MATH 31 – UNIT 0 - REVIEW
Calculus – A study of functions which can be approximated by functions which are linear.
Factoring – a2 – b2 = (a – b)(a + b)
a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2
a3 – b3 = (a – b)(a2 + ab + b2)
8x3 – 27y3 = (2x)3 – (3y)3 = (2x – 3y)(4x2 + 6xy + 9y2)
a3 + b3 = (a + b)(a2 – ab + b2)
a4 – b4 = (a2 – b2)(a2 + b2) = (a – b)(a + b)(a2 + b2)
Ex.
a) x 2  3x  18
g) 4 x 2  9
b) 2 x 2  7 x  3
h) 3x 2  5 x
c) 6 x 2  7 x  3
i) 81x 4  625
d) 10 x 2  x  2
j) x3  8
e) 56 x 2  65 x  14
k) 27 x 3  1
f) 32 x 2  40 x  8
l) 64 x3  8
Factor Theorum – If a polynomial, P(x), has (x – b) as a factor, the P(b) = 0.
Synthetic division –
x3  27 x3  0 x 2  0 x  27

x 3
x 3
3
1
1
i.e.
since
0
3
3
0
9
9
–27
27
0
Dividend = Divisor  Quotient + Remainder
(x3 – 27) =
(4x3 – 3x2 – 25x – 6) =
(2x3 + 3x2 – 32x – 48) =
1
1a
1b
When factoring fractions: To get rid of the fractions in the final term factor out the least common
multiple of the denominator and the greatest common factor from the numerator. (The new term
GCF
is obtained by multiplying the original term by the reciprocal of this factor). FACTOR=
LCM
1.
1
2
x y 
3
5
2.
2
1
1
x y 
5
7
2
3.
2
1
1
x y 
5
7
2
When taking out the common factor of the variable or a function as a common factor we take out
the least power of the common factor. ( Least = farthest left on the number line)
4.
x5 – 3x2 =
5.
x2 – 2x –3 =
When factoring we can always check by multiplying (Remember we add exponents when
multiplying)
1
2
6.
x2  x 3 
7.
2 43 1 21
x  x  x 
3
2
8.
 2 x 1 4
9.
sin 2x + sin 5x =
1
2
3
7
 4  2 x  1 4 
Read p(2,3)
Do Pg. 3 # (1 – 3) column1, All of 4
2
2a
Rationalizing: To rationalize a binomial either in the numerator or the denominator we
simply multiply by 1 in the form of the conjugate over itself.
( Conjugate = same binomial; middle sign changed)
Rationalize
3
5 2

4




 6 2
x
x4  x4
1.
Rationalize the denominator of
2.
Rationalize the numerator of
x 2
x
3.
Rationalize the numerator of
x 2
x4
3
2b
Rationalize and Simplify.
1.
x
1 1 x
2.
2
1 1

3
x
4.
x 9
x 5
x 5
2
3.
x8 3
x 1
Pg. 4 # (1, 2) col 1
4
3a
Review Properties of Exponents
1. x  x = x
a
b
xa
2.
 x a b
b
x
a+b
3. (xa)b = xab
(xayb)c = xacybc
c
 xa 
x ac

 b
y bc
y 
1
1
or  
a
x
x
x
 
 y
a
4. xa =
a
 y
 
x
 xa
a
5. x0 = 1
x1 = x
6. (x)2 = (x)(x) = x2
x2 = (x)(x)
m
7. x n  n x m or
 x
n
m
eg 1. (3xy2)3(2x)2
3
2. 4 2
3.
x 1  y 1
 x  y
1
Assignment Read pg 350 – 361
pg. 350 # 1 a, b, d, e, f, h, 2 c, d, f, g,
pg 353 # 3 b, g, j; 4 c, f, g, h
pg. 354 # 1 a, j 2 a, e
3 a, b
pg. 361 # 4 a – e, h, i,
5 a – c, e, f, i
3 a, e
5
1
xa
3b
Exponential Function
y = bx, b > 0 Y
10
x
* Look at graph
9 of y = 2
8
7
6
5
4
3
2
1
domain:
range:
horizontal asymptote:
X
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10
-2
-3
x
-4
1

x
x
-5 of y = 5 and y =   or 2 
* Look at graph
-6
 2
Y
-7
x
10
(Play with y -8
= (–2)
on calc; Zoom decimal)
9
-9
8
7
-10
reflection of y = 2x over y–axis
6
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5
y = 5x
4
3
2
1
0
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
-2
-3
-4
-5
-6
Transformations -7
-8
-9
-10
x
X
1
y =   or 2 x
 2
10
9
8
7
6
5
4
3
2
1
Y
X
1 2 3 4 5 6 7 8 9 10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
a. horizontal (phase) shift
y=2
 shifted right one
1 2 3 4 5 6 7 8 9 10
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b. vertical shift
y = 2x + 3  shifted up three
c. reflection over x and y–axis
y = 2x  reflection over y–axis
y = (2x)  reflection over x–axis
d. vertical stretch
y = 4  2x  y is increased by a factor of 4
6
3c
Logarithms
logbx = y  by = x
b > 0, b  1; x > 0
Properties of Logs
1. logb MN = logbM + logbN
2. log b
eg: log 3 + log 7 = log 21
M
 log b M  log b N
N
3. log b M N  N log b M
4. log N M 
Proof
log b M
log b N
Let log N M  k, then M  N k , log b M  log b N k etc
5. log b b x = x
6. blogb x  x (ie. blogb x  y, log b y  log b x, y  x)
log x = log 10 x
ln x = log e x
(common log; base = 10)
[lawn x] (natural log; base = e)
eg 1. Evaluate.
a. log381= x or log3 34  x or x 
log81
log 3
c. 52log5 3
b. log 1000
d. log b b
7
3d
2. Simplify.
a. 2log a +
1
log b  3log c
2
3. Solve for x.
a. log3(x + 2) = 2
b. log2(log5x) = 3
c. 22x  3 = 5x
d. log 2 x + 6log x 2 = 5
8
4a
e and Natural Logarithms
e is defined as e 2.718281828 and is used as a base for exponential functions. It is also called
Euler’s number as it is a mathematical constant discovered by the Swiss mathematician,
Leonhard Euler. It is used in calculus and we will see the development of this number as we get
into limits next unit. For now, we are treating it as we would π.
The inverse of this function is the natural log (has a base of e ) and follows the regular log laws.
It is denoted ln x.
ln x  1oge x
Ex. 1. Evaluate to 4 decimal places.
a) e4 =
b) ln 5 =
c) log3e =
d) ln e =
2. Solve each equation for x.
e3  1
x
2
a) ln (2x – 1) = 3
b) ln x = -1
c) ln x = ln5 + ln 8
d) ln(x + 6) + ln(x – 3) = ln 5 + ln 2
e) log (loge (x)) = 0
f) ln x2 = 2ln4 – 4ln2
9
4b
3. Solve to 4 decimal places
a) ln (x – 5) = 3
c) e 23 x  20
b) e5x - 1 = 12
d) 2 x  5
Assignment: Pg. 375 #3, 4, 5, 6
10
5a
Composition of Functions
Domain- The set of first members of the ordered pairs of a relation. The domain is assumed to be
the set of all real numbers for which the equation is meaningful.
eg.
State the domain:
1.
y = 2x + 1
2.
y=
3.
y=
4.
y=
5x
x+ 3
4- x
3x
\
x2 - 9
Composition of Functions: A function is composed of other functions.
eg.
Given : f(x) = 3x – 2, g(x) = sinx,
1.
f(11) =
2.
f(f(x)) =
3.
gh(x) =
4.
h(g(x)) =
5.
g(g(x)) =
and
h(x) = 5x2 + 1
Inequalities
Notation Open ( ) excludes endpoints.
Closed [ ] includes endpoints.
a. [3, 5)  3  x < 5 (show on a number line)
b. (, 2)  x < 2
c. [5, 2)  [1, )  5  x < 2 or x  1
11
5b
* Note that the point (1, 3) is in coordinate geometry but in this context (1, 3) is the open
interval including all real numbers between 1 and 3. Also note that infinity is always open.
Solving Linear Inequalities
Examples:
2. 2x  7  11
1. 8x + 11 > 5x + 20
Solving Inequalities (degree > 1 and rational inequalities) This may be done by graphing the
function and noting whether the graph is above ( y > 0) or below ( y < 0) the x–axis or by finding
the zeros of each factor in the numerator or the denominator and the finding the sign of a random
value in each interval.
Examples:
1. x2 + 2x < 15
2. (x  2)2(x + 3)  0
12
5c
3. x3(2x  5)(x + 4) < 0
4.
x 2  2x  15
0
(x  2)3
5.
x 2  3x
0
x2 1
13
5d
Intercepts
 x-int: point where the function crosses the x-axis (where y = 0)
 y-int: point where the function crosses the y-axis (where x = 0)
Examples:
1. y = x3 + 3x2  x  3
y-int:
x-int:
2. y = x3 + 2x2  15x + 14
y-int:
x-int:
3. y 
x 3  4x x(x  2)(x  2)

x 5
(x  5)
y-int:
x-int:
14
5e
Even Function

f(x) = f(x)  symmetric about the
eg f(x) = x2
f(–x) = (–x)2 = x2 = f(x)
Odd Function

f(x) = f(x)
10
9
8
7
y-axis
6
5
4
3
2
1
Y
Y
5
f(x) = cos x
X
-1p
X
0
-5p -4p
-3p -2p
1p
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10
-2
-3
-4
-5
Y
5
Y
-6
10
-7
9
4
-8
8
 symmetric about the origin
(reflect
over
both
axis
or
rotate
180)
7
-9
-5 3
6
-10
2p
3p
4p
5p
5
4
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x - http:/ / www.alentum.com/ agra
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3
2
1
f (x)  3 (x)   3 x  f (x)
1. Find the x-int(s)
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10 1 2 3 4 5 6 7 8 9 10
0
-2
-5p -4p
-3p -2p
-1p
1p
-3
-1
-4
-5
-6
-2
-7
-8 the function is even, odd, or neither.
and y-int and state whether
-3
-9
-10
-4
3
a. f(x) = 2x2 + 5
10
9
8
7
6
5
4
3
2
1
X
2x  7x
x  39
X
2p
3p
4p
5p
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Y
2 10
b. f(x) =
f(–x) = –f(x)
-5
8
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7
6
5
4
3
2
1
Y
X
X
1 2 3 4 5 6 7 8 9 10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
1 2 3 4 5 6 7 8 9 10
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5g
c. f(x) = x  3
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
-3
-4
-5
-6
-7
-8
-9
-10
d.
f(x) = x  2
10
9
8
7
6
5
4
3
2
1
Y
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1-10
-2
X
-3
-4
1 2 3 4 5 6 7 8 9 10
-5
-6
-7
-8
-9
-10
Y
X
1 2 3 4 5 6 7 8 9 10
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16
6a
Trig Review
a
  , a  arc length in radians, r  radius
r
5
 

degree  radians  X
 ie. 50 =
18
 180 

 180 
radians  degree  X
 ie.  20
9
 

sin =
r
y
r
sec =
x
x
cot =
y
y
r
csc =
x
r
y
tan =
x
cos =
eg 1. Given point or ratio, find other ratios.
a. (3, 2) is on the terminal arm of angle . Find all ratios.
cos =
sec =
sin =
csc =
tan =
cot =
b. cos = 
9
and sin is negative. Find all other ratios.
41
sin =
csc =
tan =
cot =
sec =
17
6b
Special Angles and Unit Circle
2
45
30
*
2
1
3
45
60
1
1
 
Count by 30   or 45
6
 
  , values to know are
4
3
1
(longer),
(shorter), and
2
2
11
eg 1. cos
6
2
1
or
2
2
2. csc
19 
6
3. tan 240
 4 
4. cos 2 

 3 
5. sin 90
6. tan
5

2
Y
5
Graph of Basic Functions
4
y = sinx
Y
4
y = cosx
3
3
2
2
1
1
X
X
-2p
-1p
0
1p
-2p
2p
-1p
0
-1
-1
-2
-2
-3
-3
sin(x) = sin(x)-4 odd
1p
2p
cos(x) = cos(x) -4 even
-5
-5
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5c
5
y = tanx
Y
4
3
2
1
X
-2p
-1p
0
1p
2p
-1
-2
-3
tan(x) = tan(x)-4 odd
-5
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Assignment
pg. 252 # 1 a, b, d, e,
pg. 253 # 3, 7
pg. 257 # 1 a, c, e
Sketch y = 2sin3x
2 a, c, f, 3
19
7a
Formulas:
sin2 + cos2 = 1
1 + tan2 = sec2
1 + cot2 = csc2
tan  
sin(A  B) = sinAcosB  cosAsinB
cos(A  B)  cos A cos B sin A sin B
sin(2) = 2sincos
cos(2) = cos2  sin2 = 2cos2  1 = 1  2sin2
1  cos(2)
1  cos(2)
sin 2  
, cos 2  
2
2
eg. 1. Express as a single trig function.
a. cos(7x)cos(3x)  sin(7x)sin(3x)
b. sin(3x)cos(8c)  cos(3x)sin(8c)
c. cos2(19)  sin2(19)
1  cos(10x)
2
e. 5  10sin2x
d.
f. 1  cos(4x)
g.
sin2(5A) – cos2(5A)
 4 
h. cos 2 

 3 
20
sin 
cos 
7b
2. Determine exact values of the following.
a. cos(219)cos(69) + sin(219)sin(69)
b. sin


cos
12
12
 7 
c. sin 2  
 12 
3. Solve the following equations.
a. tan2  1 = 0,     
b.
3 sin x  cos x  0 , 0  x  2
c. cos2 + cos = 0, 0    
d. sin + 3cos = 0, 0  x < 2
21
7c
e. 2sin2 + 5sin – 3 = 0, 0  x < 360
Assignment Scan 269–282 (formula) Read 287, 288
pg. 273 # 1a – d, g, h, j, 3 a, 6 a, b
pg. 279 # 2 a – g, 3
pg. 292 # 1 a, 2 a, 3 a, 9 a, b, d, h
22
7d
Arc Length
a = r ( in radians)
Sector Area
A  r 2 
 1 2
 r 
2 2
( in radians)
OR
A
1 2a 1
r    ar
2 r 2
eg 1. What is the radius of a sector with an angle of

and an area of 9.
8
2. Find the area of a sector bounded by an arc of length 15.7cm with  = 72.(1.26 rad)
23