unit 1 precalculus review & limitskosikclass.weebly.com/uploads/8/8/6/8/8868925/unit_1...1 unit...
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Unit 1 PreCalculus Review & Limits
Factoring:
Remove common factors first
2 Terms
- Difference of Squares 2 2 ( )( )a b a b a b
- Sum of Cubes
3 3 2 2a b a b a ab b
- Difference of Cubes
3 3 2 2a b a b a ab b
3 Terms
- form of 2x bx c
- form of 2ax bx c
* can use decomposition but we can also (and
should use shortcuts)
4 Terms
- Try factoring by grouping
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Ex) Factor the following fully.
a) 3 6 3 2 821 12 15m n m n m n
b) 2( 2) 11( 2) 24x x
c) 23 33 132m n mn mn
d) 9512 a
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e) 23( 4) 13( 4) 10x x
f) 3 2 2 312 25 7x y x y xy
e) 3 3( 4) ( 2)x x
g) 4 25a
Now Try
Factoring Worksheet
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Rationalizing Denominators:
Single Term Denominators
Ex) 3
10 * multiply the numerator and
denominator by the radical
found in the denominator
Two Term Denominators
Ex) 3
5 10 * multiply numerator and
denominator by the
conjugate of the denominator
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Ex) Rationalize the following.
a) 5
2 b)
9
2 6
c) 1
4
211
d) 4 1
1
x
x
e) 7
5 3 f)
4
3 2 5
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g) 6 13
6 13
h) 1
xx
x
x
i) 3
6
x j)
7 3
23
x
k) 3
3
5x l)
3
1
4 3
Now Try
Rationalizing Denominators
Worksheet
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Function Notation:
In ( )f x , f refers to the name of the function (or equation) and
x is the variable used within that function .
In ( )g a , g refers to the name of the function (or equation) and
a is the variable used within that function .
Ex) If 2( ) 5f x x , ( ) 5 2g x x , and ( )3
xh x
x
,
find the following:
a) (4) (5)f h b) ( )( 7)fh
c) ( )( )f g x d) 4 ( 1) 9h x
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e) ( )g
xh
f) ( )(3)h h
g) 2 ( 1) 5 ( )( 3) 3g x f f h x
Now Try
Function Notation
Worksheet
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Equations of Lines:
Stuff we need
+’ve slope -‘ve slope slope = 0 slope is
undefined
*line increase *line decrease *horizontal line *vertical line
y mx b 1 1( )y y m x x 2 1
2 1
y y ym
x x x
Parallel Lines Perpendicular Lines *have the same slope *have slopes that are negative
reciprocals of one another
Ex) Determine the equation of the line that passes
through 35, 120 and has a slope of 715
.
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Ex) Determine the equation of the line that is parallel to
4 7 20 0y x and has the same x-intercept as
5 3 22 0y x .
Ex) Determine the equation of the line that passes
through ( 12, 4) and (9, 5) .
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Ex) Determine the slope of the tangent to the circle 2 2( 4) ( 6) 25x y when 8x .
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Ex) Determine the equation of the tangents to the circle
given by 2 2( 3) ( 6) 100x y at the points that the
circle intersects the line 4 3 30 0x y
Now Try
Equations of Lines
Worksheet
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Solving Trigonometric Equations:
Stuff we need
S A 1
sin302
3
cos302
T C 2
sin 452
2
cos452
3
sin 602
1
cos602
sin
tancos
coscot
sin
1csc
sin
1sec
cos
*understand how reference angles can be used
siny x cosy x 360
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First Degree Equations
Ex) Solve the following, express all solutions as a
general statement in radians.
a) 3
sin2
x
b) 2sec 4 0x
c) 3cot 1 0x
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Second Degree Equations
Ex) Solve the following, express all solutions as a
general statement in radians.
a) 22sin sin 0x x
b) 24cos 1 0x
c) 22cos 7cos 4 0x x
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d) 2 23sin cos 0x x
e) 23sin cos2 2 0x x
Multiple Angle Equations
Ex) Solve the following, express all solutions as a
general statement in radians.
a) 2
sin 22
x
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b) tan 2 3 0x
c) 12cos 1 03
x
d) 22sin (2 ) sin(2 ) 1 0x x
e) 1 1 1sin cos sin 02 2 2
x x x
Now Try
Trigonometric Equations
Worksheet
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Limits:
Limits allow us to determine the value that a function gets
closer to (approaches) as x gets closer to (approaches) a
specified value
Notation lim ( )x a f x b
read: as x approaches the value of a, the value of the
function ( )f x approaches b
think: what value is ( )f x getting closer and closer to
as x gets closer and closer to a
Ex) 2
4
3 28lim 11
4x
x x
x
The limit as x approaches 4 for the function
2 3 28
( )4
x xf x
x
is equal to 11.
This means that the closer x gets to the value of 4, the
function ( )f x gets closer to the value of 11.
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Ex) Consider the following functions.
a) ( ) 2 3f x x b) 2 5 6
( )2
x xf x
x
c)
1( ) 4f x
x
as x gets as x gets as x gets
closer to 7 closer to -2 closer to
x ( )f x x ( )f x x ( )f x
Thus Thus Thus
7lim 2 3x x 2
2
5 6lim
2x
x x
x
1lim 4x
x
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Continuous and Discontinuous Functions:
A continuous function is a function that when graphed has
no breaks ( the graph consists of one piece).
Ex)
A discontinuous function is a function that when graphed
consists of 2 or more parts (is broken).
Ex)
**These points where the graph is “broken” are called the
Points of Discontinuity.
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One Sided Limits:
One sided limits allow to determine what value a function
approaches as x gets closer to a specific value from either
the right or the left side of the specified value.
2lim ( )
xf x
Means x values are approaching
a value of 2 from the right side
Ex) 2.5, 2.4, 2.1, 2.0001, etc.x
2lim ( )
xf x
Means x values are approaching
a value of 2 from the left side
Ex) 1.5, 1.8, 1.9, 1.9999, etc.x
The general Limit of 2lim ( )x f x will only exist if
2lim ( )
xf x
and 2
lim ( )x
f x equal the same value. This
means that it shouldn’t matter what side the value is
approached from to determine the value of the general
limit 2lim ( )x f x .
lim ( )x a f x b
if and only if
lim ( )x a
f x b and lim ( )
x af x b
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Ex) Given the graph of ( )y f x , determine the following
limits.
a) 6lim ( )x f x b)
2lim ( )
xf x
c) 0lim ( )x f x
d) 5
lim ( )x
f x e)
2lim ( )
xf x
f) 5lim ( )x f x
g) 2
lim ( )x
f x h) 8lim ( )x f x i)
4lim ( )
xf x
j) lim ( )x f x k) 4lim ( )x f x l) 2lim ( )x f x
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Ex) Given the function defined below, find the following
limits.
2 4 9 , if 2
( ) 3 , if 2 4
2 5 , if 4
x x x
f x x x
x x
a) 2
lim ( )x
f x b)
2lim ( )
xf x
c) 2lim ( )x f x
d) 4
lim ( )x
f x e)
4lim ( )
xf x
f) 4lim ( )x f x
Graph the function ( )y f x
Now Try
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Properties of Limits:
lim ( ) ( ) lim ( ) lim ( )x a x a x af x g x f x g x
Ex) 2 2 22 2
3 3lim 5 lim lim 5x x x
x xx x
x x
lim ( ) lim ( )x a x acf x c f x
Ex) 2 2
3 3
12 12lim 5 5lim
3 3x x
x x x x
x x
lim ( ) ( ) lim ( ) lim ( )x a x a x af x g x f x g x
Ex)
2 2
4 4 4
3 4 3 4lim 3 lim 3 lim
4 4x x x
x x x xx x
x x
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( ) lim ( )lim
( ) lim ( )
x ax a
x a
f x f x
g x g x
Ex) 6
6 2 2
6
5 lim ( 5)lim
2 1 lim 2 1
xx
x
x x
x x x x
lim ( ) lim ( )n n
x a x af x f x
Ex)
4 42 2
5 52 2
8 15 8 15lim lim
3 10 3 10x x
x x x x
x x x x
lim ( ) lim ( )n nx a x af x f x , if the root exists
Ex) 3 3
3 33 2 3 2
24 3 1 24 3 1lim lim
3 6 11 3 6 11x x
x x x x
x x x x
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Solving Limits Algebraically:
1) Solve by Direct Substitution
Many limits can be solved by substituting the value
for x directly into the function.
This can be done if the function is continuous at the
point in question.
Ex) Solve the following limits.
a) 2
5lim 2 3x x x
b) 4 2
1
5 1lim
2x
x x
x
c) 2
3limx x x
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2) Solve by Factoring
Many times if direct substitution is attempted a result
of 0
0 occurs. This happens when we attempt to find a
limit as x approaches a value where the function is
discontinuous (we are trying to find a limit at a
restriction).
Eg) 2
4
16lim
4x
x
x
Factor first, reduce, then solve
by direct substitution
Eg) Solve the following limits.
a) 2
7 2
3 28lim
10 21x
x x
x x
b)
2
0
(2 ) 4limh
h
h
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c) 3
2 2
8lim
3 2x
x
x x
3) Solve by Multiplying by a Conjugate
If direct substitutions results in 0
0, and the function
cannot be factored, try multiplying numerator and
denominator by a conjugate (especially if a radical
appears in the function).
Ex) Solve the following limits.
a) 0
1 1limx
x
x
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b) 47
47lim
7 2x
x
x
c) 2
25
23 50lim
5a
x x
x
d) 1
11
lim1
x
x
x
Now Try
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Infinite Limits:
These are limits where x approaches
To evaluate these limits, divide all terms of the numerator
and denominator by the largest power of the denominator,
then use the above identities to evaluate.
Ex) Evaluate the following limits.
a) 2
3
3lim
4 1x
x
x
lim , 0n
x x n
1lim 0 , 0x n
nx
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b) 25 2
lim7
x
x x
x
c)
3 2
3
4 2 11lim
6 5x
x x
x
d) 2
2lim
2 1n
n n
n
e)
3 4
3 4
7 5 2lim
3 12 25a
a a a
a a
f) 218 12
lim4 2
x
x x
x
g)
3 2
2 7
5 6 1lim
2 14h
h h
h h
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Infinite Sequences:
A sequence will either:
Converge Approach a specific value
Diverge Approach positive or negative
infinity or bounce between 2 or
more values
To determine if a sequence converges or diverges we find
the limit as n for the general term
Ex) Determine whether the following sequences
converge or diverge. If the sequence converges, to
what value does it converge to?
a) 1 2 3
, , , ...... , 2 3 4 1
n
n
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b) 2
2
2 6 120 , , , , ...... ,
9 19 33 2 1
n n
n
c) 5 2nt n d) 1
2
n
nt
e) 4 , 12 , 36 , ... f) 1n
nt
Now Try
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Infinite Series:
Like sequences, if the sum of a series approaches a
specific value it is called Convergent, if not then it is
Divergent
Ex) State whether the following series are convergent or
divergent.
a) 2 2 2 2 2 2 2 ...
b) 1 1
8 4 2 1 ...2 4
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Infinite Geometric Series:
If 1r , then the sum of the series will converge.
Proof: 1
1
n
n
a rS
r
sum of a geometric
series
1
lim1
n
n
a r
r
, 1r
Ex) Find the sum of each series given below.
a) 1 1
1 ...4 16
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b) 1
9 3 1 ...3
c) 3 3
3 ...5 25
Ex) Express 2.135 as a fraction.
Now Try
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Slopes of Tangent Lines:
When dealing with circles, we know that a tangent is a
line that touches the circle once and only once.
When dealing with more complicated curves, a tangent
line cannot be defined so simply.
In calculus we are often concerned with finding the slope
of a tangent line as it represents the instantaneous slope of
the graph or instantaneous rate of change in a situation
where the slope is constantly changing.
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Ex) Determine the equation of the tangent line to the
graph of 2y x at the point 2, 4 .
Finding the slope
of the tangent:
x 2y x 2 1
2 1
4
2
y y ym
x x x
Slope of the tangent
is?
Equation of the tangent
is? x 2y x 2 1
2 1
4
2
y y ym
x x x
4x
4x
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Finding Slopes of Tangents Using Limits:
Method 1:
Method 2:
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40
Ex) Determine the slope of the tangent line to the graph
of 2y x at the point 2, 4 .
Method 1: use ( ) ( )
limx a
f x f am
x a
Method 2: use 0
( ) ( )limh
f a h f am
h
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Ex) Find the slope of the tangent line to the curve 22 4 1y x x at the point 2, 15 . Use both
methods.
( ) ( )limx a
f x f am
x a
0
( ) ( )limh
f a h f am
h
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42
Ex) Determine the slope of the tangent to the curve
1xy at the point 1
2, 2
. Use both methods.
( ) ( )limx a
f x f am
x a
0
( ) ( )limh
f a h f am
h
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43
Ex) Determine the slope of the tangent to the curve
2y x when 6x . Use both methods.
( ) ( )
limx a
f x f am
x a
0
( ) ( )limh
f a h f am
h
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Ex) Determine the equation of the tangent line to the
curve 3 2y x x when 2x
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45
Ex) Determine the slope of the tangent line to the curve 2 5 2y x x at the general point whose
x-coordinate is “a”. Use this to determine the slope
of the tangent when 1 , 4 , 6x .
Now Try
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Slope as a Rate of Change:
Slope is defined as:
2 1
2 1
rise change in means
run change in
y y y ym
x x x x
Slope refers to how quickly “y” changes with respect
to each unit increase in “x”.
Ex) Slope of 5 means:
Ex) Slope of 2
3
means:
Ex) Slope of the graph below means:
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Average Rates of Change vs. Instantaneous Rates of
Change:
the slope is constantly
changing
or
the rate of change is
constantly changing
Average Rate of Change:
use two points on the
curve to find the slope,
this treats the curve like a
straight line and gives an
average rate of change
Instantaneous Rate of Change:
allows one to find the
slope or rate of change at
an exact moment
to find this determine the
slope of the tangent at a
particular point
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48
Ex) A ball is dropped from the upper observation deck
of the CN Tower, 450 m above the ground. The
distance the ball has fallen is given by 24.9d t ,
where d is the distance in metres and t is the time in
seconds.
a) Determine the average speed of the ball in the first 3
seconds.
b) Determine the speed of the ball at exactly the 3
second mark.
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Ex) The displacement, in meters of a particle moving in
a straight line is given by 2 2d t t , where t is
measured in seconds and d in metres.
a) Determine the average velocity between 3.5t s and
4.5t s.
b) Determine the instantaneous velocity at 4 s.
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Ex) A spherical balloon is being inflated. If the volume
of the balloon, V, is given by 34
3V r , where r is
the radius of the balloon in cm and V is measured in 2cm , find the following.
a) The average change in volume with respect to the
radius between 7r cm and 9r cm.
b) The instantaneous rate of change in the volume with
respect to the radius when the radius is 8 cm.
Now Try
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