Grade 12 Mathematics
General Stream
Academic Year 2021/2022 – Term 1
Unit-1
Lesson no. 1.1
Lesson name: Functions
Date: 30th August to 3rd
September
Learning objectives
• Describe subsets of real
numbers
• Identify and evaluate functions
and state their domains.
Keywords
• Set-builder notation
• Interval notation
• Function
• Function notation
• Independent variable
• Dependent variable
• Implied domain
• Piecewise-defined
function
• Relevant domain
Introduction
Subsets of Real Numbers
Any number you can think of is a Real Number, numbers are
used represent any quantity, the set of real numbers ℝ includes
the following subsets of numbers.
Subsets of Real Numbers
Describe Subsets of Real Numbers
• Real numbers can be described using set-builder notation by defining
the properties of the numbers in the set.
Describe Subsets of Real Numbers
Example 1: Use Set Builder Notation
Describe the set of numbers using set-builder notation.
a. {8, 9, 10, 11, …}
• Answer:
The numbers in the set includes
whole numbers (N).
The numbers are greater than or
equal to 8.
The set will be written as:
{x | x ≥ 8, x ∈ N}
b. x < 7
• Answer:
Page: 4
c. all multiples of three
• Answer:
The numbers can be from any
subset from the real number,
since there is no information
about the set.
The set includes all integers (ℤ)
that are multiples of three.
The set will be written as:
{x | x = 3n, n ∈ ℤ}
{x | x < 7, x ∈ ℝ}
Assume any integer is n.
Guided Practice: Use Set Builder Notation
Describe the set of numbers using set-builder notation.
1A. {1, 2, 3, 4, 5, ….}
• Answer:
The numbers in the set includes
whole numbers (W).
The numbers are greater than or
equal to 1.
The set will be written as:
{x | x ≥ 1, x ∈ ℕ}
1B. x ≤
• Answer: −3
Page: 4
1C. all multiples of 𝜋.
• Answer:
The numbers can be from any
subset from the real number,
since there is no information
about the set.
The set includes all integers (ℤ)
that are multiples of 𝜋.
The set will be written as:
{x | x = n𝜋, n ∈ ℤ}
{x | x ≤ −3, x ∈ ℝ}
Assume any integer is n.
Interval Notation
• Interval notation uses inequalities (≥, ≤, <, >) to describe subsets of real numbers using
braces [a,b], (a,b), [a,b), (a,b].
a is the lower number and b is the upper number.
• The symbols [ or ] means that number is included (≥, ≤) in the interval.
• The symbols ( or ) means that number is not included (<, >) in the interval.
• The symbols ꚙ, and -ꚙ are used to describe the unbound intervals.
Interval Notation
Example 2: Use Interval Notation
Write each set of numbers using interval notation.
a. −8 < 𝑥 ≤ 16
b. x < 11
Page: 5
c. x ≤ -16 or x > 5
• Answer:
• Answer:
• Answer:
(-8, 16]
(-ꚙ, 11)
Here we have two intervals.
(-ꚙ, -16] U (5, ꚙ)
Remember:
• 𝒐𝒓 symbolize union U
• and symbolize intersect Ո
Guided Practice: Use Interval Notation
Write each set of numbers using interval notation.
1A. −4 ≤ 𝑦 < −1
1B. 𝑎 ≥ −3
Page: 5
1C. x > 9 or x < -2
• Answer:
• Answer:
• Answer:
[-4,-1)
[-3, ꚙ)
Here we have two intervals.
(-ꚙ, -2) U (9, ꚙ)
Remember:
• 𝒐𝒓 symbolize union U
• and symbolize intersect Ո
Period 2
Identify Functions
• A relation is a rule that relates two quantities (x,y). The set of inputs is the
domain D (x) of the relation, and the set of outputs are the range R (y).
Identify Functions
• A function is a special type of relation.
• We can Identify if the relation is a function or not using the vertical line test.
Vertical Line Test
• A function is a set of ordered pairs in which no two different pairs
have the same x-value, this can be tested using the vertical line test.
Vertical Line Test
• A relation fails the vertical line
test, if x-value has more than
one corresponding y-value. (x
cannot be repeated in the
table)
• Notice that a function can
have one y-value paired
with more than one xvalue. (y can be repeated
in the table)
Example 3: Identify Relations that are Functions
Determine whether each relation represents y as a function of x.
a. The input value x is a
student’s ID number, and the
output value y is that
student’s score on a physics
exam.
b.
• Answer:
Each student (x) cannot have more
than one score (y) for the same exam.
Therefore, the relation between y and
x is a function.
• Answer:
Each x-value is assigned to exactly
one y-value. Therefore, the relation
between y and x is a function.
Page: 6
Example 3: Identify Relations that are Functions
Determine whether each relation represents y as a function of x.
c.
d. 𝒚𝟐 − 𝟐𝒙 = 𝟓
Page: 6
Remember: to remove the square,
take the square root for both sides.
• Answer:
To determine whether this equation is a function, solve the
equation for y. (make y by itself on one side, and the rest of the
equation on the other side)
• Answer:
A vertical line intersects the graph at
two points. Therefore, the graph does
not represent y as a function of x.
𝑦 2 − 2𝑥 = 5 Add 2x for both sides
+2x +2x
𝑦2
= 5 + 2𝑥
Take square root of
both sides
𝑦 2 = 5 + 2𝑥
Remember:
when taking square root
the result will be ±.
𝑦 = ± 5 + 2𝑥
This equation is not a
function because there
will be two y values,
one + and one − .
Guided Practice: Identify Relations that are Functions
Determine whether each relation represents y as a function of x.
3A. The input value x is the area
code, and the output value y is a
phone number in that
area code.
3B.
• Answer:
Each area (x) will have more than one phone
number (y) in that area. Therefore, the
relation between y and x is not a function.
Like in Abu Dhabi all phone number start with
the area code 02, where in Sharjah all phone
number start with the area code 06.
• Answer:
One x-value has two y-value.
Therefore, the relation between y
and x is not function.
Page: 6
Guided Practice: Identify Relations that are Functions
Determine whether each relation represents y as a function of x.
3C.
Page: 6
3D. 𝟑𝒚 + 𝟔𝒙 = 𝟏𝟖
• Answer:
To determine whether this equation is a function, solve the
equation for y. (make y by itself on one side, and the rest of the
equation on the other side)
• Answer:
A vertical line intersects the graph at
One point only. Therefore, the graph
represents y as a function of x.
3𝑦 + 6𝑥 = 18 subtract 6x from both sides
-6x -6x
3𝑦 = 18 − 6𝑥 Divide by 3
3
3
𝑦 = 6 − 2𝑥
This relation is a function because
there is one y values for each x
value. (linear equation)
Function Notation
• In function notation, the symbol 𝒇(𝒙) is read 𝒇 of 𝒙 and taken as the
value of the function f at 𝑥.
• 𝒇(𝒙) is the same as the y-value
𝑦 = 𝑓(𝑥).
Equation
Function notation
𝑦 = −6𝑥
𝑓(𝑥) = −6𝑥
• x in the domain can represent any value so it is called the independent variable.
A value in the range of f is represented by the dependent variable, y.
Function Notation
Example 4: Find Function Values
If 𝑔(𝑥) = 𝑥 2 + 8𝑥 − 24, find the value of 𝑔(𝑥) at the given points.
a. 𝑔(6)
b. 𝑔(−4𝑥)
• Answer:
• Answer:
To find 𝑔(6) replace all x with 6 in the
original equation.
To find 𝑔(−4𝑥) replace all x with -4x in the original
equation.
𝑔(𝑥) = 𝑥 2 + 8𝑥 − 24
𝑔(𝑥) = 𝑥 2 + 8𝑥 − 24
𝑔(6) = 6
2
The original equation
+ 8(6) − 24 Substitute x by 6
= 36 + 48 − 24
= 60
Page: 7
𝑔(−4𝑥) = −4𝑥
2
The original equation
+ 8(−4𝑥) − 24
Simplify
𝑔(−4𝑥) = 16𝑥 2 − 32𝑥 − 24
Substitute x by -4x
Example 4: Find Function Values
If 𝑔(𝑥) = 𝑥 2 + 8𝑥 − 24, find the value of 𝑔(𝑥) at the given points.
Expand 5𝑐 + 4
c. 𝑔(5𝑐 + 4)
Remember: 𝒂 + 𝒃
• Answer:
To find 𝑔(5𝑐 + 4) replace all x with 5c+4 in the original equation.
𝑔(𝑥) = 𝑥 2 + 8𝑥 − 24
𝑔(5𝑐 + 4) = 5𝑐 + 4
Page: 7
2
= 25𝑐 2 + 80𝑐 + 24
2
= 5𝑐
2
𝟐
= 𝒂𝟐 + 𝟐𝒂𝒃 + 𝒃𝟐
+ 2 5𝑐 4 + 4
= 25𝑐 2 + 40𝑐 + 16
The original equation
+ 8(5𝑐 + 4) − 24
= 25𝑐 2 + 40𝑐 + 16 + 40𝑐 + 32 − 24
Simplify
5𝑐 + 4
2
Substitute x by 5c+4
Distribute 8(5𝑐 + 4)
8 5𝑐 + 4 = 8 5𝑐 + 8(4)
= 40𝑐 + 32
2
Guided Practice: Find Function Values
If 𝑓(𝑥) =
2𝑥+3
,
𝑥 2 −2𝑥+1
find the value of 𝑓(𝑥) at the given points.
a. 𝑓(12)
b. 𝑓(6𝑥)
• Answer:
2𝑥 + 3
𝑓(𝑥) = 2
𝑥 − 2𝑥 + 1
2(12) + 3
𝑓(12) =
(12)2 −2(12) + 1
27
=
12
Page: 7
• Answer:
The original equation
Substitute x by 12
2𝑥 + 3
The original equation
𝑥 2 − 2𝑥 + 1
2(6𝑥) + 3
Substitute x by 6x
𝑓(12) =
(6𝑥)2 −2(6𝑥) + 1
𝑓(𝑥) =
Simplify
𝑓(12) =
12𝑥 + 3
36𝑥 2 − 12𝑥 + 1
Period 3
Implied Domain
• If domain is unspecified , the implied domain is the set of all real
numbers (ℝ), you must exclude values from the domain of a function
that would result in division by zero, or taking the even root of a
negative number.
Example:
𝑦 = −6𝑥
𝐷𝑜𝑚𝑎𝑖𝑛 = 𝑅
Example:
𝑥
𝑦=
𝑥−2
𝐷𝑜𝑚𝑎𝑖𝑛 = 𝑅 except 𝑥 = 2
Example 5: Find Domains Algebraically
State the domain of each function.
a. 𝑓(𝑥) =
• Answer:
2+𝑥
𝑥 2 −7𝑥
b. g(𝑡) = 𝑡 − 5
• Answer: Remember: the expression inside
When the denominator is zero, the
expression is undefined. So we need to find
the zeros of the dominator and exclude them.
𝑥 2 − 7𝑥 = 0
the even root must be positive to
be real and find the domain.
Threrfore to find the domin 𝑡 − 5 ≥ 0.
Find the zero of the dominator
𝑡−5≥0
+5 +5
𝑥(𝑥 − 7) = 0
𝑥 = 0 𝑜𝑟 𝑥 = 7
Page: 7
So the domain is all real
number except x=0 and x=7
{𝑥|𝑥 ∈ ℝ, 𝑥 ≠ 0, 𝑥 ≠ 7}
Add 5 for both sides
Therefore, the domain of 𝑔(𝑡) is all real numbers
𝑡 such that 𝑡 ≥ 5 or [5, ∞)
{𝑡|𝑡 ∈ ℝ, 𝑡 ≥ 5}
Example 5: Find Domains Algebraically
State the domain of each function.
c. ℎ(𝑥) =
1
𝑥 2 −9
• Answer:
Page: 8
Remember: the expression inside the even root must be
positive to be real & we need to exclude denominator zeros.
Interval 1
Interval 2
Interval 3
Find denominator zeros.
Add 9 for both sides
𝑥2 − 9 = 0
+9 +9
Take square root of both sides
𝑥2 = 9
𝑥 = ±3
At -5
(−5)2 −9 = 16
+
At 0
(0)2 −9 = −9
−
At 5
(5)2 −9 = 16
+
Draw the number line to find the interval where the
square root is positive. (choose a number in each interval)
Evaluate the function inside the
square root (𝑥 2 − 9) at the chosen
numbers, and find the interval
where the value is positive.
+
−
+
So the domain is the positive interval.
Domain is (−∞, −3)U (3, ∞)
Guided Practice: Find Domains Algebraically
State the domain of each function.
5A. 𝑓(𝑥) =
• Answer:
5𝑥−2
𝑥 2 +3𝑥
When the denominator
expression is undefined.
𝑥2
+ 3𝑥 = 0
5B. ℎ(𝑎) = 𝑎2 − 4
• Answer:
is
zero,
the
Find the zero of the dominator
𝑥(𝑥 + 3) = 0
𝑥 = 0 𝑜𝑟 𝑥 = −3 So the domain is all real
number except x=0 and x=−3
{𝑥|𝑥 ∈ ℝ, 𝑥 ≠ 0, 𝑥 ≠ −3}
Page: 8
Remember: the expression inside the even root
must be positive to be real and find the domain.
Find expression zeros
(inside the root).
Interval 1
Interval 2
Interval 3
𝑎2 − 4 = 0
+4 +4
At -3
Take square root of both sides
𝑎2 = 4
+
−
𝑎 = ±2
+
(−3)2 −4 = 5
At 0
(0)2 −4 = −4
At 3
(3)2 −4 = 5
+
−
+
So the domain is the positive interval.
Domain is (−∞, −2]U [2, ∞)
Piecewise-defined Function
• A function that is defined using two or more equations for different
intervals of the domain is called a piecewise-defined function.
Real-World Example 6: Evaluate a Piecewise-Defined Function
The height of children ℎ 𝑥 as a function of their parents’ heights (x) can be
modeled by the piecewise function. Find the heights of children whose
parents heights are given.
Page: 8
b. ℎ(72)
• Answer:
ℎ 𝑥 = 2𝑥 − 66
a. ℎ(67)
• Answer:
ℎ 𝑥 = 3𝑥 − 132
72 is greater 68, therefore we use
the 3rd equation (2x-66)
ℎ 72 = 2(72) − 66
67 is between 66 and 68, therefore
we use the 2nd equation (3x-132)
ℎ 67 = 3(67) − 132
= 201 − 132 = 69
Substitute x by 67
= 144 − 66 = 78
Substitute x by 67
Guided Practice: Evaluate a Piecewise-Defined Function
The speed v(t) of a vehicle in kilometers per hour can be represented by the
piecewise function. When t is the time in seconds. Find the speed of the
vehicle at each indicated time.
Page: 8
4𝑡,
60,
v t =
−6𝑡 + 1500,
0 ≤ 𝑡 ≤ 15
15 < 𝑡 < 240
240 ≤ 𝑡 ≤ 250
15 is included in the interval of the
1st equation (4t)
v 15 = 4 15 = 60
• Answer:
v 5 = 4 5 = 20
• Answer:
v t = 4t
a. v(5)
v t = 4t
b. v(15)
5 is between 0 and 15, therefore
we use the 1st equation (4t)
Substitute t by 5
Substitute x by 67
c. v(245)
• Answer:
v t = −6𝑡 + 1500
245 is included in the interval of
the 3rd equation (−6𝑡 + 1500)
v 245 = −6(245) + 1500 = 30 Substitute x by 245
Homework