itself an
x + 1 . For the square
even number of times yields a positive, not a negative. Consider f : x
root operation to yield a real number, the radicand must be non-negative (zero or positive).
x + 1 , we set the radicand greater than
Accordingly, to find the domain of the function f : x
or equal to zero and solve:
x +1 ≥ 0
x ≥ −1
Thus, the domain of f : x
x + 1 does not include all real numbers. Instead, if D is the domain
x + 1 , then D = { x : x ≥ −1, x ∈ } or in set interval notation D = [ −1, ∞ ) .
of f : x
For most operations (multiplication, exponentiation, subtraction, addition, odd roots,
etc.), we will assume the domain includes all real numbers, i.e., X = . The same assumption
does not apply to the range, however. We will assume that the co-domain includes all real
numbers, i.e., Y = , but we will not assume that the range equals the co-domain. The range of
a function is a subset of the co-domain. For example, consider the equation y = x 4 and recall
that ( −2 ) = −2 ⋅ −2 ⋅ −2 ⋅ −2 = 16 . We assume the domain includes all real numbers because the
operation of taking the fourth power is non-restrictive (any real number can be multiplied by
itself four times). Since the fourth power of all real numbers is always non-negative (equal to
zero or positive), the range of the function f : x
x 4 is restrictive and only includes nonnegative real numbers, so R = { y : y ∈ , y ≥ 0} . For another example, consider the function
4
1
, which has a range that includes all real numbers except zero. Zero is excluded
x+5
because no number divides into one zero times (remember that division by zero is undefined).
Indeed, the only way to have a quotient of zero is to have a dividend (numerator) of zero. Since
f :x
Lecture 5.3
zero is excluded, the range in set interval notation is R = ( −∞, 0 ) ∪ ( 0, ∞ ) and in set builder
notation is R = { y : y ∈ , y ≠ 0} .
In the previous two examples the range was not equal to the co-domain (Y =
).
If and
when the range is equal to the co-domain, R = Y , the function is said to map onto the co-domain
and is called a surjective function denoted f : X
Y with domain X and co-domain Y.
A surjective function (also called an onto function) is a function whose range
equals its co-domain.
Figure 1 below shows the graph of a surjective function while the function graphed in
Figure 2 is not surjective.
f :X
y
f :X
y
Y
R
x
Figure 1
x
Figure 2
Recall that functions are sets of ordered pairs that have the property that every domain
element is paired with one and only one element in the range. Consequently, each first
coordinate of a function is distinct. This is not necessarily true for the second coordinates.
Reconsider the function f : x
x 4 . This function maps −2 to 16. It also maps 2 to 16.
Consequently, the ordered pairs ( −2,16 ) and ( 2,16 ) are in the function, and it is evident that not
every second coordinate is distinct. Functions that do map distinct elements from the domain to
distinct elements in the co-domain are one-to-one functions also called injective functions.
An injective function (also called a one-to-one function) is a function that maps
distinct elements from the domain to distinct elements in the range. In other
words, an injective function is a function with the property that for every
y.
element y in the range there exists only one x in the domain such that x
If we restrict our discussion to functions that are sets of ordered pairs of real numbers that
represent points on the Cartesian plane, we can employ a simple test called the horizontal line
test to see if any function is a one-to-one function. Since each element in the range of a one-toone function is distinct, any horizontal line drawn through a one-to-one function intersects the
function only once.
Let X = Y = . Let X × Y = 2 . Let the horizontal axis of the Cartesian Plane
represent X and the vertical axis represent Y. Let F be a function graphed on the
Cartesian plane. If any horizontal line intersects the graph of F at more than one
point, then F is not one-to-one and is said to fail the horizontal line test. If any
horizontal line intersects the graph of F at no more than one point, then F is a
one-to-one or injective function and is said to pass the horizontal line test.
Lecture 5.3
Figure 3 below shows the graph of an injective function that passes the horizontal line test while
the function graphed in Figure 4 fails the horizontal line test because it is not injective.
f :X
y
R
f :X
y
R
x
Figure 3
x
Figure 4
Functions that are both surjective and injective are called bijective functions. A famous
class of bijective functions are linear functions, which are described by equations of the form
y = mx + b where m and b are real numbered constants ( m ≠ 0 ) and x and y are variables
representing the elements of the domain and co-domain respectively with both the domain and
co-domain understood to include all real numbers. The equation y = 2 x − 1 describes a particular
linear function f : x
2 x − 1 alternately denoted f = {( x, y ) : x ∈ , y ∈ , y = 2 x − 1} .
Application Exercise 5.3
Problems
#1
Let D be the domain of f. Let Y =
be the co-domain. If f : x
x+2
, what is D?
x −1
Assume that x represents elements of D.
#2
Let D be the domain of h. Let Y = be the co-domain. If h : x
Assume that x represents elements of D.
#3
Let D be the domain of g. Let Y = be the co-domain. If g : x
Assume that x represents elements of D.
#4
Let D be the domain of q. Let Y =
be the co-domain. If q : x
− x + 3 , what is D?
4
x − 3 , what is D?
x+4
, what is D? Assume
x 2 − 16
that x represents elements of D.
#5
Let X be the domain and Y the co-domain. Let X = Y =
range of f? Assume x represents elements of X.
#6
Let X = Y = . Let X be the domain and Y the co-domain. Consider the graph of f below. Is the
function surjective?
f
#7
Let X = Y = . Let X be the domain and Y the co-domain. Consider the graph of f below. Is the
function injective?
f
#8
Let X = Y =
bijective?
. Let X be the domain and Y the co-domain. Consider L : x
x . Is the function
#9
Let X = Y =
bijective?
. Let X be the domain and Y the co-domain. Consider Q : x
x 4 . Is the function
#10
Let X = Y =
bijective?
. Let X be the domain and Y the co-domain. Consider C : x
x3 . Is the function
. Consider f : x
6 . What is the
#1 D = ( −∞,1) ∪ (1, ∞ )
#2 D = ( −∞, 3]
#3 D = [3, ∞ )
#4 D = ( −∞, −4 ) ∪ ( −4, 4 ) ∪ ( 4, ∞ )
#5 R = {6}
#6 Yes.
#7 No.
#10 Yes.
#8 Yes.
#9 No.
Assignment 5.3
Problems
#1
What are two operations that imply a restriction on the domain?
#2
Let D be the domain of f. Let Y = \ be the co-domain. If f : x 6
2
, what is D? Assume
3+ x
that x represents elements of D.
#3
Let D be the domain of f. Let Y = \ be the co-domain. If f : x 6 5 − x , what is D? Assume
that x represents elements of D.
#4
Let X be the domain and Y the co-domain. Let X = Y = \ . Consider f : x 6 x 2 . What is the
range of f? Assume x represents elements of X.
#5
Let X = Y = \ . Let X be the domain and Y the co-domain of the following functions. Which
function is bijective? Assume x represents elements of X.
f : x 6 x4 − 4 x2
g : x 6 5x + 1
2
h:x 6 2
x +1