Download FUNCTIONS

1.
FUNCTIONS
INTRODUCTION
The function concept is one of the most important ideas in mathematics. Generally,
mathematics is concerned with relationships between things, and it is through the generality
of these relations that applications arise. For instance, engineers are concerned with
expressing relationships between physical quantities clearly and unambiguously. This might
be the relationship between the displacement of an oscillating object and time, or perhaps the
amplitude of an AC voltage and time. Functions can be used to describe the way quantities
change; hence, we need functions to handle practical problems analytically. In this unit, you
will learn (i) how to represent such relationships in mathematical terms, (ii) the fundamentals
of functions, and (iii) how to apply them.
LEARNING OUTCOMES
On completion of this unit, you will be able to:

Determine whether relations between two variables are functions.

Use function notation and evaluate functions.

Find the domain and range of functions.

Use transformations to sketch graphs of functions.

Determine odd and even functions.

Formulate the inverse of a function.

Identify and evaluate linear, quadratic, polynomial, and rational functions.
COMPILED BY T. PAEPAE
1.1
DEFINITION OF A FUNCTION
Many everyday phenomena involve two quantities related to each other by some rule of
correspondence. The mathematical term for such a rule of correspondence is a relation. This
section will define and develop the concept of a “function,” which is the basic mathematical
object that engineers, scientists, and mathematicians use to describe relationships between
variable quantities. Functions arise whenever one quantity depends on another. Consider the
following situations:

The area of a circle (𝐴𝐴) depends on the radius of that circle (𝑟𝑟)

The time taken for a particular journey (𝑡𝑡) depends on the average speed (𝑠𝑠)

Shoe size (𝑦𝑦) depends on how big the foot is (𝑥𝑥)

Your final mark (𝑚𝑚) depends on how well you study (𝑛𝑛)
In these examples, the two variables in each case are related, and we shall define a function
as a special type of relation between two variables, and this will lead to the powerful and
convenient functional notation which plays a vital role in mathematics.
1.1.1
Defining Functions by Equations
In mathematics, relations are often represented by mathematical equations and formulas. For
instance, suppose driving a car that averages 100 𝑘𝑘𝑘𝑘/ℎ. The distance travelled is determined
by the time travelled (𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 × 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡). This relation can be expressed by the
equation 𝐷𝐷 = 100𝑡𝑡.
For 𝑡𝑡 = 2 hours, the distance travelled will be:

𝐷𝐷 = 100(2) = 200𝑘𝑘𝑘𝑘
For each specific value of 𝑡𝑡 ≥ 0, the equation produces exactly one value for 𝐷𝐷. Such
a relation is called a function.

The variable 𝑡𝑡 is called an independent variable (or input), indicating that values can
be assigned “independently” to t. The variable D is called a dependent variable (or
output), indicating that the value of D “depends” on the value assigned to t and on the
given equation.
Because the variable t represents time in the equation D = 100t, it is reasonable to say that
t ≥ 0. This set of all allowable real number values for the independent variable t is called the
domain of the function. The set of corresponding real number values for the dependent
variable is called the range of the function.
1
1.1.2
Formal Definition of Function
In an equation with two variables, if each value of the independent variable corresponds to
exactly one value of the dependent variable, then the equation defines a function.
1.1.3
Four Ways to Represent a Function

Verbally by a sentence that describes how the input is related to the output variable.

Numerically by a table or a list of ordered pairs that matches input with output values.

Graphically by points on a graph in a coordinate plane in which the input values are
represented by the horizontal axis and the output values by the vertical axis.

Algebraically by an equation in two variables
Since an equation is just one way to represent a function, we will say “an equation defines a
function” rather than “an equation is a function.” To determine whether or not a relation is a
function, you must check whether each input value is matched with exactly one output value.
The relation is not a function if any input value is matched with two or more output values. This
is usually tested algebraically or graphically, as discussed in example 1.1.
Example 1.1
Determining if an Equation Defines a Function
Determine if each of the following equations define a function with independent variable 𝑥𝑥.
1.
𝑦𝑦 = 𝑥𝑥 2 − 4
Solution 1: (Algebraically)
For any real number 𝑥𝑥, the square of 𝑥𝑥 is a unique real number. When you subtract 4, the
result is again a unique real number. So, for any input 𝑥𝑥, there is exactly one output 𝑦𝑦, and
therefore the equation defines a function.
2.
𝑥𝑥 2 + 𝑦𝑦 2 = 16
Solution 1: (Algebraically)
In this case, it will be helpful to solve the equation for the dependent variable.
𝑥𝑥 2 + 𝑦𝑦 2 = 16
𝑦𝑦 2 = 16 − 𝑥𝑥 2
Subtract 𝑥𝑥 2 from both sides.
Take the square root of both sides.
𝑦𝑦 = ±�16 − 𝑥𝑥 2
For any 𝑥𝑥 that produces an output, there are two choices for 𝑦𝑦, one positive and one negative.
The equation has more than one output for some inputs, so it does not define a function.
2
Solution 2: Graphically (using the vertical line test)
The vertical line test can be used to determine whether a graph represents a function. If we
can draw any vertical line that intersects a graph more than once, then the graph does not
define a function because a function has only one output value for each input value.
It is very easy to determine whether an equation defines a function if you have the graph of
the equation. The two equations we considered in Example 1.1 are graphed in Figure 1.1.
(𝐴𝐴)
(𝐵𝐵)
Figure 1.1: Graphs of equations and the vertical line test
In Figure 1.1(A), any vertical line will intersect the graph of 𝑦𝑦 = 𝑥𝑥 2 − 4 exactly once. This
shows that every value of the independent variable 𝑥𝑥 corresponds to exactly one value of the
dependent variable 𝑦𝑦, and that confirms our conclusion that 𝑦𝑦 = 𝑥𝑥 2 − 4 defines a function. But
in Figure 1.1(B), there are many vertical lines that intersect the graph of 𝑥𝑥 2 + 𝑦𝑦 2 = 16 in two
points. This shows that there are values of the independent variable 𝑥𝑥 that correspond to two
different values of the dependent variable 𝑦𝑦, which confirms our conclusion that 𝑥𝑥 2 + 𝑦𝑦 2 = 16
does not define a function. These observations lead to Theorem 1.
THEOREM 1
Vertical Line Test for a Function
An equation defines a function if each vertical line in a rectangular coordinate system passes
through at most one point on the graph of the equation. If any vertical line passes through two
or more points on the graph of an equation, then the equation does not define a function.
Sometimes when a function is defined by an equation, a domain is specified, as in:
𝑦𝑦 = 2𝑥𝑥 2 + 5,
𝑥𝑥 > 0
The “𝑥𝑥 > 0” tells us that the domain is all positive real numbers. More often, a function is
defined by an equation with no domain specified. Unless a domain is specified, we will use
the following convention regarding domains and ranges for functions defined by equations.
3
1.2
CONVENTION ON DOMAINS AND RANGES
If a function is defined by an equation and the domain is not stated explicitly, then we assume
that the implied domain is the set of all real number replacements of the independent variable
that produce real values for the dependent variable. The range is the set of all values of the
dependent variable corresponding to the domain values.
1.2.1
Finding the Domain of a Function Defined by an Equation
Finding the domain of a function whose equation is provided involves remembering three
different forms. First, if the function has no denominator or an even root, consider whether the
domain could be all real numbers. Second, if there is a denominator in the function’s equation,
exclude values in the domain that force the denominator to be zero. Third, if there is an even
root, consider excluding values that would make the radicand negative.
Example 1.2
Finding the Domain of a Function
Before we begin, let us review the conventions of interval notation:

The smallest term from the interval is written first.

The largest term in the interval is written second, following a comma.

Parentheses, (or), are used to signify that an endpoint is not included, called exclusive.

Brackets, [or], are used to indicate that an endpoint is included, called inclusive.
Find the domain of each of the following functions and write the answer in interval notation:
1.
Polynomial only
TIP: whenever the function has no denominator or an even root, consider whether the domain
could be all real numbers.
1.1
𝑦𝑦 = 4𝑥𝑥 + 1
Solution:
There are no restrictions on the domain of this function. Therefore, the domain is the set of
real numbers. In interval notation, the domain is (−∞, ∞).
1.2
𝑦𝑦 = 𝑥𝑥 2 + 3𝑥𝑥 − 5
Solution:
The domain is the set of real numbers. In interval notation, the domain is (−∞, ∞).
4
2.
Fraction only
TIP: when there is a denominator in the function’s equation, exclude values in the domain that
force the denominator to be zero.
2.1
𝑦𝑦 =
5𝑥𝑥+1
𝑥𝑥 2 +5𝑥𝑥+6
Solution:
Since there is a denominator, we want to include only values of the input that do not force the
denominator to be zero. So, we will set the denominator to ≠ 0 and solve for 𝑥𝑥.
𝑥𝑥 2 + 5𝑥𝑥 + 6 ≠ 0
(𝑥𝑥 + 3)(𝑥𝑥 + 2) ≠ 0
𝑥𝑥 ≠ −3 or 𝑥𝑥 ≠ −2
Now, we will exclude −3 and −2 from the domain. To help you visualise the solution, it will be
ideal to indicate these values on the number line as follows:
−∞
−3
−2
∞
We will then use a symbol known as the union, ∪, to combine the three sets. In interval
notation, we write the solution as: (−∞, −3) ∪ (−3, −2) ∪ (−2, ∞)
2.2
𝑦𝑦 =
2𝑥𝑥−3
𝑥𝑥 2 +4
Solution:
We again set the denominator ≠ 0 and solve for 𝑥𝑥.
𝑥𝑥 2 + 4 ≠ 0
𝑥𝑥 2 ≠ −4
Remember that the domain must consist of all real numbers for which the expression is
defined. Therefore, this tells you that there is no value of 𝑥𝑥 at which the domain is undefined.
In interval notation, the domain is (−∞, ∞).
3.
Square root only
TIP: if there is an even root, consider excluding values that would make the radicand negative.
3.1
𝑦𝑦 = 2√𝑥𝑥 + 4
Solution:
5
Since there is an even root in the formula, we exclude any real numbers that result in a
negative number in the radicand. Set the radicand ≥ 0 and solve for 𝑥𝑥.
𝑥𝑥 + 4 ≥ 0
𝑥𝑥 ≥ −4
Now, we will exclude any number less than −4 from the domain. Hence, the domain in interval
notation is [−4 , ∞).
3.2
𝑦𝑦 = √3 − 2𝑥𝑥 + 5
Solution:
Since there is an even root in the formula, set the radicand ≥ 0 and solve for 𝑥𝑥.
3 − 2𝑥𝑥 ≥ 0
−2𝑥𝑥 ≥ −3
3
In interval notation, the domain is �−∞ , �.
2
3.3
𝑥𝑥 ≤
3
2
𝑦𝑦 = √𝑥𝑥 2 + 3𝑥𝑥 − 28
Solution:
Set part radicand ≥ 0 and solve for 𝑥𝑥. Then draw the number line and test the resulting
intervals.
𝑥𝑥 2 + 3𝑥𝑥 − 28 ≥ 0
(𝑥𝑥 + 7)(𝑥𝑥 − 4) ≥ 0
−∞
𝑥𝑥 ≥ −7 or 𝑥𝑥 ≥ 4
4
−7
∞
Now test these three intervals in order to make sure that indeed (𝑥𝑥 + 7)(𝑥𝑥 − 4) ≥ 0. For the
first interval, choose any value less than −7 and substitute it into (𝑥𝑥 + 7)(𝑥𝑥 − 4) ≥ 0. For
instance, let us choose −10. Substituting it results in (−10 + 7)(−10 − 4) = 42, which is
indeed ≥ 0 and therefore the first interval is part of the solution. Let us choose 0 (or any value
between −7 and 4) for the second interval. Substituting it results in (0 + 7)(0 − 4) = −28,
which is not ≥ 0 and therefore this interval is not part of the solution. For the last interval,
choose any value greater than 4 and substitute. Choosing 5 results in (5 + 7)(5 − 4) = 12,
which is indeed ≥ 0. Therefore, the final solution is (−∞, −7] ∪ [4, ∞) in interval notation.
6
4.
𝑦𝑦 =
Square root on bottom
𝑥𝑥 2 + 2𝑥𝑥 + 3
√2𝑥𝑥 − 1
Solution:
Since the square root is the denominator, we set the radicand > 0 and solve for 𝑥𝑥.
2𝑥𝑥 − 1 > 0
1
2
𝑥𝑥 >
1
2
This � � has to be excluded from the solution since it will result in division by zero. Therefore,
1
2
the solution is � , ∞� in interval notation.
5.
Square root on top
𝑦𝑦 =
√𝑥𝑥 − 4
𝑥𝑥 2 − 25
Solution:
We are going to set the denominator ≠ 0, radicand ≥ 0, and then take the intersection of the
two domains as follows:
For the numerator
𝑥𝑥 − 4 ≥ 0
𝑥𝑥 ≥ 4
For the denominator
𝑥𝑥 2 − 25 ≠ 0
(𝑥𝑥 − 5)(𝑥𝑥 + 5) ≠ 0
𝑥𝑥 ≠ 5 or 𝑥𝑥 ≠ −5
Since 𝑥𝑥 ≥ 4, the number line would look like this
−∞
4
5
Therefore, in interval notion, the solution is [4 , 5) ∪ (5 , ∞).
7
∞
6.
𝑦𝑦 =
Square root on top
√𝑥𝑥 + 3
√𝑥𝑥 2 − 16
Solution:
We are going to set the radicand in numerator ≥ 0, radicand in denominator > 0, test the
intervals for the denominator, and then find the intersection of the two domains by creating a
hybrid number line.
For the numerator
𝑥𝑥 + 3 ≥ 0
𝑥𝑥 ≥ −3
For the denominator
𝑥𝑥 2 − 16 > 0
(𝑥𝑥 − 4)(𝑥𝑥 + 4) > 0
𝑥𝑥 > 4 or 𝑥𝑥 > −4
Test these intervals the same way we did in example 3.3. You should, from the test, find that
the only valid regions are (−∞, −4) and (4, ∞). Now, since 𝑥𝑥 must be ≥ −3, this means we
must exclude the intervals (−∞, −4) and (−4, 4). Therefore, the final solution is (4, ∞).
1.2.2
Finding Domain and Range from Graphs
Another way to identify the domain and range of functions is by using graphs. Since the domain
refers to a set of possible input values, the domain of a graph consists of all the input values
shown on the 𝑥𝑥-axis. The range is the set of possible output values, which are shown on the
𝑦𝑦-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see,
the domain and range may be greater than the visible values, as shown in Figure 1.2.
Figure 1.2: Domain and range from a graph
8
We can observe that the graph extends horizontally from −5 to the right without bound, so the
domain is [−5, ∞). The vertical extent of the graph is all range values 5 and below, so the
range is (−∞, 5]. Remember that the domain and range are always written from smaller to
larger values, or from left to right for domain, and from the bottom of the graph to the top of
the graph for range.
ACTIVITY 1
1.
Determine the domain of each of the following and give solutions in interval notation.
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.3
𝑦𝑦 = √4 − 𝑥𝑥 2
[−2, 2]
𝑦𝑦 = √7 − 𝑥𝑥
𝑦𝑦 = �
(−∞, 7]
𝑥𝑥
2−𝑥𝑥
[0, 2)
𝑦𝑦 = √𝑥𝑥 + 3
𝑥𝑥−1
𝑦𝑦 =
√2−𝑥𝑥
𝑦𝑦 =
𝑥𝑥+1
2−𝑥𝑥
𝑦𝑦 =
𝑦𝑦 =
[−3, ∞)
(−∞, 2)
√𝑥𝑥+3
√1−𝑥𝑥
[−3, 1)
(−∞, 2) ∪ (2, ∞)
4𝑥𝑥+1
2𝑥𝑥−1
1
2
𝑦𝑦 = √𝑥𝑥 2 + 7𝑥𝑥 + 12
𝑦𝑦 =
𝑦𝑦 =
1
2
�−∞, � ∪ � , ∞�
(−∞, −4] ∪ [−3, ∞)
4
𝑥𝑥 2 −𝑥𝑥
(−∞, 0) ∪ (0, 1) ∪ (1, ∞)
4
𝑥𝑥 2 −9
(−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
FUNCTION NOTATION AND SUBSTITUTION
We will use letters (such as 𝑓𝑓, 𝑔𝑔, ℎ, etc.) to name functions and to provide a very important and
convenient notation for defining functions. For example, if 𝑓𝑓 is the name of the function defined
by the equation 𝑦𝑦 = 2𝑥𝑥 + 1, we will simply write:
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 1
Functional notation
The symbol 𝑓𝑓(𝑥𝑥) is read “𝑓𝑓 of 𝑥𝑥”, “𝑓𝑓 at 𝑥𝑥”, or “the value of 𝑓𝑓 at 𝑥𝑥” and represents the number
in the range of the function 𝑓𝑓 (the output) that is paired with the domain value 𝑥𝑥 (the input).
Keep in mind that if we have a function 𝑓𝑓(𝑥𝑥), then the notation 𝑓𝑓(𝑎𝑎) implies that we replace 𝑥𝑥
by 𝑎𝑎 in the equation.
9
Example 1.3
1.
Evaluating Functions
Find 𝑓𝑓(4𝑎𝑎) for 𝑓𝑓(𝑥𝑥) =
2
√𝑥𝑥−2
Solution:
𝑓𝑓(4𝑎𝑎) =
=
2
√4𝑎𝑎 − 2
2
√4 √𝑎𝑎 − 2
1
2
=
=
2 √𝑎𝑎 − 2 √𝑎𝑎 − 1
2.
Let 𝑔𝑔(𝑥𝑥) = −𝑥𝑥 2 + 4𝑥𝑥 + 1. Find each function value.
2.1
𝑔𝑔(−2)
Solution:
𝑔𝑔(−2) = −(−2)2 + 4(−2) + 1 = −11
2.2
𝑔𝑔(𝑡𝑡)
Solution:
𝑔𝑔(𝑡𝑡) = −(𝑡𝑡)2 + 4(𝑡𝑡) + 1
2.3
𝑔𝑔(𝑥𝑥 + 2)
Solution:
𝑔𝑔(𝑥𝑥 + 2) = −(𝑥𝑥 + 2)2 + 4(𝑥𝑥 + 2) + 1
= −(𝑥𝑥 2 + 4𝑥𝑥 + 4) + 4𝑥𝑥 + 8 + 1
= −𝑥𝑥 2 − 4𝑥𝑥 − 4 + 4𝑥𝑥 + 9
1.3.1
= −𝑥𝑥 2 − 5
Combinations of Functions
Just as numbers can be added, subtracted, multiplied, and divided to produce other numbers,
so functions can be added, subtracted, multiplied, and divided to produce other functions. In
this section we will discuss these operations and some others that have no analog in ordinary
arithmetic (composite functions).
10
1.3.1.1 Arithmetic Combinations of Functions
Just as two real numbers can be combined by the operations of addition, subtraction,
multiplication, and division to form other real numbers, two functions can be combined to
create new functions. For instance,
1.
Sum:
2.
Difference
3.
Product
4.
Quotient
(𝑓𝑓 + 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) + 𝑔𝑔(𝑥𝑥)
(𝑓𝑓 − 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) − 𝑔𝑔(𝑥𝑥)
(𝑓𝑓𝑓𝑓)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) . 𝑔𝑔(𝑥𝑥)
𝑓𝑓
𝑔𝑔
𝑓𝑓(𝑥𝑥)
� � (𝑥𝑥) = 𝑔𝑔(𝑥𝑥) ,
𝑔𝑔(𝑥𝑥) ≠ 0
The domain of an arithmetic combination of functions 𝑓𝑓 and 𝑔𝑔 consists of all real numbers
that are common to the domains of 𝑓𝑓 and 𝑔𝑔. That is, if the domain of 𝑓𝑓 is 𝐴𝐴 and the domain of
𝑔𝑔 is 𝐵𝐵, then the domain of 𝑓𝑓 + 𝑔𝑔 is the intersection 𝐴𝐴 ∩ 𝐵𝐵 because both 𝑓𝑓 (𝑥𝑥) and 𝑔𝑔(𝑥𝑥) have
to be defined.
Example 1.4
1.
Combinations of Functions and their Domains
Let 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 1 and 𝑔𝑔(𝑥𝑥) = 𝑥𝑥 2 + 2𝑥𝑥 − 1.
1.1 Find the functions (𝑓𝑓 + 𝑔𝑔)(𝑥𝑥) , (𝑓𝑓 − 𝑔𝑔)(𝑥𝑥).
Solution:
(𝑓𝑓 + 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) + 𝑔𝑔(𝑥𝑥)
= (2𝑥𝑥 + 1) + (𝑥𝑥 2 + 2𝑥𝑥 − 1)
= 4𝑥𝑥 + 𝑥𝑥 2
(𝑓𝑓 − 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓(𝑥𝑥) − 𝑔𝑔(𝑥𝑥)
= (2𝑥𝑥 + 1) − (𝑥𝑥 2 + 2𝑥𝑥 − 1)
= 2𝑥𝑥 + 1 − 𝑥𝑥 2 − 2𝑥𝑥 + 1
= 2 − 𝑥𝑥 2
1.2 Find (𝑓𝑓 + 𝑔𝑔)(2) and (𝑓𝑓 − 𝑔𝑔)(−2)
Solution:
(𝑓𝑓 + 𝑔𝑔)(2) = 4(2) + (2)2 = 12
(𝑓𝑓 − 𝑔𝑔)(−2) = 2 − (−2)2 = −2
11
2.
Let 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 and 𝑔𝑔(𝑥𝑥) = √4 − 𝑥𝑥 2
𝑓𝑓
𝑔𝑔
𝑔𝑔
𝑓𝑓
Find the functions � � (𝑥𝑥) , � � (𝑥𝑥) and their domains.
Solution:
𝑓𝑓(𝑥𝑥)
𝑓𝑓
√𝑥𝑥
=
� � (𝑥𝑥) =
𝑔𝑔(𝑥𝑥) √4 − 𝑥𝑥 2
𝑔𝑔
𝑔𝑔(𝑥𝑥)
𝑔𝑔
� � (𝑥𝑥) =
𝑓𝑓(𝑥𝑥)
𝑓𝑓
=
√4 − 𝑥𝑥 2
√𝑥𝑥
The domain of 𝑓𝑓 is [0, ∞) and the domain of 𝑔𝑔 is [−2, 2]. The intersection of these domains is
𝑓𝑓
𝑔𝑔
𝑔𝑔
𝑓𝑓
[0, 2]. So, the domains of � � and � � are as follows.
𝑓𝑓
𝑔𝑔
Domain of � � = [0, 2)
1.3.2
𝑔𝑔
𝑓𝑓
Domain of � � = (0, 2]
and
Composition of Functions
Another way of combining two functions is to form the composition of one with the other. For
instance, given any two functions 𝑓𝑓 and 𝑔𝑔, we start with a number 𝑥𝑥 in the domain of 𝑔𝑔 and
find its output 𝑔𝑔(𝑥𝑥). If this number 𝑔𝑔(𝑥𝑥) is in the domain of 𝑓𝑓, then we can calculate the value
of 𝑓𝑓(𝑔𝑔(𝑥𝑥)). This new function ℎ(𝑥𝑥) = 𝑓𝑓(𝑔𝑔(𝑥𝑥)) is obtained by substituting 𝑔𝑔 into 𝑓𝑓. It is called
the composition of 𝑓𝑓 and 𝑔𝑔 and is denoted by 𝑓𝑓 ∘ 𝑔𝑔, (read “𝑓𝑓 composed with 𝑔𝑔”). That is
(𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓�𝑔𝑔(𝑥𝑥)�
The domain of (𝑓𝑓 ∘ 𝑔𝑔) is the set of all 𝑥𝑥 in the domain of 𝑔𝑔 such that 𝑔𝑔(𝑥𝑥) is in the domain of
𝑓𝑓. In other words, (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) is defined whenever both 𝑔𝑔(𝑥𝑥) and 𝑓𝑓�𝑔𝑔(𝑥𝑥)� are defined.
When computing 𝑓𝑓 ∘ 𝑔𝑔, it is important to keep in mind that the first function that appears in the
notation (𝑓𝑓, in this case) is usually the second function that is applied. For this reason, 𝑓𝑓 ∘ 𝑔𝑔 is
sometimes read as “𝑓𝑓 following 𝑔𝑔”.
Example 1.5
1.
Finding the Composition of Functions and the Domain
Given 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 + 2 and 𝑔𝑔(𝑥𝑥) = 4 − 𝑥𝑥 2 , find the following:
1.1 (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥)
12
Solution:
(𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = 𝑔𝑔(𝑥𝑥) + 2
= 𝑓𝑓(4 − 𝑥𝑥 2 )
= (4 − 𝑥𝑥 2 ) + 2
= 6 − 𝑥𝑥 2
1.2 (𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥)
Solution:
(𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = 4 − (𝑓𝑓(𝑥𝑥))2
= 𝑔𝑔(𝑥𝑥 + 2)
= 4 − (𝑥𝑥 + 2)2
= 4 − (𝑥𝑥 2 + 4𝑥𝑥 + 4)
= −𝑥𝑥 2 − 4𝑥𝑥
1.3 (𝑔𝑔 ∘ 𝑓𝑓)(−2)
Solution:
(𝑔𝑔 ∘ 𝑓𝑓)(−2) = −(−2)2 − 4(−2) = 4
2.
Given 𝑓𝑓(𝑥𝑥) = √4 − 𝑥𝑥 2 and 𝑔𝑔(𝑥𝑥) = √3 − 𝑥𝑥, find the composition (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥). Then find the
domain of (𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥).
Solution:
(𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥) = 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = �4 − (𝑔𝑔(𝑥𝑥))2
= 𝑓𝑓�√3 − 𝑥𝑥�
2
= �4 − �√3 − 𝑥𝑥�
= �4 − (3 − 𝑥𝑥)
2
remember that �√𝑎𝑎� = 𝑎𝑎 as long as 𝑎𝑎 ≥ 0.
= √1 + 𝑥𝑥
Although √1 + 𝑥𝑥 is defined for all 𝑥𝑥 ≥ −1, we must restrict the domain of (𝑓𝑓 ∘ 𝑔𝑔) to those
values that are also in the domain of 𝑔𝑔. Therefore, the domain of (𝑓𝑓 ∘ 𝑔𝑔) is [−1, 3].
It is possible to take the composition of three or more functions. For instance, the composite
function (𝑓𝑓 ∘ 𝑔𝑔 ∘ ℎ)(𝑥𝑥) is found by first applying ℎ, then 𝑔𝑔, and then 𝑓𝑓 as follows:
(𝑓𝑓 ∘ 𝑔𝑔 ∘ ℎ)(𝑥𝑥) = 𝑓𝑓�𝑔𝑔((ℎ(𝑥𝑥))�
13
Example 1.6
A Composition of Three Functions
If 𝑓𝑓(𝑥𝑥) = 1 − 2𝑥𝑥 2, 𝑔𝑔(𝑥𝑥) = √3 − 𝑥𝑥 , and ℎ(𝑥𝑥) = 𝑥𝑥 2 , find (𝑓𝑓 ∘ 𝑔𝑔 ∘ ℎ)(𝑥𝑥).
Solution:
(𝑓𝑓 ∘ 𝑔𝑔 ∘ ℎ)(𝑥𝑥) = 𝑓𝑓 �𝑔𝑔�ℎ(𝑥𝑥)��
= 𝑓𝑓�𝑔𝑔(𝑥𝑥 2 )�
= 𝑓𝑓 ��3 − 𝑥𝑥 2 �
2
= 1 − 2 ��3 − 𝑥𝑥 2 �
= 1 − 2(3 − 𝑥𝑥 2 )
= 2𝑥𝑥 2 − 5
ACTIVITY 2
1.
Let 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 1; 𝑔𝑔(𝑥𝑥) = 4𝑥𝑥 + 2 and ℎ(𝑥𝑥) = 4𝑥𝑥 2 + 4𝑥𝑥 + 1. Find each of the following:
1.1
1.2
1.3
1.4
1.5
2.
2.2
2.3
2.4
[98]
(𝑓𝑓 − 𝑔𝑔)(−1)
[1]
ℎ
𝑔𝑔
3
� � (1)
� 2�
(𝑔𝑔 − ℎ)(4)
[−63]
𝐹𝐹(10)
[89]
𝐹𝐹(5𝑢𝑢)
[25𝑢𝑢2 − 5𝑢𝑢 − 1]
𝐹𝐹(𝑢𝑢2 )
[𝑢𝑢4 − 𝑢𝑢2 − 1]
[5𝑢𝑢2 − 5𝑢𝑢 − 5]
5𝐹𝐹(𝑢𝑢)
Let 𝐹𝐹(𝑤𝑤) = −𝑤𝑤 2 + 2𝑤𝑤. Find:
3.1
3.2
3.3
3.4
4.
(𝑓𝑓 ⋅ 𝑔𝑔)(3)
Let 𝐹𝐹(𝑢𝑢) = 𝑢𝑢2 − 𝑢𝑢 − 1. Find:
2.1
3.
[15]
(𝑓𝑓 + 𝑔𝑔)(2)
𝐹𝐹(−4)
[−24]
𝐹𝐹(2 − 𝑎𝑎)
[2𝑎𝑎 − 𝑎𝑎2 ]
[−(𝑎𝑎2 + 6𝑎𝑎 + 8)]
𝐹𝐹(4 + 𝑎𝑎)
If 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 , show that:
4.1
[−8]
𝐹𝐹(4)
𝑓𝑓(𝑥𝑥 + 3) − 𝑓𝑓(𝑥𝑥 − 1) =
15
𝑓𝑓(𝑥𝑥)
2
14
4.2
5.
5.2
7.
8.
7
2
𝑓𝑓(2 − ℎ) = 𝑓𝑓(2 + ℎ)
1
𝑥𝑥
𝑎𝑎𝑎𝑎
�.
𝑏𝑏−𝑎𝑎
If 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 − 𝑥𝑥, show that 𝑓𝑓(𝑥𝑥 + 1) = 𝑓𝑓(−𝑥𝑥).
Let 𝑓𝑓(𝑡𝑡) = (𝑡𝑡 − 2)2 and 𝑔𝑔(𝑡𝑡) = 𝑡𝑡 + 1. Find:
8.2
(𝑓𝑓 ∘ 𝑔𝑔)(𝑡𝑡)
[𝑡𝑡 2 − 2𝑡𝑡 + 1]
[𝑡𝑡 2 − 4𝑡𝑡 + 5]
(𝑔𝑔 ∘ 𝑓𝑓)(𝑡𝑡)
If 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 + 2𝑥𝑥 − 1 and 𝑔𝑔(𝑥𝑥) = 2𝑥𝑥 − 3, find each of the following functions.
9.1
9.2
9.3
10.
1
2
𝑓𝑓 � � = 𝑓𝑓 � �
If 𝑓𝑓(𝑥𝑥) = , show that 𝑓𝑓(𝑎𝑎) − 𝑓𝑓(𝑏𝑏) = 𝑓𝑓 �
8.1
9.
= 𝑓𝑓(4)
If 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 − 4𝑥𝑥 + 6, show that:
5.1
6.
𝑓𝑓(𝑥𝑥+3)
𝑓𝑓(𝑥𝑥−1)
(𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥)
[4𝑥𝑥 2 − 8𝑥𝑥 + 2]
(𝑔𝑔 ∘ 𝑔𝑔 ∘ 𝑔𝑔)(𝑥𝑥)
[8𝑥𝑥 − 21]
(𝑓𝑓 ∘ 𝑔𝑔)(𝑥𝑥)
Domain
� √2 − 𝑥𝑥 �
(𝑔𝑔 ∘ 𝑓𝑓)(𝑥𝑥)
��2 − √𝑥𝑥�
[2𝑎𝑎2 + 4𝑎𝑎 − 5]
(𝑔𝑔 ∘ 𝑓𝑓)(𝑎𝑎)
If 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 and 𝑔𝑔(𝑥𝑥) = √2 − 𝑥𝑥, find the following functions and their domains.
10.1
10.2
4
(−∞, 2]
[0, 4]
Domain
10.3
(𝑓𝑓 ∘ 𝑓𝑓)(𝑥𝑥)
4
� √𝑥𝑥 �
(𝑔𝑔 ∘ 𝑔𝑔)(𝑥𝑥)
��2 − √2 − 𝑥𝑥�
[0, ∞]
Domain
10.4
[−2, 2]
Domain
15
1.4
TRANSFORMATION OF FUNCTIONS
The graph of a function can provide valuable insight into the information provided by that
function. However, there is an endless variety of functions and it seems like an insurmountable
task to learn about so many different graphs. Nonetheless, the graphs shown in Figure 1.3
represent the most commonly used functions (usually called parent functions) in algebra.
Familiarity with the basic characteristics of these simple graphs will help you analyze the
shapes of more complicated graphs.
Figure 1.3: Parent functions
1.4.1
Vertical and Horizontal Shifts
If a new function is formed by performing an operation on a given function, then the graph of
the new function is called a transformation of the graph of the original function. Many
functions have graphs that are simple transformations of the parent graphs. Thus, knowing
the graphs of these common functions and how to shift, reflect, and stretch them can help you
sketch a wide variety of simple functions by hand. For instance, you can obtain the graph of
ℎ(𝑥𝑥) = 𝑥𝑥 2 + 2
by shifting the graph of 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 upward two units, as shown in Figure 1.4(A). In function
notation, ℎ and 𝑓𝑓 are related as follows
ℎ(𝑥𝑥) = 𝑥𝑥 2 + 2 = 𝑓𝑓(𝑥𝑥) + 2
Similarly, you can obtain the graph of
𝑔𝑔(𝑥𝑥) = (𝑥𝑥 − 2)2
by shifting the graph of 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 to the right two units, as shown in Figure 1.4(B). In this
case, the functions 𝑔𝑔 and 𝑓𝑓 have the following relationship
𝑔𝑔(𝑥𝑥) = (𝑥𝑥 − 2)2 = 𝑓𝑓(𝑥𝑥 − 2)
16
(𝐵𝐵)
(𝐴𝐴)
Figure 1.4: A shifted quadratic function
We can summarize this discussion about horizontal and vertical shifts as follows.
1.4.1.1 Vertical shifting
To summarise Figure 1.4(A), adding a constant to a function shifts its graph vertically: upward
if the constant is positive and downward if it is negative. That is, if we add or subtract a constant
𝑐𝑐 to 𝑓𝑓(𝑥𝑥), then the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) is transformed as follows
𝑦𝑦 = 𝑓𝑓(𝑥𝑥) + 𝑐𝑐
and
𝑦𝑦 = 𝑓𝑓(𝑥𝑥) − 𝑐𝑐
(𝑐𝑐 > 0)
The 𝑦𝑦-coordinate of each point on the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) + 𝑐𝑐 is 𝑐𝑐 units above the 𝑦𝑦-coordinate
of the corresponding point on the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) . Thus, we obtain the graph of 𝑦𝑦 =
𝑓𝑓(𝑥𝑥) + 𝑐𝑐 simply by shifting the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) upward 𝑐𝑐 units. Similarly, we obtain the
graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) − 𝑐𝑐 by shifting the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) downward 𝑐𝑐 units, as shown below.
1.4.1.2 Horizontal shifting
To summarize our observation from Figure 1.4(B), this is how we obtain the graphs of
𝑦𝑦 = 𝑓𝑓(𝑥𝑥 + 𝑐𝑐)
and
𝑦𝑦 = 𝑓𝑓(𝑥𝑥 − 𝑐𝑐)
(𝑐𝑐 > 0)
knowing the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥). The value of 𝑓𝑓(𝑥𝑥 − 𝑐𝑐) at 𝑥𝑥 is the same as the value of 𝑓𝑓(𝑥𝑥) at
𝑥𝑥 − 𝑐𝑐. Since 𝑥𝑥 − 𝑐𝑐 is 𝑐𝑐 units to the left of 𝑥𝑥, it follows that the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥 − 𝑐𝑐) is just the
graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) shifted to the right 𝑐𝑐 units. Similar reasoning shows that the graph of 𝑦𝑦 =
𝑓𝑓(𝑥𝑥 + 𝑐𝑐) is the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) shifted to the left 𝑐𝑐 units, as shown below.
17
Example 1.7
1.
Combining Horizontal and Vertical Shifts
Sketch a graph of 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 − 3 + 4.
Solution:
We start with the parent function 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 and shift it to the right 3 units (since this is a
change on the inside of the function) to obtain the graph of 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 − 3. We then shift the
resulting graph 4 units upward to obtain the graph of 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 − 3 + 4, as shown below.
2.
Sketch a graph of 𝑦𝑦 = |𝑥𝑥 + 1| − 3.
Solution:
We again start with the parent function 𝑦𝑦 = |𝑥𝑥| and shift it 1 unit to the left to obtain the graph
of 𝑦𝑦 = |𝑥𝑥 + 1|. We then shift the resulting graph 3 units downward to obtain the graph of 𝑦𝑦 =
|𝑥𝑥 + 1| − 3, as shown in the Figure below.
18
To find the 𝑥𝑥 intercepts in an absolute value equation such as |𝑥𝑥 + 1| = 3, remember that the
absolute value will be equal to 3 if the quantity inside the absolute value is 3 or −3. This leads
to two different equations (|𝑥𝑥 + 1| = 3 and |𝑥𝑥 + 1| = −3) that we can solve independently.
1.4.2
Graphing Functions Using Reflections about the Axes
The second common type of transformation is a reflection. For example, if you consider the 𝑥𝑥axis to be a mirror, the graph of
ℎ(𝑥𝑥) = −𝑥𝑥 2
is the mirror image (or reflection) of the graph of
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2
as shown in the Figure below
We can summarise the three common reflections as follows:
(i)
The graph of 𝑦𝑦 = −𝑓𝑓(𝑥𝑥) can be obtained from the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) by changing the
sign of each 𝑦𝑦 coordinate. This has the effect of moving every point to the opposite side
of the 𝑥𝑥 axis. So, the graph of 𝑦𝑦 = −𝑓𝑓(𝑥𝑥) is the reflection through the 𝒙𝒙 axis of the
(ii)
graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) [Figure 1.5(a)].
The graph of 𝑦𝑦 = 𝑓𝑓(−𝑥𝑥) can be obtained from the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) by changing the
sign of each 𝑥𝑥 coordinate. This has the effect of moving every point to the opposite side
of the 𝑦𝑦 axis. So, the graph of 𝑦𝑦 = 𝑓𝑓(−𝑥𝑥) is the reflection through the 𝒚𝒚 axis of the
(iii)
graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) [Figure 1.5(b)].
The graph of 𝑦𝑦 = −𝑓𝑓(−𝑥𝑥) can be obtained from the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) by changing the
sign of each 𝑥𝑥 and 𝑦𝑦 coordinate. So, the graph of 𝑦𝑦 = −𝑓𝑓(−𝑥𝑥) is the reflection through
the origin of the graph of 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) [Figure 1.5(c)].
19
Figure 1.5: Refection of a function
Example 1.8
Reflecting Graphs
Sketch the graph of each of the following functions.
1.
2.
3.
4.
𝑦𝑦 = −√𝑥𝑥
𝑦𝑦 = √−𝑥𝑥
𝑦𝑦 = −√−𝑥𝑥
𝑦𝑦 = −√2 − 𝑥𝑥 + 4
Solution:
Using the parent function 𝑦𝑦 = √𝑥𝑥 as a reference, a good way to remember which sides the
transformation arms will go is to look at the signs. For instance, in 𝑦𝑦 = −√𝑥𝑥, 𝑥𝑥 is positive and
𝑦𝑦 negative, and therefore the arm will point towards the third quadrant. Also, notice that the
arms ‘in most cases’ move away from the 𝑦𝑦-axis.
2.59
1.4.3
𝑦𝑦 = −√𝑥𝑥
𝑦𝑦 = √−𝑥𝑥
𝑦𝑦 = −√−𝑥𝑥
Even and Odd Functions
(2,4)
−14
𝑦𝑦 = −√2 − 𝑥𝑥 + 4
Certain transformations leave the graphs of some functions unchanged, as discussed below.
1.4.3.1 Function symmetry
As shown in the figure below, some functions exhibit symmetry (with respect to one of the
coordinate axes or with respect to the origin) so that reflections result in the original graph.
20
For example, reflecting the graph of 𝑦𝑦 = 𝑥𝑥 2 through the 𝑦𝑦-axis does not change the graph.
Functions whose graphs are symmetric about the 𝑦𝑦-axis are called even functions. Similarly,
3
reflecting the graph of 𝑦𝑦 = √𝑥𝑥 through the origin does not change the graph. Functions with
this property are called odd functions.
1.4.3.2 Determining even and odd functions algebraically
From the figure above, we can show that 𝑓𝑓(−𝑥𝑥) = 𝑓𝑓(𝑥𝑥) for even functions and 𝑓𝑓(−𝑥𝑥) = −𝑓𝑓(𝑥𝑥)
for odd functions. More formally, we have the following definitions:


If 𝑓𝑓(−𝑥𝑥) = 𝑓𝑓(𝑥𝑥) for all 𝑥𝑥 in the domain of 𝑓𝑓, then 𝑓𝑓 is an even function.
If 𝑓𝑓(−𝑥𝑥) = −𝑓𝑓(𝑥𝑥) for all 𝑥𝑥 in the domain of 𝑓𝑓, then 𝑓𝑓 is an odd function.
Note: a function can also be neither even nor odd if it does not exhibit either symmetry.
Example 1.9
Even and Odd Functions
Determine whether the functions are even, odd, or neither.
1.
𝑔𝑔(𝑥𝑥) = 𝑥𝑥 4 + 1
Solution:
𝑔𝑔(−𝑥𝑥) = (−𝑥𝑥)4 + 1
= 𝑥𝑥 4 + 1
= 𝑔𝑔(𝑥𝑥)
This function is even because 𝑔𝑔(−𝑥𝑥) = 𝑔𝑔(𝑥𝑥).
2.
ℎ(𝑥𝑥) = 𝑥𝑥 5 − 𝑥𝑥
Solution:
ℎ(−𝑥𝑥) = (−𝑥𝑥)5 − (−𝑥𝑥)
= −𝑥𝑥 5 + 𝑥𝑥
= −(𝑥𝑥 5 − 𝑥𝑥) = −ℎ(𝑥𝑥)
This function is odd because ℎ(−𝑥𝑥) = −ℎ(𝑥𝑥).
21
3.
𝑓𝑓(𝑥𝑥) =
𝑥𝑥
1−𝑥𝑥
Solution:
−𝑥𝑥
1 − (−𝑥𝑥)
−𝑥𝑥
=
1 + 𝑥𝑥
𝑓𝑓(−𝑥𝑥) =
≠ 𝑓𝑓(𝑥𝑥) ≠ −𝑓𝑓(𝑥𝑥)
The function is neither even nor odd.
4.
1
3
𝑘𝑘(𝑥𝑥) = + √𝑥𝑥
𝑥𝑥
Solution:
1
3
+ √−𝑥𝑥
−𝑥𝑥
1 3
1 3
= − − √𝑥𝑥 = − � + √𝑥𝑥 � = −𝑘𝑘(𝑥𝑥)
𝑥𝑥
𝑥𝑥
𝑘𝑘(−𝑥𝑥) =
This function is odd because 𝑘𝑘(−𝑥𝑥) = −𝑘𝑘(𝑥𝑥).
1.4.4
Graphing Functions using Stretches and Compressions
Adding a constant to the inputs or outputs of a function changed the position of a graph with
respect to the axes, but it did not affect the shape of a graph. We now explore the effects of
multiplying the inside (input values) or the outside (output values) by some quantity.
1.4.4.1 Vertical stretches and compressions
When we multiply a function by a positive constant, we get a function whose graph is stretched
or compressed vertically in relation to the graph of the original function. That is, given a
function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), a new function 𝑦𝑦 = 𝑎𝑎𝑎𝑎(𝑥𝑥), where is 𝑎𝑎 constant, is a vertical stretch or
vertical compression of the function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), as follows:


If 𝑎𝑎 > 1, then the graph will be vertically stretched.
If 0 < 𝑎𝑎 < 1, then the graph will be vertically compressed.
22
1.4.4.2 Horizontal stretches and compressions
Now we consider changes to the inside of a function. When we multiply a function’s input by
a positive constant, we get a function whose graph is stretched or compressed horizontally in
relation to the graph of the original function. That is, given a function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), a new function
𝑦𝑦 = 𝑓𝑓(𝑏𝑏𝑏𝑏), where 𝑏𝑏 is a constant, is a horizontal stretch or horizontal compression of the
function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), as follows:


If 𝑏𝑏 > 1, then the graph will be horizontally compressed by 1⁄𝑏𝑏.
If 0 < 𝑏𝑏 < 1, then the graph will be horizontally stretched by 1⁄𝑏𝑏.
Example 1.10
Combining All Transformations
In the following, identify the parent function, describe the transformation, and sketch the graph
of each function.
1.
𝑦𝑦 = √4𝑥𝑥 + 2
Solution:
Parent graph:
Square root function
Transformation:
Horizontal shift, 0.5 units to the left
𝑦𝑦
1.41
𝑦𝑦 = √4𝑥𝑥 + 2
𝑥𝑥
−0.5
23
2.
𝑦𝑦 = 4 − 2(𝑥𝑥 + 3)2
Solution:
Rewrite the function
𝑦𝑦 = −2(𝑥𝑥 + 3)2 + 4
Parent function:
Quadratic function
Transformation:
Reflection across the 𝑥𝑥-axis and a vertical stretch by a factor of 2.
Horizontal shift to the left by 3 units
Vertical shift, 4 units up
(−3, 4)
−4.41
3.
𝑦𝑦
−1.59
𝑥𝑥
−14
𝑦𝑦 = √3𝑥𝑥 − 6 + 3
Solution:
Separate a stretch/shrink from a shift:
𝑦𝑦 = �3(𝑥𝑥 − 2) + 3
Parent function:
Square root function
Transformation:
Horizontal shrink/compression by factor 1⁄3
Horizontal shift, 2 units to the right
Vertical shift, 3 units up
𝑦𝑦
(2, 3)
24
𝑥𝑥
ACTIVITY 3
1.
Describe the transformation and sketch the graph of each of the following functions.
1.1
1.2
1.3
1.4
1.5
1.5
𝑦𝑦 = (𝑥𝑥 + 3)2 − 5
𝑦𝑦 = 4 − √3 − 𝑥𝑥
𝑦𝑦 = 2|𝑥𝑥 + 3| − 1
𝑦𝑦 = √2𝑥𝑥 + 5 − 1
𝑦𝑦 = |4 − 3𝑥𝑥| + 2
INVERSE FUNCTIONS
Let 𝑓𝑓 and 𝑓𝑓 −1 be inverse functions defined on all real numbers. Let us then say that we have
the following situation:
𝑓𝑓(2) = 5
𝑓𝑓 −1 (3) = 7
𝑓𝑓 −1 �𝑓𝑓(2)� = 2
𝑓𝑓�𝑓𝑓 −1 (3)� = 3
𝑓𝑓 −1 (5) = 2
𝑓𝑓(7) = 3
Notice that in both these situations, the function and the inverse cancel each other. That is, if
we start with 𝑥𝑥, apply 𝑓𝑓, and then apply 𝑓𝑓 −1 , we arrive back at 𝑥𝑥, where we started. Similarly,
𝑓𝑓 undoes what 𝑓𝑓 −1 does. In general, any function that reverses the effect of 𝑓𝑓 in this way must
be the inverse of 𝑓𝑓. We express these observations precisely as follows:
𝑓𝑓 −1 �𝑓𝑓(𝑥𝑥)� = 𝑥𝑥
𝑓𝑓�𝑓𝑓 −1 (𝑥𝑥)� = 𝑥𝑥
This property is called the inverse function property or “cancellation property”. The domain of
𝑓𝑓 must be equal to the range of 𝑓𝑓 −1, and the range of 𝑓𝑓 must be equal to the domain of 𝑓𝑓 −1.
Generally, given a function 𝑓𝑓(𝑥𝑥), we represent its inverse as 𝑓𝑓 −1 (𝑥𝑥), read as “𝑓𝑓 inverse of 𝑥𝑥.”
The raised −1 is part of the notation and not an exponent; it does not imply a power of −1.
That is, 𝑓𝑓 −1 (𝑥𝑥) does not mean
Example 1.11
1
𝑓𝑓(𝑥𝑥)
since
1
𝑓𝑓(𝑥𝑥)
is the reciprocal of 𝑓𝑓 and not the inverse.
Verifying Inverse Functions
1
2
Verify that 𝑓𝑓 −1 (𝑥𝑥) = 𝑥𝑥 − 2 is the inverse of 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 + 4.
Solution:
Show that (𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑥𝑥 and (𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑥𝑥.
(𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑓𝑓(𝑓𝑓 −1 (𝑥𝑥))
25
(𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑓𝑓 −1 (𝑓𝑓(𝑥𝑥))
= 𝑓𝑓 −1 (2𝑥𝑥 + 4)
= 𝑓𝑓(0.5𝑥𝑥 − 2) + 4
= 2(0.5𝑥𝑥 − 2) + 4
= 0.5(2𝑥𝑥 + 4) − 2
= 𝑥𝑥 − 4 + 4
= 𝑥𝑥 + 2 − 2
= 𝑥𝑥
= 𝑥𝑥
As mentioned above, inverse “undoes” or reverses what the function has done. However, not
all functions have inverse functions; those that do are called one-to-one.
1.5.1
One-to-One Functions
One-to-one functions are important because they are precisely the functions that possess
inverse functions. Consider two functions 𝑓𝑓 and 𝑔𝑔 whose arrow diagrams are shown in the
Figure below.
Note that 𝑓𝑓 never takes on the same value twice (any two numbers in A have different images),
whereas 𝑔𝑔 does take on the same value twice (both 2 and 3 have the same image, 4). In
symbols, 𝑔𝑔(2) = 𝑔𝑔(3) but 𝑓𝑓(𝑥𝑥1 ) ≠ 𝑓𝑓(𝑥𝑥2 ) whenever 𝑥𝑥1 ≠ 𝑥𝑥2 . Functions that have this latter
property are called one-to-one.
An equivalent way of writing the condition for a one-to-one function is this:
If 𝑓𝑓(𝑥𝑥1 ) = 𝑓𝑓(𝑥𝑥2 ), then 𝑥𝑥1 = 𝑥𝑥2
If a horizontal line intersects the graph of 𝑓𝑓 at more than one point, then we see from the
Figure below that there are numbers 𝑥𝑥1 ≠ 𝑥𝑥2 such that 𝑓𝑓(𝑥𝑥1 ) = 𝑓𝑓(𝑥𝑥2 ). This means that 𝑓𝑓 is not
one-to-one.
This geometric method for determining whether a function is one-to-one is commonly called a
horizontal line test. Formally, a function is one-to-one if and only if no horizontal line
intersects its graph more than once.
26
1.5.2
Finding the Inverse of a Function
Let us now examine how we compute the inverse of a function. Given a function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), the
first coordinates of points on the graph are represented by 𝑥𝑥, and the second coordinates are
represented by 𝑦𝑦. Finding the inverse by reversing the order of the coordinates would then
correspond to switching the variables 𝑥𝑥 and 𝑦𝑦. This leads us to the following procedure, which
can be applied whenever it is possible to solve 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) for 𝑥𝑥 in terms of 𝑦𝑦.
If the function is written with function notation, like 𝑓𝑓(𝑥𝑥), replace the function
Step 1:
symbol with the letter 𝑦𝑦.
Interchange 𝑥𝑥 (independent variable) and 𝑦𝑦 (dependent variable).
Step 2:
Solve the resulting equation for 𝑦𝑦.
Step 3:
Replace 𝑦𝑦 by 𝑓𝑓 −1 (𝑥𝑥).
Step 4:
Verify that 𝑓𝑓 and 𝑓𝑓 −1 are inverse functions of each other.
Step 5:
Example 1.12
1.
Finding the Inverse of a Function
3
2
Find the inverse of 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 + 2 and verify the result.
Solution:
We will use the strategy given above to find 𝑓𝑓 −1. We will then verify the result by showing that
(𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑥𝑥 and (𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑥𝑥.
Step 1: Replace 𝑓𝑓(𝑥𝑥) with 𝑦𝑦.
3
2
𝑦𝑦 = 𝑥𝑥 + 2
Step 2: Interchange 𝑥𝑥 and 𝑦𝑦.
3
𝑥𝑥 = 𝑦𝑦 + 2
2
Step 3: Solve the equation for 𝑦𝑦.
2𝑥𝑥 = 3𝑦𝑦 + 4
𝑦𝑦 =
Multiply both sides by 2
2𝑥𝑥−4
3
Step 4: Replace 𝑦𝑦 by 𝑓𝑓 −1 (𝑥𝑥).
𝑓𝑓 −1 (𝑥𝑥) =
2𝑥𝑥−4
3
3
2
To verify that 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 + 2 and 𝑓𝑓 −1 (𝑥𝑥) =
2𝑥𝑥−4
3
are inverse functions of each other, we must
show that (𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑥𝑥 and (𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑥𝑥. This is left as an exercise.
27
2.
Find 𝑓𝑓 −1 for 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 − 1 and verify the result.
Solution:
𝑦𝑦 = √𝑥𝑥 − 1
𝑥𝑥 = �𝑦𝑦 − 1
𝑥𝑥 2 = 𝑦𝑦 − 1
𝑦𝑦 = 𝑥𝑥 2 + 1
𝑓𝑓 −1 (𝑥𝑥) = 𝑥𝑥 2 + 1
We use the inverse function property to verify our answer:
(𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑓𝑓(𝑓𝑓 −1 (𝑥𝑥))
(𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥) = 𝑓𝑓 −1 (𝑓𝑓(𝑥𝑥))
= 𝑓𝑓(𝑥𝑥 2 + 1)
= 𝑓𝑓 −1 �√𝑥𝑥 − 1�
2
= �(𝑥𝑥 2 + 1) − 1
= �√𝑥𝑥 − 1� + 1
= √𝑥𝑥 2
= 𝑥𝑥 − 1 + 1
= 𝑥𝑥
= 𝑥𝑥
Therefore, 𝑓𝑓(𝑥𝑥) and 𝑓𝑓 −1 (𝑥𝑥) are inverse functions since (𝑓𝑓 ∘ 𝑓𝑓 −1 )(𝑥𝑥) = 𝑥𝑥 = (𝑓𝑓 −1 ∘ 𝑓𝑓)(𝑥𝑥).
1.5.3
Graphing the Inverse of a Function
The principle of interchanging 𝑥𝑥 and 𝑦𝑦 to find the inverse function also gives us a method for
obtaining the graph of 𝑓𝑓 −1 from the graph of 𝑓𝑓. If 𝑓𝑓(𝑎𝑎) = 𝑏𝑏, then 𝑓𝑓 −1 (𝑏𝑏) = 𝑎𝑎. Thus, the point
(𝑎𝑎, 𝑏𝑏) is on the graph of 𝑓𝑓 if and only if the point (𝑏𝑏, 𝑎𝑎) is on the graph of 𝑓𝑓 −1 . But we get the
point (𝑏𝑏, 𝑎𝑎) from the point (𝑎𝑎, 𝑏𝑏) by reflecting in the line 𝑦𝑦 = 𝑥𝑥 (see Figure 1.6(A). Therefore,
as Figure 1.6(B) illustrates, the following is true:
The graph of 𝑓𝑓 −1 is obtained by reflecting the graph of 𝑓𝑓 in the line 𝑦𝑦 = 𝑥𝑥.
(𝐵𝐵)
(𝐴𝐴)
Figure 1.6: Plotting inverse functions
28
Example 1.13
1.
Graphing the Inverse of a Function
Sketch the graph of 𝑦𝑦 = √𝑥𝑥 − 2.
Solution:
Using the transformations, we sketch the graph of 𝑦𝑦 = √𝑥𝑥 − 2 by plotting the parent function
𝑦𝑦 = √𝑥𝑥 and shifting it to the right 2 units, as discussed before.
2.
Use the graph of 𝑓𝑓 to sketch the graph of 𝑓𝑓 −1.
Solution:
The graph of 𝑓𝑓 −1 is obtained from the graph of 𝑓𝑓 (question 1) by reflecting it in the line 𝑦𝑦 = 𝑥𝑥,
as shown in the Figure below.
ACTIVITY 4
1.
Find the inverse of the following functions:
1.1
1.2
1.3
1.4
1.5
ℎ(𝑥𝑥) =
𝑓𝑓(𝑥𝑥) =
5
𝑥𝑥−2
5
�𝑥𝑥 + 2�
2𝑥𝑥−1
𝑥𝑥+3
3𝑥𝑥+1
� 2−𝑥𝑥 �
3
ℎ(𝑥𝑥) = 𝑥𝑥 3 + 3
� √𝑥𝑥 − 3�
[3 − 𝑥𝑥 2 ]
𝑓𝑓(𝑥𝑥) = √3 − 𝑥𝑥
𝑓𝑓(𝑥𝑥) =
2𝑥𝑥+3
𝑥𝑥−1
𝑥𝑥+3
�𝑥𝑥−2�
1.6
LINEAR FUNCTIONS
1.6.1
The Slope-Intercept Form of a Line
The simplest mathematical model for relating two variables is the linear equation in two
variables 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑏𝑏. The equation (in which the variable occurs to the first power only) is
29
called linear because its graph is a line. By letting 𝑥𝑥 = 0, you can see that the line crosses the
𝑦𝑦-axis at 𝑦𝑦 = 𝑏𝑏, as shown in Figure 1.7(A). In other words, the 𝑦𝑦-intercept is (0, 𝑏𝑏). The
steepness or slope of the line is 𝑚𝑚. The slope of a nonvertical line is the number of units the
line rises (or falls) vertically for each unit of horizontal change from left to right, as shown in
Figure 1.7(A) and Figure 1.7(B).
(𝐴𝐴)
(𝐵𝐵)
Figure 1.7: Increasing and decreasing functions
A linear equation that is written in the form 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑏𝑏 is said to be written in slope-intercept
form because it identifies the slope and the 𝑦𝑦-intercept. The slope determines if the function
is an increasing linear function, a decreasing linear function, or a constant function as follows.



𝑓𝑓(𝑥𝑥) = 𝑚𝑚𝑚𝑚 + 𝑏𝑏 is an increasing function if 𝑚𝑚 > 0.
𝑓𝑓(𝑥𝑥) = 𝑚𝑚𝑚𝑚 + 𝑏𝑏 is a decreasing function if 𝑚𝑚 < 0.
𝑓𝑓(𝑥𝑥) = 𝑚𝑚𝑚𝑚 + 𝑏𝑏 is a constant function if 𝑚𝑚 = 0.
1.6.2
Finding the Slope of a Line
If a line passes through two distinct points 𝑃𝑃1 = (𝑥𝑥1 , 𝑦𝑦1 ) and 𝑃𝑃2 = (𝑥𝑥2 , 𝑦𝑦2 ), we can calculate its
slope 𝑚𝑚, as follows:
𝑚𝑚 =
rise
vertical change
∆𝑦𝑦 𝑦𝑦2 − 𝑦𝑦1
=
=
=
run horizontal change ∆𝑥𝑥 𝑥𝑥2 − 𝑥𝑥1
30
1.6.3
The Point-Slope Form of a Linear Equation
Suppose a line has slope 𝑚𝑚 and passes through the point (𝑥𝑥1 , 𝑦𝑦1 ). If (𝑥𝑥, 𝑦𝑦) is any other point
on the line (see the Figure below),
then:
𝑦𝑦−𝑦𝑦1
𝑥𝑥−𝑥𝑥1
= 𝑚𝑚
Multiplying both sides by (𝑥𝑥 − 𝑥𝑥1 ) yields:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
This is called the point-slope form of an equation of a line. Keep in mind that the slope-intercept
form and the point-slope form can be used to describe the same function. We can move from
one form to another using basic algebra.
Example 1.14
Using the Point-Slope Form
Find an equation for the line that has slope
final answer in slope-intercept form.
2
3
and passes through the point (−2, 1). Write the
Solution:
2
3
Since 𝑚𝑚 = , 𝑥𝑥1 = −2 and 𝑦𝑦1 = 1, then:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
2
𝑦𝑦 − 1 = (𝑥𝑥 − (−2))
3
3(𝑦𝑦 − 1) = 2(𝑥𝑥 + 2)
1.6.4
7
2
𝑦𝑦 = 𝑥𝑥 +
3
3
Parallel and Perpendicular Lines
From geometry, we know that two vertical lines are parallel to each other and that a horizontal
line and a vertical line are perpendicular to each other. To tell when two lines are parallel or
perpendicular, the theorem, which is stated without proof, provides a convenient procedure.
Given two nonvertical lines 𝐿𝐿1 and 𝐿𝐿2 with slopes 𝑚𝑚1 and 𝑚𝑚2 , respectively, then:
31
𝐿𝐿1 ‖ 𝐿𝐿2
𝑚𝑚1 = 𝑚𝑚2
if and only if
𝐿𝐿1 ⊥ 𝐿𝐿2
𝑚𝑚1 𝑚𝑚2 = −1
if and only if
The symbols ‖ and ⊥ mean, respectively, “is parallel to” and “is perpendicular to”. In the case
of perpendicularity, the condition 𝑚𝑚1 𝑚𝑚2 = −1 also can be written as:
𝑚𝑚1 = −
Example 1.15
1
𝑚𝑚2
or
𝑚𝑚2 = −
1
𝑚𝑚1
Parallel and Perpendicular Lines
Given the line 𝐿𝐿: 3𝑥𝑥 − 2𝑦𝑦 = 5 and the point 𝑃𝑃 = (−3, 5), find an equation of a line (in slope-
intercept form) through 𝑃𝑃 that is:
1.
Parallel to 𝐿𝐿
Solution:
First, we need to find the slope of 𝐿𝐿 by writing 3𝑥𝑥 − 2𝑦𝑦 = 5 in the equivalent slope-intercept
form 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐:
3𝑥𝑥 − 2𝑦𝑦 = 5
−2𝑦𝑦 = −3𝑥𝑥 + 5
3
5
𝑦𝑦 = 𝑥𝑥 −
2
2
3
2
3
2
So, the slope of 𝐿𝐿 is . The slope of a line parallel to 𝐿𝐿 is the same , hence:
2.
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
3
𝑦𝑦 − 5 = (𝑥𝑥 + 3)
2
19
3
𝑦𝑦 = 𝑥𝑥 +
2
2
Perpendicular to 𝐿𝐿
Solution:
3
2
2
3
Since the slope of 𝐿𝐿 is , the line perpendicular to 𝐿𝐿 is − . Hence:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
2
𝑦𝑦 − 5 = − (𝑥𝑥 + 3)
3
2
𝑦𝑦 = − 𝑥𝑥 + 3
3
32
1.6.5
Modeling Real-World Problems with Linear Functions
In the real world, problems are not always explicitly stated in terms of a function or represented
with a graph. Fortunately, we can analyze the problem by first representing it as a linear
function and then interpreting the components of the function. As long as we know or can
figure out the initial value and the rate of change (slope) of a linear function, we can solve
many different kinds of real-world problems.
Example 1.16
1.
Modeling with Linear Functions
Water is being pumped into a swimming pool at the rate of 20 liters per min. Initially, the
pool contains 760 liters of water.
1.1
Find a linear function 𝑉𝑉 that models the volume of water in the pool at any time 𝑡𝑡.
Solution:
We need to find a linear function
𝑉𝑉(𝑡𝑡) = 𝑚𝑚𝑚𝑚 + 𝑏𝑏
that models the volume 𝑉𝑉(𝑡𝑡) of water in the pool after 𝑡𝑡 minutes. The rate of change of volume
is 20 liters per min, so 𝑚𝑚 = 20. Since the pool contains 760 liters to begin with (at 𝑡𝑡 = 0), we
have 𝑉𝑉(0) = (0 . 𝑡𝑡) + 𝑏𝑏 = 760, so 𝑏𝑏 = 760. Now that we know 𝑚𝑚 and 𝑏𝑏, we get the model
1.2
𝑉𝑉(𝑡𝑡) = 20𝑡𝑡 + 760
If the pool has a capacity of 2200 liters, how long does it take to completely fill the pool?
Solution:
We want to find the time 𝑡𝑡 at which 𝑉𝑉(𝑡𝑡) = 2200. So, we need to solve the equation
2200 = 20𝑡𝑡 + 760
Solving for 𝑡𝑡, we get 𝑡𝑡 = 72. Therefore, it takes 72 min to fill the pool.
2.
Working as an insurance salesperson, Themba earns a base salary plus a commission
on each new policy. Thus, Themba’s weekly income, 𝐼𝐼, depends on the number of new
policies, 𝑛𝑛, he sells during the week. Last week he sold 3 new policies and earned R760
for the week. The week before, he sold 5 new policies and earned R920. Find an equation
for 𝐼𝐼(𝑛𝑛) and interpret the meaning of the components of the equation.
Solution:
The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding
the rate of change.
33
920 − 760
5−3
𝑅𝑅 160
𝑚𝑚 =
2 policies
𝑚𝑚 =
𝑚𝑚 = 𝑅𝑅 80 per policy
Keeping track of units can help us interpret this quantity. Income increased by R160 when the
number of policies increased by 2, so the rate of change is R80 per policy. Therefore, Themba
earns a commission of R80 for each policy sold during the week. We can then solve for the
initial value as follows.
𝐼𝐼(𝑛𝑛) = 80𝑛𝑛 + 𝑏𝑏
760 = 80(3) + 𝑏𝑏
𝑏𝑏 = 520
when 𝑛𝑛 = 3, 𝐼𝐼(3) = 760
The value of 𝑏𝑏 is the starting value for the function and represents Themba’s income when
𝑛𝑛 = 0 or when no new policies are sold. We can interpret this as Themba’s base salary for the
week, which does not depend upon the number of policies sold. Therefore:
𝐼𝐼(𝑛𝑛) = 80𝑛𝑛 + 520
Our final interpretation is that Themba’s base salary is R520 per week and he earns an
additional R80 commission for each policy sold.
ACTIVITY 5
1.
2.
3.
4.
5.
6.
1
Find the equation of the line going through (1, 3) and (3, 6).
�𝑦𝑦 = 2 (3𝑥𝑥 + 3)�
the line 1 = 𝑦𝑦 − 3𝑥𝑥
�𝑦𝑦 = − 3 𝑥𝑥 + 1�
3𝑥𝑥 + 4𝑦𝑦 = 2.
�𝑦𝑦 = 4 (−3𝑥𝑥 + 13)�
𝑘𝑘.
[𝑘𝑘 = 3]
Find the equation of a line that passes through the point (3, 0) and is perpendicular to
1
Find an equation of the line 𝐿𝐿 through (−1, 4) and parallel to the line 𝑀𝑀 with the equation
1
If the point (3, 𝑘𝑘) lies on the line with slope 𝑚𝑚 = −2 passing through the point (2, 5), find
If the point (2, 𝑘𝑘) lies on the line with slope 𝑚𝑚 = 3 passing through the point (1,6), find 𝑘𝑘.
[𝑘𝑘 = 9]
Determine whether the following pairs of lines are parallel, perpendicular, or neither:
6.1
𝑦𝑦 = 3𝑥𝑥 + 2 and 𝑦𝑦 = 3𝑥𝑥 − 2
[Parallel]
34
6.2
6.3
6.4
[Perpendicular]
6𝑥𝑥 + 3𝑦𝑦 = 1 and 4𝑥𝑥 + 2𝑦𝑦 = 3
[Parallel]
7.1
Have slope 1
7.2
Have 𝑦𝑦 intercept 2
[𝑘𝑘 = 3]
�𝑘𝑘 = − 2�
Be parallel to the line 2𝑥𝑥 − 4𝑦𝑦 = 1
�𝑘𝑘 = 2�
For what values of 𝑘𝑘 will the line 𝑘𝑘𝑘𝑘 − 3𝑦𝑦 = 4𝑘𝑘 have the following properties:
7.4
7.5
Pass through the point (2,4)
Be perpendicular to the line 𝑥𝑥 − 6𝑦𝑦 = 2
3
[𝑘𝑘 = −6]
3
[𝑘𝑘 = −18]
The sales per share for Starbucks Corporation were R6.97 in 2001 and R8.47 in 2002.
8.1
8.2
9.
3𝑥𝑥 − 2𝑦𝑦 = 5 and 2𝑥𝑥 + 3𝑦𝑦 = 4
[Perpendicular]
7.3
8.
[Neither]
𝑥𝑥 = 3 and 𝑦𝑦 = −4
6.5
7.
𝑦𝑦 = 2𝑥𝑥 − 4 and 𝑦𝑦 = 3𝑥𝑥 + 5
Write a linear equation that gives the sales per share in terms of the year.
Then predict the sales per share for 2003.
[𝑦𝑦 = 1.5𝑡𝑡 + 5.47]
[𝑦𝑦(3) = 𝑅𝑅9.97]
A hot dog vendor pays R25 per day to rent a pushcart and R1.25 for the ingredients in
one hot dog.
9.2
Find the cost of selling 𝑥𝑥 hot dogs in 1 day.
9.3
If the daily cost is R355, how many hot dogs were sold that day?
9.1
10.
What is the cost of selling 200 hot dogs in 1 day?
[𝐶𝐶(𝑥𝑥) = 1.25𝑥𝑥 + 25]
[𝐶𝐶(200) = 𝑅𝑅275]
[264]
UJ purchased exercise equipment worth R12,000 for the campus fitness center. The
equipment has a useful life of 8 years. The salvage value at the end of 8 years is R2000.
Write a linear equation that describes the book value (the difference between the original
value and the total amount of depreciation accumulated to date) of the equipment each
[𝑉𝑉 = −1250𝑡𝑡 + 12000]
year.
11.
Marcus currently has 200 songs in his music collection. Every month, he adds 15 new
songs.
11.1
Write a formula for the number of songs, in his collection as a function of time,
the number of months.
11.2
How many songs will he own in a year?
35
[𝑁𝑁(𝑡𝑡) = 15𝑡𝑡 + 200]
[𝑁𝑁(12) = 380]
1.7
POLYNOMIAL AND RATIONAL FUNCTIONS
Functions defined by polynomial expressions are called polynomial functions. Unlike the linear
functions whose graphs do not change direction, the graphs of polynomial functions can have
one or many peaks and valleys (or turning points). This property makes them suitable models
for many real-world situations. For example, a factory owner notices that if she increases the
number of workers, productivity increases, but if there are too many workers, productivity
begins to decrease. This situation is modeled by a polynomial function of degree 2 (a quadratic
function). The growth of many animal species follows a predictable pattern, beginning with a
period of rapid growth, followed by a period of slow growth and then a final growth spurt. This
variability in growth is modeled by a polynomial of degree 3 (a cubic function).
1.7.1
Quadratic Functions
To be completed.
1.7.2
Polynomial Functions
To be completed.
1.7.3
Dividing Polynomials
To be completed.
1.7.4
Rational Functions
To be completed.
ACTIVITY 6
To be completed.
36
1.8
CONIC SECTIONS
To be completed.
ACTIVITY 1
To be completed.
37