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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

2

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

3

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

4

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

5

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

6

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

7

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

8

e) ( )g

xh

f) ( )(3)h h

g) 2 ( 1) 5 ( )( 3) 3g x f f h x

Now Try

Function Notation

Worksheet

9

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 .

12

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

13

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

17

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

18

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

Page 27 # 1 to 3, 5 to 11

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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

28

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

29

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

Page 50 # 1 to 8

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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

36

b) 1

9 3 1 ...3

c) 3 3

3 ...5 25

Ex) Express 2.135 as a fraction.

Now Try

Page 56 # 1 to 5

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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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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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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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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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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

Page 9 # 1, 2, 6, 7,

Page 35 # 8, 9 10

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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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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

Page 43 # 1, 2, 3, 7, 8