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Chapter One
1
Prerequisites for Calculus
PREREQUISITES FOR CALCULUS
Sets and Intervals:
DEFINITIONS:
Set: is a collection of things under certain conditions.
Elements: are the things which set up the set.
Example 1: A={1,2,5,7,10}: A is a set ; 1,2,5,7,10 are elements.
Real Numbers (R): is a set of all rational and irrational numbers. R= {-∞, +∞}
-∞
+∞
0
Real Number line
Integer Numbers (I): a set of all irrational numbers.
I = {-∞,----,-3,-2,-1,0,1,2,3,----,+∞} negative and positive numbers only
Natural Numbers (N): consist of zero and positive integer numbers only.
N ={0,1,2,3,----,+∞}
Interval: is a set of all real numbers between two points on the real number
line. (it is a subset of real numbers)
-∞
+∞
A
x
B
1. Open interval: is a set of all real numbers between A&B excluded (A&B are
not elements in the set). {x: A < x < B} or (A, B)
-∞
+∞
A
x
B
2. Closed interval: is a set of all real numbers between A&B included (A&B are
elements in the set). {x: A ≤ x ≤ B} or [A, B]
-∞
+∞
A
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x
B
……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
3.Half-Open interval (Half-Close): is a set
of all real numbers between A & B with one
-∞
+∞
A
x
B
of the end-points as an element in the set.
a) (A, B]= {x: A < x ≤ B}
-∞
+∞
b) [A, B)= {x: A ≤ x < B}
A
x
B
Types of intervals
Notation
Set description
Type
(a,b)
{x: a < x < b}
Open
[a,b]
{x: a ≤ x ≤ b}
Closed
[a,b)
{x: a ≤ x < b}
Half-opened
(a,b]
{x: a < x ≤ b}
Half-opened
(a,∞)
{x: x > a}
Open
[a,∞)
{x: x ≥ a}
Closed
(-∞,b)
{x: x < b}
Open
(-∞,b]
{x: x ≤ b}
R (ser of all real
numbers)
Closed
Both open
and closed
Finite:
Infinite:
(-∞,∞)
Picture
a
b
a
b
a
b
a
b
a
a
b
b
Union and Intersections of intervals: if A and B are two sets then:
The union is the set whose numbers belong to A or B (or both) and denoted by
(A  B), and intersection is the set whose numbers belong to both A and B and
denoted by (A∩B).
Examples:
1. Solve for x the following and show their solution set {x: x2=4}
Sol: x2 = 4  x = ±2
 the
-2
set is {-2,+2}
2
2. Solve for x the following and show their solution set {x: x2≤4}
Sol: x2 ≤ 4  x2-4≤ 0  (x-2)(x+2) ≤ 0
Let (x-2)(x+2) =0  either (x-2)=0  x=2
or (x+2)=0  x=-2
---------------- 2 +++ +∞
-2
+∞
++++++++
Sign of (x+2) -∞ -------2
2
------- +++++ +∞
Sign of (x-2)(x+2) -∞ +++++
……………………. Mathematics \1st class
Sign of (x-2) -∞
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Chapter One
 the
Prerequisites for Calculus
interval is {x: -2 ≤ x ≤ 2} or [-2,2]
3. Solve for x the following and show their solution set {x: x2-x-30≤0}
Sol: x2-x-30 ≤ 0  (x-6)(x+5) ≤ 0
Let (x-6)(x+5) =0  either (x-6)=0  x=6
or
(x+5)=0  x=-5
 the
6
-------∞
+++ +∞
Sign of (x-6)
-5
++++++++ +∞
Sign of (x+5) -∞ - - - - -5
6
Sign of (x-6)(x+5) -∞ +++++ - - - - - +++++ +∞
interval is {x: -5≤x≤6} or [-5,6]
4. Solve for x the following and show their solution set {x: x2-x-30>0}
Sol: from Ex.3 the interval is {x: x<-5}  {x: x>6}
or (-∞,-5)  (6, ∞)
5. Solve for x the following and show their solution set {x:
Sol:
+1
----------------++++++ +∞
-2
Sign of (x+2) -∞ ------+++++++++++++ +∞
Sign of (x-1) -∞
Sign of (x-3) -∞ -------------------------
( x  1)( x  2)
Sign of
x  3
 the
( x  1)( x  2)
0}
x3
-∞ ---------
+3
+++ +∞
-2 ++++ +1------ +3++++
+∞
interval is { x: -2 ≤ x ≤ +1}  {x: x >+3} or [-2,+1]  (+3, ∞)
6. Solve for x: { x: 0 < x < 5}  { x: 1< x < 7}
Sol: from the line of numbers:
(0,5)
-∞
(1,7)
-∞
+∞
0
5
1
(0,5)  (1,7) -∞
0
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+∞
7
+∞
7
……………………. Mathematics \1st class
Chapter One

Prerequisites for Calculus
the interval is {0<x<7} or (0,5)  (1,7)=(0,7)
7. Solve for x: { x: x<1}  { x: x  0}
Sol: from the line of numbers:
(-∞,1)
-∞
[0, ∞)
-∞
+∞
1
+∞
0
+∞
(-∞,1)∩ [0,∞)-∞
0
 the
1
interval is {x: 0≤ x <1} or [0,1)
8. Solve for x: { x: x<0}  { x: x >0}
Sol: from the line of numbers:
(-∞,0)
-∞
(0, ∞)
-∞
+∞
0
0

(-∞,0)  (0,∞)= -∞
+∞
+∞
0
Inequalities:
Rules for Inequalities
Let a, b, and c are real numbers, then:
1. if a < b and b < c then a < c
2. a < b  a + c < b + c
3. a < b  a - c < b - c
4. a < b and c > o  ac < bc (c is positive)
5. a < b and c < o  ac > bc (c is negative)
special case a < b  -a > -b
6. if a < b and c < d then a+c < b+d
7. a > 0 
1
0
a
8. If a and b are both positive or both negative, and a < b 
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1 1

a b
……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Examples: Solve for x the following:
1. 3+7x ≤ 2x-9

Sol: 3+7x-3≤2x-9-3

5x≤-12 
So the interval (-∞,
x≤
7x≤2x-12

7x-2x≤2x-12-2x
 12
5
 12
]
5
+∞
-∞
 12
5
2. 7≤ 2-5x < 9
Sol: 7-2≤2-5 x -2<9-2

-1≥x>-7/5

5≤-5x<7


-7/5<x≤-1
5/(-5)≤-5x/(-5)<7/(-5)
+∞
-∞
So the interval (-7/5,-1]
-7/5
3. x2-3x > 10
-1
sol: x2-3x-10>0  (x-5)( x+2)>0
-2
+++++++++++++ +∞
Sign of (x+2) -∞ ------+5
++++++ +∞
Sign of (x-5) -∞ -----------------2
-------- +5 ++++++ +∞
++++++
∞
Sign of (x-5)(x+2)
So the interval (-∞,-2)  (5, ∞)
Homework:
1. Explain the following intervals on the line numbers:
(a) -3 < x < 5
(b) 2 ≤ x ≤ 6
(d) x >5
(e) x ≤ 2
(c) -4 < x ≤ 0
2. Solve the following inequalities:
(a) {x: x3- 3x ≤ 5x2 - 9x}
(c) { x :
2x
 0}
( x  2)( x  1)
(b) {x: 6x- x2 > 0}
(d) { x :
x2 1
0}
x2  1
3. Solve the following inequalities and express the solution in terms of intervals:
(a) 3x < 5x – 8
(b) 12 ≥ 5x – 3 > -7
(c) 3x2 + 5x – 2 < 0
(d) 2x2 + 9x +4 ≥ 0
(e)
3x  2
0
2x  7
(f)
3
2

x 9 x  2
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Chapter One
Prerequisites for Calculus
Analytical Geometry (Coordinate in the Plane)
Each point in the plane can be represented
with a pair of real numbers (a,b), the number a is
the horizontal distance from the origin to point P,
while b is the vertical distance from the origin to
point P.
The origin divides the x-axis into positive xaxis to the right and the negative x-axis to the left,
also, the origin divides the y-axis into positive
y-axis upward and the negative y-axis downward.
The axes divide the plane into four regions called quadrants, numbered I,
II, III and IV.
Distance between Points and (Mid-Point Formula):
Distance between points in the plane is
calculated with a formula that comes from
Pythagorean Theorem:
Distance Formula for Points in the Plane
yo
The distance between P(x1,y1) and Q(x2,y2)
d  (x) 2  (y ) 2  ( x2  x1 ) 2  ( y2  y1 ) 2
xo
and the mid-point formula:
xo 
x1  x2
;
2
yo 
y1  y2
2
Example 1: Calculating distance between P(-1,2) and Q(3,4):
Sol.:
d = (3  (1)) 2  (4  2) 2  (4) 2  (2) 2  20  2 5
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Chapter One
Prerequisites for Calculus
Slope and Equation of Line
DEFENITION: Slope
The constant
m
rise y y2  y1


run x x2  x1
is the slope of nonvertical line P1P2
Note1: Horizontal line have m=0 (y=0), and the
vertical line has no slope or the slope of vertical
line is undefined (x=0).
y
LineL1
Line L2
Slope m1
Slope m2


Note2: Parallel lines have the same slope
x
If m1= m2 then and the lines are parallel.
Note3: If two non-vertical lines L1and L2 are
perpendicular, their slopes m1 and m2 satisfy m1. m2=-1, so each slope is the
negative reciprocal of the other
m1  
1
,
m2
m2  
1
m1
To see this, notice by inspecting similar triangles that
m1=a/h and m2=-h/a.
Hence m1. m2= (a/h). (-h/a)= -1
Point-Slope Equation:
We can write an equation for a non-vertical
straight line L if we know its slope m and the coordinate
P(x,y)
of one point P1(x1,y1) on it. If P(x,y) is any other point on
L, then we can use two points P1 and P to compute the
P1(x1,y1)
slope,
m
y  y1
x  x1
So that
y - y1 =m (x - x1)
or
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y = y1 + m (x - x1)
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Chapter One
Prerequisites for Calculus
The equation
y = y1 + m (x - x1)
is the point-slope equation of the line that passes through the point P1(x1,y1) and
has slope m.
Example 2: write an equation for the line through the point (2,3) with slope -3/2.
Sol.: we substitute x1 = 2, y1 = 3, and m = -3/2 into the point-slope equation and
obtain
3
y  3  ( x  2),
2
or
3
y   x6
2
When x = 0, y = 6 so the line intersects the y-axis at y = 6.
Example 3: A line through two points: write an equation for the line through
(-2,-1) and (3,4)
Sol.: The line's slope is
m
1 4  5

1
23 5
We can use this slope with either of the two given points in the point-slope
equation;
With (x1, y1) = (–2, –1)
with (x1,y1) = (3,4)
y = –1 + 1 . (x–(–2))
y = 4 + 1 . (x–3)
y = –1 + x + 2
y = 4 + x –3
y=x+1
y=x+1
either way, y = x + 1 is an equation for the line
The y-coordinate of the point where a non-vertical
line intersects the y-axis is called the y-intercept of the
line. Similarly, the x-intercept of a non-horizontal line
is the x-coordinate of the point where it crosses the xaxis. A line with slope m and y-intercept b passes
through the point (0, b), so it has equation:
y=b+m(x-0), or, more simply,
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y = mx + b
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Chapter One
Prerequisites for Calculus
The equation:
y = mx + b
is called the slope-intercept equation of the line with slope m and y-intercept b
Note: The general form of straight line equation is Ax + By + C= 0
Example: finding the slope and y-Intercept for the line 8x + 5y = 20.
Sol.: solve the equation for y to put it in slope-intercept form:
8
8x + 5y = 20  5y = –8x + 20  y   x  4
5
The slope m = –8/5. the y-intercept is b = 4.
The Distance from a Point to a Line:
d
To calculate the distance from a certain
point P(x1,y1) to a line L:
P(x1,y1)
Q(x2,y2)
1. Find an equation for the line L` that pass
L`
through a point P(x1,y1) and perpendicular to
the line L.
2. Find the point Q1(x2,y2) where L` meet with L.
3. Calculate the distance between P and Q.
4. to find d use the distance between two points formula.
Or you can use the following formula:
The distance (d) between the line L is Ax + By + C= 0 and the point P(x1,y1)
d
Ax1  By1  C
A2  B 2
Example 4: Find the distance from the point P(2,1) to the line y = x + 2
Sol.: the slope of L is m=1 (from slope-intercept form y=mx+b),
so the slope of L` is m`=-1/1=-1
y - y1 = m (x - x1) 
from point P(2,1) & m=-1
 y - 1= -1 (x -2)
 y = -x +2+1
 y =- x + 3
To find Q (the point of L and L` intersection put yL= yL`)
x+2=-x+3 
2x=1
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
x=1/2 
y=5/2
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Chapter One
Prerequisites for Calculus

the coordinate of the point Q is (1/2, 5/2)
3
3
unit length
d  ( x2  x1 ) 2  ( y2  y1 ) 2  (2  1 ) 2  (1  5 ) 2 
2
2
2
2
2
Another solution:
Put the equation of the line L in the general form Ax+By+C=0
y=x+2


x - y+ 2= 0 by comparing with the general form
A=1, B=-1 and C=2 also x1=2 and y1=1
d
Ax1  By1  C
A2  B 2

1* (2)  (1) * (1)  2
12  12

3
unit length
2
Angles between Two Lines:
L1,m1
y
To find the angle between the lines L1 and L2 which
have slopes m1 and m2 respectively, draw horizontal line


passes through the point of their intersections as shown.
L2,m2

So

Where: m1=tan
tan  


tan = tan(-)
tan   tan 

1  tan  . tan 
0
m2=tan
m1  m2
1  m1m2
Example 5: Find the following:
1. The slope of the line 2x+3y=5?
2. The distance from the above line to the point P(-1,0).
Sol.:
1. Put the equation of the line in the point-slope equation form (y=mx+b)
2
5
3y=5-2x   y   x 
so m=-2/3
3
3
&
b=5/3
2. Put the equation of the line in the general form (Ax+By+C=0)
2x+3y-5=0

A=2, B=3 and C=-5, but x1=-1 and y1=0
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x
Chapter One
d
Prerequisites for Calculus
Ax1  By1  C
A B
2
2

2 * (1)  3 * (0)  (5)
2 3
2
2
7
13

y
Example 6: Plot the given pair of points and the
equation for the line determined by them.
The points (1,1) & (2,1)
Sol.: m 
(1,1)
y2  y1 1  1

0
x2  x1 2  1
(2,1)
0
Put the equation on the slope-point equation
form choosing any of the two points P1(1,1)
y = y1 + m (x - x1)
y

y = 1 + 0 (x - 1)
=1
Example 7: Find the slope of the line 3x-4y=-8 and then find the distance from
the point P(3,-2) to this line.
Sol.:
a) Put the equation of the line in the point-slope equation form (y=mx+b)
3x-4y=-8  4y=3x +8  4y=3/4x +2
 m=-3/4 &
b=2
b) Put the equation of the line in the general form (Ax+By+C=0)
 3x-4y+8=0 A=3, B=-4 and C=8, but x1=3 and y1=-2
3x-4y=-8
d
Ax1  By1  C
A B
2
2

3 * (3)  (2) * (4)  8
3  (4)
2
2

988
9  16

25
25
 5 units length
Homework:
1. Find the line L passes through the point (1,2) and parallel the line L`:
x + 2y = 3.
2. Find the line L passes through the point (-2,2) and perpendicular to the line
L` : 2x + y = 4.
3. If A, B, C and C` are constants, show that:
(a) the lines:
Ax + By + C = 0
Ax + By + C` = 0
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are parallel
(12) ……………………. Mathematics \1st class
x
Chapter One
(b) the lines:
Prerequisites for Calculus
Ax + By + C = 0
Bx - Ay + C` = 0
are perpendicular
4. The perpendicular distance ON from the origin to the line L is  and ON
makes an angle  with the positive x-axis. Show that L has an equation
y sin +x cos = 
5. Find the line that passes through the point P(1,2) and the point of
intersection of two lines x +2y = 3 and 2x - 3y = -1.
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Chapter One
Prerequisites for Calculus
Functions
DEFINITION: Function
A function from a set D (domain) to a set R (range) is a rule that assigns to
unique (single) element f(x)  R to each element x D
f: x
f(x) it means that f sends x to f(x)
1
it means that f sends x to
x2
f: x
f ( x)  y 
1
x2
R
 The set of x is called the "Domain" of the function (Df).
 The set of y is called the "Range" of the function (Rf).
 x & y are variables.
 x is independent variable.
 y is dependent variable.
Domain (Df): is the set of all possible inputs (x-values).
Range (Rf): is the set of all possible outputs (y-values).
To find Domain (Df) and the Range (Rf) the following points must be
noticed:
1. The denominator in a function must not equal zero (never divide by zero).
2. The values under even roots must be positive.
Examples: Find the Domain (Df) and Range (Rf) of the following functions:
1. y  f ( x) 
1
x
Sol: denominator must not equal zero  x  0  Df ={x: x ≠ 0}.
To find Rf : we must convert the function from y=f(x) into x=f(y).
x 
1
y
 Rf={y: y ≠ 0}.
2. y  f ( x)  9  x 2
Sol: The values under even roots must be positive
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Chapter One
Prerequisites for Calculus
 9-x2 ≥ 0  (3-x) (3+x) ≥ 0

+3
------
+∞
-3
+++++++++++++
+∞
Sign of (3-x) -∞ ++++++++++++++
Df ={x: -3≤x ≤ 3}.
To find Rf: we must
----------
Sign of (3+x) -∞
convert the function from
+3
-3
-∞ ------------- ++++++ ---------- +∞
Sign of (3-x) (3+x)
y=f(x) into x=f(y).
y  9  x2  y 2  9  x2
 x2  9  y2  x   9  y 2
So the values under even
9-y2 ≥ 0  (3-y) (3+y) ≥ 0

+3
------
+∞
-3
+++++++++++++
+∞
Sign of (3-y) -∞ ++++++++++++++++
roots must be positive
Sign of (3+y) -∞
Rf ={y: -3≤y ≤ 3}.
But the values of y must be
Sign of (3-y) (3+y)
----------
+3
-3
0
---------- +∞
++++++
-∞ ------------//////////////////////////
always positive, we must
exclude negative values, 
Rf ={y: 0 ≤ y ≤ 3}.
1
3. y  f ( x) 
9  x2
Sol: The values under even roots must be positive and the denominator must not
equal zero, so:
9-x2 > 0  (3-x) (3+x) > 0

Df ={x: -3 < x < 3}.
To find Rf : we must convert the function from y=f(x) into x=f(y).
y

x2 
1
9  x2

9y2 1

y2
y2 
1
9  x2
x

9y2 1
y2
9  x2 

1
y2
x

x2  9 
1
y2
9y2 1
y
The values under even roots must be positive and the denominator must not
equal zero, so: 9y2-1 ≥ 0

Sign of (3y-1) -∞ ---------------------------
(3y-1) (3y+1) ≥ 0
Sign of (3y+1) -∞
----------
+1/3
++++++ +∞
-1/3
+++++++++++++
+∞
+1/3
0
+++++++-1/3------------ ++++++ +∞
Sign of (3y-1) (3y+1) -∞ ///////////////////////////
University of Kufa\Civil Eng. ………………
(15) ……………………. Mathematics \1st class
Chapter One

Prerequisites for Calculus
Rf ={y: -∞≤ y ≤ -1/3}  {y: 1/3≤ y ≤ ∞}.
denominator must not equal zero  y  0
But the values of y must be always positive; we must exclude negative values,

Rf ={y: 1/3≤ y ≤ ∞}
4. y  f ( x) 
x 1
x 1
Sol: The values under even
roots must be positive and the
Sign of (x-1) -∞ ---------------------------
denominator must not equal
Sign of (x+1) -∞
zero, so:
x-1≠0
and

x≠1
Sign of
x 1
0
x 1
x  1 -∞
x  1
----------
+1
+++++ +∞
-1
+++++++++++++
+∞
++++++ -1----------- +1
+++++++ +∞
Df = (-∞, -1]  (1,+ ∞)
To find Rf : we must convert the function from y=f(x) into x=f(y).
y
x 1
x 1


y2 
x 1
x 1
y2 x  x  y2  1

y 2 ( x  1)  x  1

y2x  y2  x 1

x( y 2  1)  y 2  1

x
y2  1
y2 1
You can see that there is no root in the form x = f(y), so only the denominator
must not equal zero so:
y2-1≠0 
y2≠1

y≠ ±1 
Rf =R\{-1,+1}
But the values of y must be always positive; we must exclude negative values,

Rf =[0,+∞)\{+1}
5. y  f ( x)   1  x 2
Sol: The values under even
+1
Sign of (1-x) -∞ ++++++++++++++++ ------------- +∞
roots must be positive:
Sign of (1+x) -∞
1-x ≥0 
2
 Df=[-1,+1]
(1-x) (1+x) ≥ 0
----------
-1
+++++++++++++
+∞
------------ -1+++++++ +1
------------ +∞
Sign of (1-x)(1+x) -∞
To find Rf: we must convert the function from y=f(x) into x=f(y).
University of Kufa\Civil Eng. ………………
(16) ……………………. Mathematics \1st class
Chapter One
y   1  x2 
Prerequisites for Calculus
y2  1  x2

x2  1  y2

x   1  y2
The values under even roots must be positive:
1- y2 ≥ 0

(1-y) (1+y) ≥ 0
 Rf=[-1,+1]
+1
Sign of (1-y) -∞ ++++++++++++++++ ------------- +∞
But the values of y must be
Sign of (1+y) -∞
always negative; we must
exclude positive values,

----------
-1
+++++++++++++
+1
0
------------ -1+++++++
------------ +∞
//////////////////////////
Sign of (1-y)(1+y) -∞
Rf =[-1,0]
Homework: Find the domains and ranges of the following functions.
1. y 
x2 1
x2  1
+∞
2. y 
2x
( x  2)( x  1)
4. y  x 2  4
5. y  x 2  4
7. y  9  x
8. y 
1
3 x
University of Kufa\Civil Eng. ………………
3. y 
2x
( x  2)( x  1)
6. y 
2x
2 x
9. x 2  xy  y 2  3  0
(17) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Absolute Value Function: it is defined as:
 x if
y  x  x2  
 x if
x0
x0
Absolute Value Properties
1. |-a | = | a|
A number and its additive inverse or negative have
the same absolute value.
2. |ab | = | a||b |
The absolute value of a product is the product of the
absolute values.
3.
a
a

b
b
The absolute value of a quotient is the quotient of the
absolute values.
4. |a +b| ≤ |a | +| b| The triangle inequality. The absolute value of the sum
of two numbers is less than or equal to the sum of
their absolute values.
Absolute Values and Intervals
If a is any positive number, then
5. | x | = a
if and only if x = ±a
6. | x | < a
if and only if -a < x < a
7. | x | > a
if and only if x > a or x < - a
8. | x | ≤ a
if and only if -a ≤ x ≤ a
9. | x | ≥ a
if and only if x ≥ a or x ≤ - a
Examples: Solve the following for x?
1. x  3
Sol.: So
x=3
2. x  3
Sol.: So
a: x < 3
&
x=-3 Df={3,-3}

Da=(-∞,3)
Da
-∞
and
b: -x < 3 (multiply by -1)
Db
-∞
 x > -3  Db = (-3,∞)
 Df = Da  Db = (-3,3)
University of Kufa\Civil Eng. ………………
Da  Db -∞
+3
-3
-3
+∞
+∞
+3
(18) ……………………. Mathematics \1st class
+∞
Chapter One
Prerequisites for Calculus
3. x  3
Sol.: Note (solution must consist the remaining part of the real numbers line of
the previous example)
a: either x ≥ 3
So

Da=[3,∞)
b: or -x ≥ 3 (multiply by -1)  x ≤ -3

Db=(-∞,-3]
 Df = Da  Db = (-∞,-3]  [3, ∞)
=R\(-3,3)
Da
-∞
Db
-∞
Da  Db -∞
4. 2 x  3  7
Sol.:  7  2x  3  7
 4  2x  10
+3
-3
-3
+∞
+∞
+3
+∞
(6th property)

2  x  5
 Df = [-2,5]
5. x  9  3
Sol.: x  9  3
 x  12
or x  9  3 (7th property)
or x  6
 Df = (-∞,6)  (12, ∞)
6. x  3  x  1
Sol.:
x3
x 1
1

x3
1
x 1

1 
(divided by x  1 )
(3rd property)
x3
 1 (6th property)
x 1
a.
x3
1
x 1

x  3  ( x  1)
0
x 1

x  3  x 1
0 
x 1

x3
1  0
x 1
2
0
x 1
 Da = (1, ∞)
b.
Sign of (-2)
x3
x3
 1 
1  0
x 1
x 1
-∞
-2
---------------------------------
+∞
+1
+++++ +∞
Sign of (x-1) -∞ -------------------------2
+1
-2
+++++++++++++++
Sign of
---------- +∞
∞
x  1
+1
+++++++++++++ +∞
Sign of (x-1) -∞ ------+2
++++++ +∞
Sign of (x-2) -∞ ----------------+2
+1
-------++++++
∞
++++++ +∞
Sign of (x-1)(x-2)
University of Kufa\Civil Eng. ………………
(19) ……………………. Mathematics \1st class
Chapter One

Prerequisites for Calculus
x  3  x 1
0
x 1

 Db = (- ∞,1)  (2, ∞)
2x  4
2( x  2 )
x2
0 
0 
0
x 1
x 1
x 1
+1
Da -∞
a.

x4
2
x3
x4
2
x3



2 x
0
x3

+∞
x4
2  0
x3
x  4  2x  6
0
x3
b.
+∞
x4
 2
x3
or
x  4  2( x  3)
0
x3
 Da = [2,3)
+2
+2
Df=Da  Db-∞
x4
2
7.
x3
Sol.: either
+1
Db -∞
 Df = Da  Db = (2, ∞)
+∞
Sign of (2-x) -∞ +++++++
+2
------------------------- +∞
+3
-----------------++++++ +∞
Sign of (x-3) -∞
2  x -∞ ---------- +2+++++ +3-------------- +∞
Sign of
x3
(x≠3 because denominator should not equal zero.)
x4
 2 
x3
x4
2 0
x 3
x  4  2( x  3)
0
x3
x  4  2x  6
0
x3


+10/3
+++++++ +∞
Sign of (3x-10) -∞-----------------------+3
++++++++++++++ +∞
Sign of (x-3) -∞ --------+3-------- +10/3
++++++ +∞
Sign of (3x-10)(x-3) -∞ ++++++
3 x  10
0
x3
Da -∞
 Db = (3, 10/3] (x≠3)
Db -∞
So Df = Da  Db = [2,10/3]\{3}
Da  Db-∞
+2
+3
+∞
+2
+3 +10/3
+∞
+2
+10/3
+∞
+3
Homework: Find the values of x that satisfies the inequality.
1. |x-5| < 9
2.
3x  1
1
2
4. |x-3| ≤ |2x-6|
5.
x3
2
6  5x
University of Kufa\Civil Eng. ………………
3.
x
1  1
2
(20) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
The Greatest Integer Function (Stepped Function):
The function whose values at any number x is the greatest integer less
than or equal to x is called greatest integer
function. It is denoted x , or in some books [x] or
[[x]] or int x
The greatest integer function:
[x] ≤ x <[x] +1
Example 1: Find the integer of the following:
[2.4]=2,
[1.9]=1,
[0.1]=0,
[-1.2]=-2,
[-0.3]=-1,
[-2.0]=-2
[0.0]=0,
Example2: Find the interval of the following functions:
1. [x]=2
Sol. 2  x  3
2. [2x]=1
Sol. 1  2 x  2 
1
 x 1
2
1 
3.  x   2
3 
1
3
Sol. 2  x  3  6  x  9
4.
2x  1
1
2
Sol.  1  2 x  0    x  0
Properties of greatest integer value:
1. [[[[x]]]] = [x]
2. [x+n] =[x] + n where n is integer
3. [x-n] =[x] - n where n is integer
4. -[x] ≠ [-x]
University of Kufa\Civil Eng. ………………
(21) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Function Defined in Pieces:
While
some
functions
defined
by
single
formulas, others are defined by applying different
formulas to different parts of their domains. One
example is the absolute value function
 x if
y  x  x2  
 x if
x0
x0
Example: the values of the function
 x if

f ( x)   x 2 if
 1 if

x0
0  x 1
x 1
Example: Graph the function
y  f ( x)  x  3  x  2
Sol. Recall the definition of absolute value:
 x if
y  x  x2  
 x if
x0
x0
 ( x  3) if
 ( x  3) if
( x  3)  0
( x  3)  0
It follows that: x  3  
 x  3 if

 x  3 if
Similarly
 ( x  2) if
x2 
 ( x  2) if
 x  2 if

 x  2 if
x3
x3
( x  2)  0
( x  2)  0
x  2
x  2
These expressions show that we must consider three cases:
x < -2,
-2 ≤ x < 3
Case I: if x < -2 we have
and
x≥3
f ( x)  x  3  x  2
f ( x)   x  3  x  2
f ( x)  2 x  1
University of Kufa\Civil Eng. ………………
(22) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Case II: if -2 ≤x < 3 we have f ( x)  x  3  x  2
y
f ( x)   x  3  x  2
12
11
f ( x)  5
10
f(x)=-2x+1
f(x)=2x-1
9
8
Case III: if x ≥ 3 we have
f ( x)  x  3  x  2
7
6
f(x)=5
5
f ( x)  x  3  x  2
4
3
f ( x)  2 x  1
2
1
Thus
 2 x  1, if

f ( x)   5,
if
 2 x  1, if

x  2
2  x  3
x3
University of Kufa\Civil Eng. ………………
x
0
-6
-5
-4
-3
-2
-1
-1
0
1
2
3
4
-2
-3
(23) ……………………. Mathematics \1st class
5
Chapter One
Prerequisites for Calculus
Sums, Difference, Product and Quotients of Functions:
Definition: If f and g are functions, then we define the functions
 (f+g)(x)= f(x)+g(x)
Sum
Difference  (f-g)(x)= f(x)-g(x) or (g-f)(x)= g(x)-f(x)
Product
 (f.g)(x)= f(x).g(x)
Quotient
 (f/g)(x)= f(x)/g(x); where g(x) ≠0
…(1)
or (g/f)(x)= g(x)/f(x); where f(x) ≠0 …(2)
are also functions of (x), defined for any value of x that lies in both Df and Dg
( x  D f  Dg ), except the points which g(x) =0 in eq.(1) or f(x) =0 in eq.(2)
Example 1: Combining Functions Algebraically:
The function defined by the formulas
f ( x)  x
and g ( x)  1  x
have domains D( f )  [0, ) and D( g )  (,1] . The points common to these
domains are the points
[0, )  (,1]  [0,1]
The following table summarizes the formulas and domains for the various
algebraic combinations of the two functions. We also write f * g for the product
function fg.
Function Formula
Domain
f g
( f  g )( x)  x  1  x
[0,1]  D( f )  D( g )
f g
( f  g )( x)  x  1  x
[0,1]
g f
( g  f )( x)  1  x  x
[0,1]
f *g
( f * g )( x)  f ( x) g ( x)  x(1  x)  x  x 2
[0,1]
f
g
f
f ( x)
x
( x) 

g
g ( x)
1 x
f
g
g ( x)
1 x
( x) 

f
f ( x)
x
g
University of Kufa\Civil Eng. ………………
[0,1) ( x  1 excluded )
(0,1] ( x  0 excluded )
(24) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Example 2: Give the domain of f(x) and g(x) and the corresponding domains of
where f ( x)  4  x 2 and g(x) = 3x+1
f+g, f-g, g-f, f.g, f /g and g /f
Sol. The domain of f(x) (Df) is:
4 - x2 ≥ 0

 (2 – x) (2 + x) ≥ 0
Df=[-2,+2]
Sign of (2-x) -∞
The domain of g(x) (Dg=R)
So
Sign of (2+x) -∞
D f  Dg  [2,2]
Sign of (2-x) (2+x)
Function
Formula
-∞
+++++++++++++
+2
---------- +∞
-2
+++++++++++++ +∞
+2
0
--------- -2 ++++++
--------- +∞
----------
Domain
f+g (x) =
4  x 2  (3x  1)
[-2,+2]
f-g (x) =
4  x 2  (3x  1)
[-2,+2]
g-f (x) =
(3x  1)  4  x 2
[-2,+2]
f.g (x) =
4  x 2 .(3x  1)
[-2,+2]
f
(x) =
g
4  x2
3x  1
[-2,+2]\{-1/3}
g
(x) =
f
3x  1
4  x2
(-2,+2)
Homework: Give the domains of f and g and the corresponding domains of f+g,
f-g, f.g, f/g, and g/f for the following:
1. f ( x)  3x 2 ;
2. f ( x)  x 
1
;
x
g ( x) 
1
.
2x  3
1
g ( x)  x  .
x
3. f ( x)  x  3 ;
g ( x)  x  3 .
4. f ( x)  x3  3x ;
g ( x)  3x 2  1 .
5. f ( x)  x 2  4 ;
g ( x)  7 x 2  1
University of Kufa\Civil Eng. ………………
(25) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Composition of Functions:
Definition: If f and g are functions, the composite f o g "f composed with g"
or g o f "g composed with f" are defined by:
(f o g)(x) =f (g(x))
and
(g o f)(x) =g (f(x)) respectively.
Examples 1: Find the formula for f(g(x)) and g(f(x)) if g(x) = x2 and f(x) =x-7, then
find the value of f(g(2)) and g(f(2))
Sol.: a: For f(g(x))=?,
f(x) = x-7
f(g(x)) = (g(x))-7 = x2-7 
b: For g(f(x))=?,
f(g(2)) = 22-7 = 4-7 = -3
g(x) = x2
g(f(x)) = (f(x))2 = (x-7)2 
g(f(2)) = (2-7)2= (-5)2= 25
Examples 2: If f(x) =x2+1 and g ( x)  x , find
a: (f o g)(x) and (g o f)(x)?
b: the domains of (f o g)(x) and (g o f)(x)?
Sol.: (f o g)(x)= f(g(x))  f ( x )   x   1  x  1

D fo g  Dg  {x : x  0}
(go f)(x)=g(f(x))  g ( x 2  1)  x 2  1

Dgo f  D f  R
2
Examples 3: Finding formulas for composites:
If f ( x)  x
and g ( x)  x  1, find
(a) (f o g)(x)
(b) (g o f)(x)
(c) (f o f)(x)
(d) (g o g)(x)
Sol.:
Composite
(a) (f o g)(x)
Domain
[-1, )
 f ( g ( x))  g ( x)  x  1
[0, )
(b) (g o f)(x)  g ( f ( x))  f ( x)  1  x  1
[0, )
1
(c) (f o f)(x)  f ( f ( x))  f ( x) 
x x 4
(d) (g o g)(x)  g ( g ( x))  g ( x)  1  ( x  1)  1  x  2
(-, )
Example 4: If f ( x)  x and g ( x)  2  x , find each function and its domain
(a) fog
(b) gof
c) fof
(d) gog
Sol.:
(a)
( f o g )( x)  f ( g ( x))  f ( 2  x ) 
University of Kufa\Civil Eng. ………………
2 x  4 2 x
(26) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
The domain of fog ={x: 2-x ≥ 0}={x :x ≤ 2}=(-∞,2].
(b)
( go f )( x)  g ( f ( x))  g ( x )  2  x
For
x to be defined we must have x ≥ 0. For
have 2  x  0 , that is,
2  x to be defined we must
x  2 , or x  4 . Thus, we have 0 ≤ x ≤ 4, so the domain
of gof is the closed interval [0,4].
(c)
( f o f )( x)  f ( f ( x))  f ( x ) 
x 4 x
The domain of fof is [0,∞)
(d)
( go g )( x)  g ( g ( x))  g ( 2  x )  2  2  x
This expression is defined when 2 – x ≥ 0, that is x ≤ 2, and 2  2  x  0 . This
latter inequality is equivalent to 2  x  2 , or 2 – x ≤ 4, that is, x ≥ -2. Thus,
-2
≤ x ≤ 2, so the domain of gog is the closed interval [-2,2].
Homework: Find (fog), (gof), (fof) and (gog) of the following functions:
1. f ( x)  x 3 ;
g ( x)  x 2  3 .
2. f ( x)  3x 2  2 ;
g ( x) 
3. f ( x)  2 x  1 ;
g ( x)  x 2  3 .
4. f ( x)  7 ;
g ( x)  4 .
5. f ( x)  x 3 ;
g ( x)  sin x  3 .
6. f ( x) 
1
;
1 x
1
.
3x  2
2
g ( x)  3 x .
7. f ( x)  3x 2  x ;
g ( x)  2 x  1 .
8. If f ( x)  x  5 and g ( x)  x 2  3 ; Find
(a) f (g (0))
(b) g ( f (0))
University of Kufa\Civil Eng. ………………
(c) f ( f (5))
(d) g (g (2))
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Chapter One
Prerequisites for Calculus
Graph of Functions (Graph of Curves):
To graph the curve of a function, we can follow the following steps:
1. Find the domain and range of the function.
2. Check the symmetry of the function
3. Find (if any found) points of intersection with x-axis and y-axis.
4. Choose some another points on the curve.
5. Draw s smooth line through the above points.
Symmetry Tests for Graphs:
If f(x,y) = 0 is any function then:
1. Symmetry about x-axis: If f(x,-y)= f(x,y)
2. Symmetry about y-axis: If f(-x,y)=f(x,y) It is called an even function.
3. Symmetry about the origin: If f(-x,-y)=f(x,y) It is called an odd function
Examples 1: Check the symmetry of the graph of the following curves:
1. y = f(x)=x2
sol.: f(x,y) = x2- y =0

. f(x,-y)= x2+ y =0 ≠ f(x,y) not o.k.
(ii) f(-x,y)= (-x)2- y =0

. f(-x,y)= x2- y =0= f(x,y) o.k.
(iii) f(-x,-y)= (-x)2-(-y)=0

. f(-x,-y)= x2+ y =0 ≠ f(x,y) not o.k.
 (i) f(x,-y)= x2- (-y) =0
So the function has symmetry only about y-axis. It is called an even function.
2. y= f(x)=x3
Sol. f(x,y) = x3- y =0

. f(x,-y)= x3+ y =0 ≠ f(x,y) not o.k.
(ii) f(-x,y)= (-x)3- y =0

. f(-x,y)= -x3- y =0 ≠ f(x,y) not o.k.
(iii) f(-x,-y)= (-x)3-(-y)=0

. f(-x,-y)= -x3+ y =0 (multiply by -1)
 (i) f(x,-y)= x3- (-y) =0

f(-x,-y)= x3- y =0= f(x,y) o.k.
So the function has symmetry only about the origin. It is called an odd function.
3. x2 =y2 + 4
University of Kufa\Civil Eng. ………………
(28) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Sol. f(x,y) = y2 –x2+4=0
 (i) f(x,-y)= (-y)2 –x2+4=0

. f(x,-y)= y2 –x2+4=0= f(x,y) o.k.
(ii) f(-x,y)= y2 –(-x)2+4=0

. f(-x,y)= y2 –x2+4=0= f(x,y) o.k.
(iii) f(-x,-y)= (-y)2 –(-x)2+4=0

. f(-x,-y)= y2 –x2+4=0= f(x,y) o.k.
So the function has symmetry about x-axis, y-axis and the origin.
DEFINITIONS
Even Function, Odd Function
A function y = f(x) is an
even function of x if f(-x) = f(x) symmetry about y-axis
odd function of x if f(-x) = -f(x) symmetry about origin
for every x in the function's domain.
Examples 2: Recognizing Even and Odd functions
o
f(x)= x2Even function:
o
f(x)= x2 + 1
(-x)2= x2 for all x; symmetry about y-axis.
Even function:
(-x)2 +1= x2 +1 for all x; symmetry about y-
axis.
o
f(x)= x Odd function:
o
f(x)= x + 1
(-x)= x for all x; symmetry about the origin.
Not odd:
f(-x)=-x+1, but -f(x)=-x-1. The two are not equal.
Not even:
f(-x)=-x+1, but f(x)=x+1. for all x ≠ 0.
Example 3: Sketch the graph of the curve y = f(x) = x2-1
Sol.: Step 1: Find Df, Rf of the function?
University of Kufa\Civil Eng. ………………
(29) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Df=(-∞,∞);
To find Rf : we must convert the function from y=f(x) into x=f(y).
y = x2-1 
So
x   y 1
x2= y+1

y +1 ≥ 0

y ≥ -1
R=[-1,∞)
Step 2: Find x and y intercept?
To find x-intercept put y=0 
x2-1=0 
x2=±1
So x-intercept are (-1,0) and (+1,0).
To find y-intercept put x=0 
y = 0-1 
y = -1
So y-intercept is (0,-1).
y
8
Step 3: check the symmetry:
(3,8)
7
f(-x) = (-x)2-1= x2-1= f(x)
6
5
- f(x) =-(x2-1)= - x2+1≠ f(x)
4
So it is an even function ( it is symmetric
(2,3)
3
2
about y-axis).
1
(-1,0)
Step 4: Choose some another point on the
-3
-2
(1,0)
0
-1
0
-1
1
x
2
3
(0,-1)
-2
curve.
x y
2 3
3 8
Step 5: Draw smooth line through the above points.
Homework: Draw the following functions:
1. y  f ( x)  3x 2  2
2. x 2  y 2  1
3. y 2  4 x  1
4. x  y 3
5. y  [x] ; for -3 ≤ x ≤ 3
6. y  x  [x] ; for -2 ≤ x ≤ 2
7. y  4  x
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(30) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Shifting, Shrinking and Stretching:
Shift formulas: (for c > 0)
Vertical shifts
y= f(x)+c
or
y-c= f(x)
y= f(x)-c
or
y+c= f(x)
Horizontal shifts
shifts the graph of f up by c units.
shifts the graph of f down by c units.
shifts the graph of f left by c units.
shifts the graph of f right by c units.
y= f(x+c)
y= f(x-c)
Shrinking, Stretching and Reflecting Formulas:
(for c > 1)
y=c f(x)
Stretches the graph of f c units along y-axis.
y
1
f ( x)
c
y= f(cx)
x
y f( )
c
(for c = -1)
y=- f(x)
y= f(-x)
Shrinks the graph of f c units along y-axis.
Shrinks the graph of f c units along x-axis.
Stretches the graph of f c units along x-axis.
Reflects the graph of f across the x-axis.
Reflects the graph of f across the y-axis.
Example 1: The graph of
y x
is a
reflection of y  x across the xaxis, and y   x is a reflection
across the y-axis.
Example 2: Shift the graph of the function
f(x) = x2 ; if Df={x: -2 ≤ x ≤ 3} and
Rg={y: 0 ≤ y ≤ 9}.
(a) one unit right.
(b) two units left.
(c) one unit up.
(d) two units down.
Sol.: (a) Shifting the function f(x) one unit right:
g(x) = f(x-1) = (x-1)2 and Dg={x: -2 ≤ x-1 ≤ 3}={x: -1 ≤ x ≤ 4}
Note: In case of horizontal shifts, the range of the function will not be changed.
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Chapter One
Prerequisites for Calculus
x
y=f(x)=x2
x-1
y=g(x)=(x-1)2
-2
-1
0
1
2
3
4
4
1
0
1
4
9
-
-2
-1
0
1
2
3
4
1
0
1
4
9
y
g(x)=f(x-1)
9
f(x)
8
7
6
5
4
3
2
(b) Shifting the function f(x) two units left:
1
-2
h(x) = f(x+2) = (x+2) and Dh={x: -2 ≤ x+2 ≤ 3}={x: -4 ≤ x ≤ 1}
2
0
-1
x
0
1
2
3
4
5
-1
-2
Note: In case of horizontal shifts, the range of the function will not be changed.
x
-4
-3
-2
-1
0
1
2
3
y=f(x)=x2
4
1
0
1
4
9
x+2
-2
-1
0
1
2
3
-
y=h(x)=(x+2)2
4
1
0
1
4
9
-
y
h(x)=f(x+2)
9
f(x)
8
7
6
5
4
3
2
1
(c) Shifting the function f(x) one unit up:
-4
-3
-2
-1
0
x
0
1
2
3
4
5
-1
-2
w(x) = f(x)+1 = x2 +1 and Rw={y: 0 ≤ y-1 ≤ 9}={y: 1 ≤ y ≤ 10}
Note: In case of vertical shifts, the domain of the function will not be changed.
y
x
-2
-1
0
1
2
3
y=f(x)=x2
4
1
0
1
4
9
y=w(x)= x2 +1
5
2
1
2
5
10
10
w(x)=f(x)+1
9
f(x)
8
7
6
5
4
3
2
1
(d) Shifting the function f(x) two units down:
-3
q(x) = f(x)-2 = x2 -2 and Rq={y: 0 ≤ y+2 ≤ 9}={y: -2 ≤ y ≤ 7}
University of Kufa\Civil Eng. ………………
-2
-1
0
x
0
1
2
3
-1
-2
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4
Chapter One
Prerequisites for Calculus
y
y=f(x)=x2
4
1
0
1
4
9
x
-2
-1
0
1
2
3
y=q(x)= x2 -29
8
2
7
-1
6
5
-2
4
-1
3
2
2
1
7
0
-3
-2
-1
f(x)
q(x)=f(x)-2
x
0
1
2
3
4
-1
-2
Example 3: Sketch the graph of the curve y= f(x) = |x|
-3
-4
Sol.: Step1: Find Df, Rf of the function?
 x if
y  f ( x)  x  
 x if

Df=(-∞,∞)
and
x0
x0
Rf= [0,∞);
Step2: Find x and y intercept?
To find x-intercept put y=0 
x=0
To find y-intercept put x=0 
y=0
So x- and y-intercept is (0,0).
Step 3: check the symmetry:
f(-x) = |-x|=|x|= f(x)
- f(x) =-|x|≠ f(x)
So it is an even function (it is symmetric about y-
y
f(x)=-x when x<0
f(x)=x when x>=0
3
axis).
2
Step 4: Choose some another point on the curve.
x y
1 1
2 2
Step 5: Draw smooth line through the above
points.
University of Kufa\Civil Eng. ………………
1
x
0
-3
-2
-1
0
1
2
-1
(33) ……………………. Mathematics \1st class
3
Chapter One
Prerequisites for Calculus
Example 4: Use graph of the function y=|x| to sketch the graph of the following
functions, then show their domains and range
(a) y=|x+1|
y
f(x)=-(x+1) when x+1<0
f(x)=x+1 when x+1>=0
Sol.
4
 ( x  1) if
y  x 1  
 ( x  1) if
( x  1) if

 x  1 if
( x  1)  0
( x  1)  0
3
2
x  1
x  1
1
Df=(-∞,∞) and Rf=[0,∞)
(b)
x
0
Shifting the function y=|x| one unit left.
-5
-4
-3
-2
-1
0
1
2
3
4
(-1,0)
y
-1
f(x)=-x+2 when x<0
f(x)=x+2 when x>=0
7
y=|x|+2
6
 ( x)  2 if
( x)  2 if
Sol. y  x  2  
( x)  0
( x)  0
5
4
3
Shifting the function y=|x| two up.
(0,2) 2
1
Df=(-∞,∞) and Rf=[2,∞)
-5
(c) y=-|x|
-4
-3
-2
-1
0 y
0
x
1
2
3
4
5
-1
1
 ( x)   x if
 ( x)  x if
Sol. y  f ( x)   x  
( x)  0
( x)  0
-2
x
0
-3
-2
-1
0
Reflecting the graph of the function y=|x|
1
2
-1
f(x)=-x when x>=0
f(x)=x when x<0
across x-axis.
3
-2
Df=(-∞,∞) and Rf=(-∞,0]
-3
y
(d) y=2-|1-x|
3
Sol. y=2-|1-x|=-|1-x|+2|=-|x-1|+2
if
  ( x  1)  2

 (( x  1))  2 if
 x  3 if

 x  1 if
( x  1)  0
x 1  0
x 1
x 1
(1,2)
2
f(x)=x+1 when x<1 1
f(x)=-x+3 when x>=1
x
0
-3
Reflecting the graph of the function y=|x| across x-
-2
-1
0
1
2
3
-1
-2
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4
5
Chapter One
Prerequisites for Calculus
axis, then shifting it one unit right and two units up.
Df=(-∞,∞) and Rf=(-∞,2]
(e) y=1-|x+1|
y
Sol. y=1-|x+1|=-|x+1|+1
2
  ( x  1)  1 if

 (( x  1))  1 if
  x if

 x  2 if
(-1,1)
( x  1)  0
( x  1)  0
x
0
-5
x  1
x  1
1
-4
-3
-2
-1
0
1
-1
f(x)=x+2 when x<-1
2
3
4
f(x)=-x when x>=-1
-2
Reflecting the graph of the function y=|x|
-3
across x-axis, then shifting it one unit left
-4
and one unit up.
Df=(-∞,∞) and Rf=(-∞,1]
Example 5: If f ( x)  4  x 2 which has Df=[-2,2] and Rf=[0,2], shrink and stretch it
horizontally by two units and then
sketch
the
original
and
y
resulting
4
functions
3
Sol.: (a) shrinking:
g ( x)  f (c.x) 
4  ( 2 x) 2 
f(x)=sqrt(4-x^2)
2
4  4x  2 1  x2
g(x)=f(2x)=sqrt(4-(2x)^2)
1
Dg={x: -2 ≤ 2x ≤ 2}={x: -1≤ x ≤ 1}
x
0
-2
-1
0
Note: In case of horizontal shrinks, the range
-1
of the function will not be changed.
-2
(b) stretching:
1
2
y
4
x
x
x2
16  x 2 1
g ( x)  f ( )  4  ( ) 2  4 


16  x 2
c
2
4
4
2
Dg={x: -2 ≤ x/2 ≤ 2}={x: -4≤ x ≤ 4}
3
g(x)=f(x/2)=sqrt(4-(x/2)^2)
2
Note: In case of horizontal stretches, the range
1
f(x)=sqrt(4-x^2)
of the function will not be changed.
x
0
-4
-3
-2
-1
0
1
2
3
-1
-2
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(35) ……………………. Mathematics \1st class
4
Chapter One
Prerequisites for Calculus
y
Example 6: Repeat the above example but here shrink
4
and stretch the function vertically.
3
Sol.: (a) shrinking:
f(x)=sqrt(4-x^2)
2
1
1
g ( x)  f (. x) 
4  x2
c
2
g(x)=(1/2)f(x)=(1/2)sqrt(4-x^2)
1
Rg={y: 0 ≤ 2y ≤ 2}={y: 0 ≤ y ≤ 1}
x
0
-2
-1
0
Note: In case of vertical shrinks, the domain of
-1
the function will not be changed.
-2
1
2
y
(b) stretching:
g(x)=2*f(x)=2*sqrt(4-x^2)
4
g ( x)  cf (.x)  2 4  x 2
3
f(x)=sqrt(4-x^2)
2
Rg={y: 0 ≤ y/2 ≤ 2}={y: 0 ≤ y ≤ 4}
1
Note: In case of vertical stretches, the domain of the
x
0
function will not be changed.
-2
-1
0
1
2
-1
-2
Example 7: Use the graph of the function
y
1.5
y  f ( x)  1  x 2 to sketch the graph of
1.0
y=f(x)
the following functions:
y=g(x)=f(2x)
1. y  g ( x)  1  4 x
0.5
2
Sol.: y  1  4 x 2  1  (2 x) 2
-1.0
x
0.0
0.0
-0.5
This function may be obtained by shrinking
0.5
1.0
-0.5
the function f ( x)  1  x 2 by two units horizontally ( g ( x)  f (2 x) ).
4.5
2. y  h( x)  1 
Sol.: y  1 
2
x
9
4.0
3.5
3.0
2.5 y
x2
x
 1  ( )2
9
3
2.0
1.5
This function may be obtained by stretching
the function f ( x)  1  x 2 by three units
1.0
0.5
y=h(x)=f(x/3)
y=f(x)
y=g(x)=f(2x)
0.0
-3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0
-0.5
x
x
3
horizontally ( h( x)  f ( ) ).
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Chapter One
3. y  w( x) 
Prerequisites for Calculus
1
1  x2
3
Sol.: y  w( x) 
1
1  x2
3
This function may be obtained by shrinking the
y
4.5
function f ( x)  1  x 2 by three units vertically
( h( x ) 
4.0
y=q(x)=4.f(x/2)
3.5
1
f ( x) ).
3
3.0
2.5
2.0
x2
4. y  q( x)  4 1 
4
1.5
y=f(x)
1.0
y=w(x)=(1/3)f(x)
0.5
x2
x
Sol.: y  q( x)  4 1   4 1  ( )2
4
2
-2.0
-1.5
x
0.0
-0.5 0.0
-0.5
-1.0
0.5
1.0
1.5
2.0
This function may be obtained by stretching the function f ( x)  1  x 2 by
x
2
two units horizontally and four units vertically ( q( x)  4. f ( ) ).
Homework:
y
1. Sketch the graph of the following curves by shifting, reflecting,
9
8
7
shrinking and stretching the graph of the given functions
6
5
4
appropriately.
3
2
y=x2
1
(a) The given function y=x2
x
0
-3
-2
-1
0
1
2
3
-1
(i) y=1+(x-2)2
(ii) y=2-(x+1)2
(iii) y=-2(x+1)2-3
(iv) y=(1/2)(x-3)2+2
(v) y=x2 + 6x
-2
(vi) y=x2 + 6x -10
y
(b) The given function y  x
(i) y  3  x  1
(ii) y  1  x  4
1
x 1
(iii) y 
2
4
3
(iv) y   3x
1
x
0
-2
-1
0
y
1
(c) The given function y 
x
1
(i) y 
x3
1
(iii) y  2 
x 1
y x
2
1
2
3
4
-1
40
-2
30
1
(ii) y 
1 x
x 1
(iv) y 
x
20
y=1/x
10
x
0
-3
-2
-1
0
1
2
3
-10
-20
-30
-40
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Chapter One
Prerequisites for Calculus
y
(d) The given function y=|x|
(i) y=|x+2|-2
3
(ii) y=1-|x-3|
2
(iv ) y  x  4 x  4
2
1
(iii) y=|2x-1|+2
(e) The given function y  3 x
(i) y  1  23 x
(iii) y  2  3 x  1
 ( x  2) 2  x  2
y=|x|
x
0
-3
-2
-1
0
1
2
3
-1
(ii) y  x  1  3
(iv) y  3 x  2
3
2. Shrink and stretch the following functions along both xaxis and y-axis by (3/2) units then sketch the resulting function.
(a) x2 + y2 = 4,
2
2
(b) 2x + y /2 = 6,
Df={x: -2 ≤ x ≤ 2}
Rf={y: -2 ≤ y ≤ 2}
Df={x: -2 ≤ x ≤ 3}
Rf={y: -2 ≤ y ≤ 2 6 }
(c) y=3x2 – 2x +1,
Df={x: -1 ≤ x ≤ 2}
Rf={y:
University of Kufa\Civil Eng. ………………
6
≤ y ≤ 9}
9
(38) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Trigonometric Functions
We measure angles in degrees, but in calculus it is usually best to use
radians.
1. Degree measure: One degree (1o) is the measure of an angle generated
by 1/360 of revolution.
2. Radian measure: The radian measure of the angle ACB at the center of
the unit circle (circle with radius equals one unit)
equals to the length of the arc that the angle cuts
from the unit circle.
If angle ACB cuts an arc A`B` from a second circle
centered at C, then circular sector A`CB` will be similar to
circular sector ACB. In particular,
Length of arc A`B`
Length of arc AB

Radius of sec ond circle Radius of first circle
In notations
s 
 
r 1


s
r
or
s=r
To find the relation between degree measure and radian measure, you know that
one circle equals 360o in degree and 2 in radians so:

2 radians= 360o
 1o 

180
 radians= 180o
rad  0.01745rad
o
180
and 1rad     57o17'44.8"
  
The six basic trigonometric functions:
sine: sin  
y
,
r
cosecant: csc 
x
r
cosine: cos  ,
tangent: tan  
secant: sec  
1
r
 ,
sin  y
1
r

cos x
sin 
y
1
cos  x
 , cotangent: cot  


cos  x
tan  sin 
y
from Pythagoras theorem:
University of Kufa\Civil Eng. ………………
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Chapter One
2
2
x +y =r

Prerequisites for Calculus
x2  y 2
1
r2

2
cos2sin2

x2 y 2

1
r2 r2
true for all values of  


When we divided eq.(1) by cos2yields:
cos 2 sin 2
1


2
2
cos  cos  cos 2

1tan2 sec2
And when we divided eq.(1) by sin2yields:
cos 2 sin 2
1


2
2
sin  sin  sin 2

cot2 +1 csc2
Identities:
- Periodicity: A function is periodic with period p if
f(x+p)= f(x)
for every value of x.
cos(±2) =cos
sin(±2) =sin
tan(±2) =tan
cot(±2) =cot
sec(±2) =sec
csc(±2) =csc
- Symmetry:
Even functions
Odd functions
cos(-x)=cosx
sec(-x)=secx
sin(-x)=-sinx
tan(-x)=-tanx
cot(-x)=-cotx
csc(-x)=-cscx
- Shift formulas:
sin(x + /2) = cos(x);
cos(x + /2) = -sin(x)
sin(x - /2) = -cos(x);
cos(x - /2) = sin(x)
- Addition formulas:
cos(A+B)=cosA cosB – sinA sinB
sin(A+B)=sinA cosB + cosA sinB
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Chapter One
Prerequisites for Calculus
- Double angle formulas:
cos2=cos2sin2
sin2 2.sin .cos
- Half angle formulas:
cos 2  
1  cos 2
2
sin 2  
1  cos 2
2
Graph of trigonometric functions:
1. y = sin x
 Domain and Range of the function
v
Df=(-∞,∞)
u
From Figure nearby, we conclude that:
-r ≤ v ≤ r
1 
v
1
r
(divide the inequality by r)

1  sin   1
Rf=[-1,1]
 Symmetry:
f(-x) = sin(-x) =-sinx = - f(x)
≠ f(x)
So it is an odd function (it is symmetric
about the origin).
 Additional points:
0
0
x
y

1

0
3
-1
2
0
2. y = cos x
 Domain and Range of the function
Df=(-∞,∞)
From Figure, we conclude that:
-r ≤ u ≤ r
1 
u
1
r
(divide the inequality by r)

1  cos  1
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Chapter One
Prerequisites for Calculus
Rf=[-1,1]
 Symmetry:
f(-x) = cos(-x) =cosx = f(x)
≠- f(x)
So it is an even function (it is symmetric
about the y-axis).
 Additional points:
0
1
x
y
3. y = tan x =

0

-1
3
0
2
1
sin x
cos x
 Domain and Range of the function
cosx ≠ 0 
Df= R\{ x  n

2
xn

2
; n=±1, ±3, ±5
; n=±1, ±3, ±5}
Rf=(-∞,∞)
 Symmetry:
f(-x) = tan(-x) =
sin(  x)  sin x
=-tanx ≠ f(x)

cos(  x)
cos x
=- f(x)
So it is an odd function (it is symmetric about the origin).
 Asymptotes:
To find vertical asymptote put the denominator equal to zero.

cosx = 0
xn

2
; n=±1, ±3, ±5
 Additional points:
x
y
0
0

±∞

0
3
±∞
University of Kufa\Civil Eng. ………………
2
0
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Chapter One
Prerequisites for Calculus
4. y = cot x =
cos x
sin x
 Domain and Range of the function
x=n ; n=0, ±1, ±2, ±3

sinx ≠ 0
Df= R\{x=n; n=0, ±1, ±2, ±3}
Rf=(-∞,∞)
 Symmetry:
cos(  x)
cos x
=-cotx ≠ f(x)

sin(  x)  sin x
f(-x) = cot(-x) =
=- f(x)
So it is an odd function (it is symmetric about the origin).
 Asymptotes:
To find vertical asymptote put the denominator equal to zero.
x≠n ; n=0, ±1, ±2, ±3

sinx = 0
 Additional points:

0
0
±∞
x
y
5. y = sec x =

±∞
3
0
2
±∞
1
cos x
 Domain and Range of the function
cosx ≠ 0
xn

Df= R\{ x  n

2

2
; n=±1, ±3, ±5
; n=±1, ±3, ±5}
From Figure, we conclude that:
-r ≤ u ≤ r
1 
(divide the inequality by r)
u
1
r
 sec  1 

1  cos  1
sec   1
or

cos  1

1
1
cos 
sec   1
Rf= R\(-1,1)
 Symmetry:
University of Kufa\Civil Eng. ………………
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Chapter One
Prerequisites for Calculus
f(-x) = sec(-x) =
1
1
=sec x = f(x)

cos(  x) cos x
≠- f(x)
So it is an even function (it is symmetric about the y-axis).
 Asymptotes:
To find vertical asymptote put the denominator equal to
zero.
cosx = 0
xn


2
; n=±1, ±3, ±5
 Additional points:
x
y
6. y = csc x =

±∞
0
1

-1
3
±∞
2
1
sin   1

1
sin x
 Domain and Range of the function
sinx ≠ 0

x≠n; n=0, ±1, ±2, ±3
Df= R\{x=n; n=0, ±1, ±2, ±3}
From Figure, we conclude that:
-r ≤ v ≤ r
1 
(divide the inequality by r)
v
1
r
 csc  1 

1  sin   1
csc  1
or

1
1
sin 
csc  1
Rf= R\(-1,1)
 Symmetry:
f(-x) = csc(-x) =
1
1
=-cscx ≠ f(x)

sin(  x)  sin x
=- f(x)
So it is an odd function (it is symmetric about the
origin).
 Asymptotes:
To find vertical asymptote put the denominator
equal to zero.
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Chapter One
Prerequisites for Calculus

sinx = 0
x=n ; n=0, ±1, ±2, ±3
 Additional points:
x
y
0
±∞

1

±∞
3
-1
2
±∞
Example: Sketch the graphs of the following functions.
1. y=sin2x
Sol.: We obtain the graph of y=sin2x from that of y=sinx by shrinking it two
units horizontally.
Dg=(-∞,∞);
Rg=[-1,1]
2. y=2cosx
Sol.: In order to get the graph of y=2cosx we
multiply the y-coordinate of each point on the
graph of y=cosx by 2. This means that the
graph of y=cosx gets stretched vertically by a
factor of 2.
Dg=(-∞,∞);
Rg={y: -1 ≤
y
≤1}
2
={y: -2 ≤ y ≤2} = [-2, 2]
3. y=1-sinx
Sol.: To obtain the graph of
y=1-sinx, we again start with
y=sinx. We reflect across the
x-axis to get the graph of and then we shift 1 unit upward to get y=1-sinx.
Dg=(-∞,∞);
Rg={y: -1 ≤ y-1 ≤1}
={y: 0 ≤ y ≤2} = [0, 2]
4. y= 2cos3x,
-2 ≤ 3x ≤ 2
Sol.: We obtain the graph of y=g(x)=2cos3x from that of y=f(x)=cosx by
shrinking it three units horizontally and stretching it two units vertically.
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Chapter One
Prerequisites for Calculus
Dg={x: -2 ≤ 3x ≤ 2} ={x: 
=[ 
2
2
≤x≤
}
3
3
y
2
2 2
,
]
3 3
f(x)=cosx
-6.29
Rg={y: -1 ≤
5. y=sin(x+
-4.71
1
-3.14
0
0.00
-1.57
y
≤1}={y: -2 ≤ y ≤2} = [-2, 2]
2

),
4
g(x)=2f(3x)
x
1.57
3.14
4.71
-1
-2
-2 ≤ x ≤ 2

4
Sol.: We obtain the graph of y=g(x) =sin(x+ ) from that of y=f(x)=sinx by shifting
it

units left.
4
Dg={x: -2 ≤ x+
=[ 
y
9
7

≤ 2} ={x: 
≤x≤
}
4
4
4
9 7
, ]
4 4
g(x)=f(x+0.785)
1
f(x)=cosx
0
x
-6.29 -4.71 -3.14 -1.57 0.00 1.57 3.14 4.71
-1
While the range will not be changed, Rg=[-1,1]
-2
6. y = sin|x|
if
sin( x)
sin(  x)   sin x if
Sol.: y  sin x  
( x)  0
( x)  0
y
g(x)=sin|x|
2
1
Dg = (-∞,∞);
Rg=[-1,1]
-6.29 -4.71 -3.14 -1.57
0
0.00
x
1.57
3.14
4.71
f(x)=sinx-1
-2
7. y= |cosx|
y
 (cos x) if
 (cos x) if
Sol.: y  cos x  
Dg=(-∞,∞);
(cos x)  0
(cos x)  0
2
1
Rg=[0,1]
-6.29 -4.71 -3.14 -1.57
f(x)=cosx
0
0.00
g(x)=|cosx|
x
1.57
3.14
4.71
-1
-2
University of Kufa\Civil Eng. ………………
(46) ……………………. Mathematics \1st class
Chapter One
Prerequisites for Calculus
Homework: Graph the following functions and show domains and ranges of
them.
1. y  sin 2 x
2. y  2 tan x
5. y  x 2  1
6. y  sin
9. y  2 sec x
10. y  (2 x) 2
13. y  2 x  1
14. y 
x
3
3x  x
x
3. y  cot 3x
7. y  2 cos
11. y 
15. y 
x
3
x x
2
x
x
2
4. y 
1  cos 2 x
2
8. y  x 2  4
12. y 
16. y 
cos x  cos x
2
sin x
cos x
x3  x
17. y 
x
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(47) ……………………. Mathematics \1st class