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CSEC May 2017 Solution
CSEC Mathematics
Section 7 – Relations, Functions and Graphs
Relation
A relation is defined as a set of ordered pairs that obeys a particular rule.
The notation x x + 3, is a relation, which
means ‘x is mapped onto x + 3.
We can write the set of ordered pairs for the
notation, given {x: – 2 x 3}as:
The set of first values in the set of ordered pairs are called the domain elements, i.e.
{– 2, – 1, 0, 1, 2, 3}. These are the x values
or the set of x-coordinates. The set of second
values in the set of ordered pairs are called the image or the range elements, i.e.
{1, 2, 3, 4, 5, 6}. These are the y values or the set of y-coordinates.
The relation x x + 3 can be represented
using an arrow diagram (or relation diagram or mapping diagram) as shown below.
The Co-domain consists of the set: {0, 1, 2, 3, 4, 5, 6, 7}.
What is the difference between the range and the co-domain?
The relation x x + 3 can be represented as:
f : x x + 3 or f(x) = x + 3 or y = x + 3.
Answer the following.
1. 4 5 6 7 3 8 9 10 11 12 13 Domain Co-domain
a) List the set of elements for the:
i) Domain ii) Co-domain iii) Range
iv) Pre-image
b) Write down the set of ordered pairs for the
mapping.
2.
3. a) Draw an arrow diagram to represent the ordered pairs
{(0, - 2), (1, 3), (2, 8), (3, 13 ), (4, 18)}. b) List the set of elements of the:
i. Domain ii. Co-domain iii.
Range/Image 4.
Prepared by H. Walker March 30, 2016 6 7 8 9 10 3 13 15 17 19 21 23 10 11 12 13 14 7 Domain Co-domain
a) List the set of elements for the:
i) Domain ii) Co-domain iii) Range
iv) Pre-image
b) Write down the set of ordered pairs for the
mapping.
c) Find the rule that governs this relation.
a b c d e u v w Domain Co-domain
a) List the set of elements for the:
i) Domain ii) Co-domain iii) Range
iv) Pre-image
b) Write down the set of ordered pairs for the
5. Describe each of the following statements in words.
i. g : x 7 – 5x ii. h(x) = 8x
iii. y = 4x + 6
Types of Relations
There are four types of relations:
1. One to one relation – a relation is one to one if, for each element of X, there is one corresponding element of Y. For example,
9
10
11
e
36
40
44
2. Many to one relation – a relation is many to one if more than one element of X maps onto one corresponding element of Y. For example,
3. One to many relation – a relation is one to many, if for each element of X maps onto more than one corresponding element of Y. For example,
Prepared by H. Walker March 30, 2016
4. Many to many relation – a relation is many to many if for at least one element of X, there corresponds more than one element of Y and for more than one element of X
there corresponds at least one element of Y. For example,
Functions
A function is defined as a relation in which each element in the domain is mapped onto one and only one element in the range.
NB
A one to many relation and many to many relation is not a function.
All functions are relations but not all relations are functions.
Exercise
Answer the following.
1. Determine whether or not each of the following relations is a function.
a. x 1 – 3xDomain {-3, - 2, - 1, 0, 1, 2}
b. f(x) = 3x2 + 2x – 5
Domain {-3, - 2, - 1, 0, 1, 2}
c. y = Domain { 0, 1, 4, 9}
2.Draw a relation diagram to show a many to one mapping using the domain
{16, 25, 36, 49, 64}.
The Inverse of a Function
The notation f – 1 (x) means ‘the inverse of
the function f of x’ or ‘f inverse x’.
A function can only have an inverse if it is both one to one and its range is the same as the co-domain. The inverse of a function is a
reflection of the function in the line y = x.
If a function is a many to one mapping, then its inverse will not be a function. This is so, since the inverse of the function will be a one to many mapping which is not a
function.
Domain Co-domain
a
b
c
f Domain Co-domain
c
a
many to one relation (a function)
one to many relation (not a function)
In expressing the inverse of a function, the
co-domain of f becomes the domain of f – 1
and the domain of f becomes the co-domain of f – 1.
Exercise
Given the function f(x) = { (1, 4), (2, 7), (3,
10), (4, 13), (5, 16), (6, 19) }. a. State the set of:
Prepared by H. Walker March 30, 2016 Domain Co-domain
a f(a
) f
i) image/range (y-coordinate)
ii) pre-image (x-coordinate)
b. What is the rule of f(x) ?
c. Draw a mapping diagram to represent f(x).
d. What is the inverse of f(x) ?
e. Draw a mapping diagram to show f – 1(x).
Steps in Finding the Inverse of a Linear Function
Given the function f:x ax + b, where a and
b are real numbers and x is a variable, we
define the equation as f(x) = ax + b.
On the L.H.S. of the equation, replace f(x)
with x and on the R.H.S. of the equation
replace x with f – 1(x). That is x = af – 1(x)+b.
af (x) = x – b
f – 1(x) = which is the inverse of f(x).
Hint: The inverse of f(x) can be identified by
first stating a and b, then your result will x
minus b over a.
Given that f(x) = 3x + 1 then the inverse of
f(x) is f – 1(x) =
When x = 2, f(2) = 3(2) + 1 = 7 (2, 7).
Now the inverse of (2, 7) is (7, 2).
That is, f – 1(7) = .
Prepared by H. Walker March 30, 2016 Domain Co-domain
a f(a
) f
f – 1
Domain Co-domain
2 f(2
) = 7 f
When x = 4, f(4) = 3(4) + 1 = 13(4, 13). Now the inverse of (4, 13) is (13, 4).
That is, f – 1(13) = .
Find: a. f – 1(13) b. f – 1(0)
c. f – 1( ½ ) d. f – 1(1) e. f – 1(2)
f. f – 1(- 4) g. f – 1(- 3)
Note: If f(x) = a then x = f – 1(a)
Exercise
Answer the following.
1. Given f(x) = – 4x + 2, find:
Domain Co-domain
4 f(4
) =1 3 f
a. f(– 3) Ans: f(- 3) = 14
b. f – 1(x) Ans:
c. the value of x for which f(x) = 10.
Ans: x = - 2
2. Given f(x) = – x – 5, find:
a. f(0) Ans: f(0) = - 5
b. f – 1(x) Ans:
c. the value of x for which f(x) = 8.
Ans: x = - 13
3. Given y = 2x – 3, find:
a. y when x = – 5 Ans: y = - 13
b. f – 1(x) Ans:
15
c. the value of x for which y = 6.
Ans: x = 4.5
4. Given f(x) = 6 – 5x , find:
a. f(– 2) Ans: f(- 2) = 16
b. f – 1(x) Ans:
c. the value of x for which f(x) = – 7
Ans: x =
5. Given f(x) = – 9 – 3x, find:
a. f(– 3) Ans: f(- 3) = 0
b. f – 1(x) Ans:
c. the value of x for which f(x) = 10.
6. Given f(x) = – 2 – 5x, find the value of x
for which – 2 – 5x = 13. Ans: x = - 3
7. Given y = – 3x + 4, find the value of x for
which – 3x + 4 = – 16. Ans: x =
Exercise
Answer the following.
Find the inverse of the following functions.
1. y = 2x – 5
2. f(x) = 5x + 8
3. f:x 4 + x
4. y = 6 – 3x
5. f(x) = – 7 – x
Finding the Inverse of a Non-linear
Function of the Form
Given that , then the inverse is
as follows:
Replace f(x) with x on the LHS of the
equation and replace x with f – 1(x) on the
RHS of the equation. That is, .
Then make f – 1(x) the subject of the equation
‘distribute x to each
term inside the bracket on LHS’
‘group like terms’
‘factorize f – 1(x) from
‘divide both sides by
2x – 3’
which is the inverse of f(x).
Exercise
Answer the following.
1. Determine the inverse of the following functions.
a. Ans:
b. Ans:
c. Ans:
d. Ans:
e. Ans:
2. Find the inverse of and hence
solve the equation for the value of x.
Hint: f(x) = 7, thus x = f – 1(7)
Ans: and x = 4
3. Find the inverse of and hence
solve the equation for the value of x.
Ans: and x = 3
4. Find the inverse of and hence
5. Find the inverse of and hence
solve the equation .
6. Given ,
a. determine f – 1(x)
b. Calculate the value of x, if f(x) = 15.
7. Given the function ,
a. find g – 1(x)
b. determine the value of x, if g – 1(x) = .
Writing a Quadratic Equation
y = ax2 + bx + c in the form a(x + h)2 + k
We can express a quadratic function
21
y = ax2 + bx + c in the form y = a(x + h)2 + k
where h = and k = c – or
k =
Exercise
Answer the following.
Express the following quadratic functions in
the form y = a(x + h)2 + k or
y = k – a(x + h)2.
1. y = 3x2 – 6x + 3 2. y = 4x2 + 8x – 12
3. y = x2 – 5x + 2 4. y = 2x2 – x + 3
5. y = 5x2 + x – 2
Exercise
Find the inverse of the following quadratic functions.
1. y = x2 – 7 2. y = 8x2 – 5
3. y = 3 – 4x2 4. y = 2 + 3x2
5. y = 6 – x2
Completing The Squares
We can find the inverse of a function using ‘Completing the squares’.
Example
Find the inverse of y = 2x2 – 4x + 6.
Solution: y = 2(x2 – 2x + 3)
y = 2[(x – 1)2 + 3 – 12]
y = 2[x – 1)2 + 2]
23
y = 2(x – 1)2 + 4
Replace y with x and replace x with f – 1(x)
x = 2[f – 1(x) – 1]2 + 4
Make f – 1(x) the subject
x – 4 = 2[f – 1(x) – 1]2
Square root both sides
The root cancels the square
Add 1 to both sides
Ans:
y = ax + bx + c in the form y = a(x + h) + k where
h = and k = c –
Exercise
Answer the following.
Find the inverse of the following quadratic functions.
1. y = 3x2 – 6x + 3 2. y = 4x2 + 8x – 12
3. y = x2 – 5x + 2 4. y = 2x2 – x + 3 5.
y = 5x2 + x – 2
Composite Functions
Given that f(x) = ax + b and g(x) = cx + d,
where a, b, c and d are real numbers and x is
a variable.
Then fg(x) = ag(x) + b, thus
fg(x) = a(cx + d) + b.
We obtained the composite function fg(x) by
substituting g(x) for x in f(x).
Also gf(x) = cf(x) + d, thus gf(x) = c(ax + b)
+ d.
We obtained the composite function gf(x) by
substituting f(x) for x in g(x).
NOTE: fg(x) gf(x). That is, composite
functions are non-commutative.
Consider the functions f(x) = 2x + 1 and
g(x) = .
Now gf(x) = g(2x + 1) or gf(x) =
and gf(x) = = .
Thus, gf(5) = .
Note: fg(x) ≠ gf(x)
(fg)– 1 = g– 1f– 1(x) and (gf)– 1 = f– 1g– 1(x) Exercise
Answer the following.
Determine the following composite functions.
1. a) gf(x) where f(x) = x + 2 and
g(x) =
b) fg(x)
Prepared by H. Walker March 30, 2016
5 11
f
7
2. a) gf(x) where f(x) = 3x – 1 and
g(x) = 2x + 5
b) fg(– 3) c) gf(– 2) d) f2(x)
e) gg(1).
3. Given f(x) = 2x – 1, g(x) = x + 2 and
h(x) = 3x +1 find:
a) fh(x) Ans: fh(x) = 6x + 1
b) gf(2) Ans: gf(2) = 5
c) gg(– 1) Ans: gg(– 1) = 3
d) h2(– 3) Ans: h2(– 3) = - 23
e) f2(0) Ans: f2(0) = - 3
g) fgh(x) Ans: fgh(x) = 6x + 5
h) gfh(– 2) Ans: gfh(– 2) = - 9
i) hgf(x) Ans: hgf(x) = 6x + 7
4. Find the following functions, given
f(x) = 2x – 1, g(x) = x + 2 and h(x) = 3x +1.
a) h – 1(x) b) f – 1(x) c) g – 1(x)
d) h – 1 g – 1(x) e) f – 1 h – 1(x)
f) g – 1h – 1(6) g) h – 1 f – 1( – 2)
i) (gh) – 1 j) (hf) – 1 k) (hg) – 1(6)
l) (fh) – 1 ( – 2)
Linear Functions
A linear function is of the form y = mx + c
or y = m(x – x1) + y1 , where m is the
gradient of the function and c is the
y-intercept, that is, when x = 0, y = c. A linear function can also be expressed in the
form ax + by = c.
Gradient and Intercepts of Linear Functions
Given the line segment AB, where A(x1, y1)
and B(x2, y2):
The gradient of the line AB, is:
gradient (m) = .
The equation of the line AB is
y = mx + c or y = m(x – x1) + y1.
The y-intercept of a linear function is at
where the line meets the y-axis. Give diagram
The x-intercept of a linear function is at
the point where y = 0. That is, the point
where the line meets the x-axis. Give
diagram
The mid-point of the line AB is
(x, y) = where A(x1, y1)
and B(x2, y2).
The length of the line AB is
The Gradient of Parallel and Perpendicular Lines
If two lines are parallel, they have the same
If two lines are perpendicular (meet or
intersect at 900/right angle), then the product
of their gradients is – 1. That is,
m1 m2 = – 1. Thus, m1 = or
m2 = . Hence, the gradient of m1 is the
negative reciprocal of m2 and vice-versa.
We can find the equation of a line by using a
point, (x1, y1), on the line, the gradient of the
line and the equation y = m(x – x1) + y1 or
y = mx + c.
Exercise
1. The equation of a line is 3x – 2y = 7,
determine the: a. y–intercept
b. x–intercept
c. gradient
2. Write down the gradient, x-intercept and
y-intercept of each of the following lines.
a. 4(x + 6) = 2y
b. 6x – 3y + 81 = 27
c. 15 – 5y = 6x
d. 7 – 8x – 2y = 0
e. y = 3x + 5
f. 2x – 2y – 7 = 0
3. A line segment has coordinates R(3, – 1) and T(– 2, 4). Find the:
a. mid-point of the line RT Ans: (0.5, 1.5)
b. length of the line RT Ans: 7.07 units
c. gradient of RT Ans: - 1
d. equation of the line RT Ans: y = - x + 2
e. equation of the line which is parallel to RT and passes through the origin.
f. equation of the line which is
perpendicular to RT and passes through the
point (– 4, – 2). Ans: y = x + 2
Give January 2016 – 6b
4. (i) A line JK has equation 2y = 5x + 6.
Determine the gradient of JK. Ans: 2.5
(ii) Another line GH is perpendicular to JK and passes through the point (5, – 1). State
the gradient of the line GH. Ans: - 0.4
(iii) Determine the equation of the line GH.
Ans: y = -0.4x + 1
5. Find the equation of the line which is
parallel to the line 3y = 9x – 7 and passes
through the point (1, 5). Ans: y = 3x + 2
6. Find the equation of the straight line which passes through (2, – 3) and is
perpendicular to the line 4x + 3y – 5 = 0.
Ans: y = 3/4x – 9/2 or y = 0.75x – 4.5
7. Find the equation of the line
perpendicular to x + 3y = 6 and passes
through the point (0, 6).
8. Find the equation of the straight line
which is parallel to the line 3y – 12x + 3 = 0
and passes through the point (5, – 2).
9. Find the equation of the line
perpendicular to the line 7y – 3x + 28 = 0
10. Find the equation of the line which is
parallel to the line 4x + 3y = 1 and passes
through the point (5, 4).
11. Find the equation to the line which is
perpendicular to the line 2x – 4y = 6 and
passes through the point (1, 3).
Graph of Linear Functions Exercise
Do the following questions on the same graph.
Note: A line can be drawn using two points.
1 a. Draw on graph paper a representation of the line y = 4.
b. Write down the coordinates of any two points on the line.
c. Use the coordinates of the two points which you have identified to find the
gradient of the line.
2 a. Draw on graph paper a representation
of the line x = – 3.
b. Write down the coordinates of any two points on the line.
c. Use the coordinates of the two points which you have identified to find the
Note: The gradient of all horizontal lines is zero. The gradient of all vertical lines is
infinity or undefined.
3 a. Draw the graph of the line
L1: y = 2x – 3 and L2: y = 2x + 5.
b. What do you notice about the two lines?
c. State the gradient of the each line.
d. Write down the coordinates of any two points on each line.
e. Use the coordinates of the two points which you have identified for each line to find the gradient of the line respectively.
4 a. Draw the graph of the line
L3: 2y + x = 8.
b. What do you notice about this line?
c. State the gradient of L3.
d. Write down the coordinates of any two
points on L3.
e. Use the coordinates of the two points
which you have identified on the L3 to find
its gradient.
5 a. Draw the graph of y = – 2x + 1
6 a. Copy and complete the table below for
y = 3 – 2x.
x – 3 – 2 – 1 0 1 2 3 4
y
b. Using 1 cm to represent 1 unit on the x
-axis and 1 cm to represent 1 unit on the
y-axis. Draw the graph of y = 3 – 2x for
– 3 x 4.
c. Using your graph determine the value of
x for which y = 3 – 2x = – 4.
Graph of the Quadratic Curve (y = ax2 + bx + c)
For f(x) = ax2 + bx + c if a < 0 then the curve
is a maximum curve and if a > 0, the
curve is a minimum curve.
For any curve the y-intercept is when x = 0
and the x-intercept is when y = 0
The turning point/stationary point/maximum
point/minimum point, occurs at x = .
The maximum or minimum value of a
quadratic curve is the largest or smallest y-value on the curve, respectively. It occurs
at x = .
The line or axis of symmetry of a quadratic
curve is x = .
A minimum curve A maximum curve Turning point
The root(s) of a quadratic curve is the value
of x when y = 0. It is the point where the
curve meets the x-axis.
Exercise
1. Determine the axis of symmetry, maximum/minimum value, the
maximum/minimum point and root(s) for each of the following quadratic functions.
a. y = x2 + 2x – 3
b. y = – x2 + x + 2
c. f(x) = 2x2 – 8x + 8
d. f(x) = – 3x2 + 6x + 6
e. y = –3x2 – 3x +6
2. The table below is designed to show
values of x and y for the function
y = x2 – 2x – 3 for integer values of x
from – 2 to 4. MAY 2015 – QUESTION 4
x – 2 – 1 0 1 2 3 4
y 5 – 3 – 4 – 3 5
a. Complete the table for the function
y = x2 – 2x – 3.
b. On graph paper plot the graph of
y = x2 – 2x – 3 using a scale of 2 cm to
represent 1 unit on the x-axis and 1 cm to
represent 1 unit on the y-axis.
c. Use your graph to answer the following:
i. The values of x for which x2 – 2x – 3 = 0.
ii. The minimum value of y = x2 – 2x – 3.
iii. The equation of the line of symmetry of
iv. The roots of y = x – 2x – 3.
Linear Programming (Vol. 2, Page 1018 – R. Toolsie)
In linear programming, we are given a
number of conditions and we have to form a set of suitable inequations to represent these conditions. We normally have to find the maximum or minimum value of an
expression using the set of linear inequations.
The common region where the inequations intersect is normally a polygon. The
maximum or minimum value is normally satisfied by a vertex of the polygon or by all points along a side of the polygon.
1. JANUARY 2014 QUESTION 4
f) Find the equation of the line parallel to line 2 and it passes through the point
(- 4, -7). Ans: y = x – 3
g) Find the maximum and minimum value
for which P = 4x – 3y – 2 for region S.
Ans: Maximum value = 0 Minimum value = – 8
2. JANUARY 2016 QUESTION 9
3. MAY 2013 QUESTION 9a
4. MAY 2012 QUESTION 9b
5. JANUARY 2012 QUESTION 9b
