Grants Tutoring 2011 MATH 1500 Midterm PDF

Title Grants Tutoring 2011 MATH 1500 Midterm
Author Stephanie Ag
Course Introduction To Calculus
Institution University of Manitoba
Pages 26
File Size 1.5 MB
File Type PDF
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Download Grants Tutoring 2011 MATH 1500 Midterm PDF


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INTRO CALCULUS (DIFFERENTIATION and APPLICATIONS)

Volume 1 of 2 September 2011 edition

This volume covers the topics on a typical midterm exam.

© 1997-2011 Grant Skene for Grant’s Tutoring (www.grantstutoring.com) DO NOT RECOPY Grant’s Tutoring is a private tutoring organization and is in no way affiliated with the University of Manitoba.

While studying this book, why not hear Grant explain it to you? Contact Grant for info about purchasing Grant’s Audio Lectures. Some concepts make better sense when you hear them explained. Better still, see Grant explain the key concepts in person. Sign up for Grant’s Weekly Tutoring or attend Grant’s Exam Prep Seminars. Text or Grant (204) 489-2884 or go to www.grantstutoring.com to find out more about all of Grant’s services. Seminar Dates will be finalized no later than Sep. 25 for first term and Jan. 25 for second term.

HOW TO USE THIS BOOK I have broken the course up into lessons.

Study each lesson until you can do all of my

lecture problems from start to finish without any help. lesson.

Then do the Practise Problems for that

If you are able to solve all the Practise Problems I have given you, then you should have

nothing to fear about your Midterm or Final Exam.

I have presented the course in what I consider to be the most logical order. Although my books are designed to follow the course syllabus, it is possible your prof will teach the course in a different order or omit a topic.

Make

sure

you

are

It is also possible he/she will introduce a topic I do not cover.

attending

your

class

regularly!

Stay

current

with

the

material, and be aware of what topics are on your exam. Never forget, it is your prof that decides what will be on the exam, so pay attention. If you have any questions or difficulties while studying this book, or if you believe you have found a mistake, do not hesitate to contact me.

at the bottom of every page in this book.

My phone number and website are noted

“Grant’s Tutoring” is also in the phone book.

I welcome your input and questions. Wishing you much success,

Grant Skene Owner of Grant’s Tutoring

© 1997-2011 Grant Skene for Grant’s Tutoring (text or call (204) 489-2884) DO NOT RECOPY Grant’s Tutoring is a private tutoring organization and is in no way affiliated with the University of Manitoba.

INTRO CALCULUS

1

Formulas and Definitions to Memorize The Definition of Continuity: f  x  is continuous at x=a if and only if

f  x  

The Definition of Derivative:

lim

h0

lim

x a

f  x   f  a .

f x  h  f x  h

x n   nx n

1

The Power Rule:

 f  g 

The Product Rule:

The Quotient Rule:

 f g  f g

 T  T  B  T B    B B 2

 f u    f  u  u

The Chain Rule:

un   nun

The Chain Rule Version of Power Rule:

1

 u

Derivatives of Trigonometric Functions:

sin u   cos u u 

tan u   sec u  u 

sec u 

cos u 

cot u 

csc u    csc u cot u  u

 sin u u

2

 csc u u 2

 sec u tan u u 

Derivatives of Exponential and Logarithmic Functions:

e u   eu  u  a u   a u  u 

ln u   ln

a

u u

log a u  

u

u ln a

Derivative of an Inverse Function:

 f   a 

Fundamental Theorem of Calculus:

 u f t dt  f u u          a 

© 1997-2011 Grant Skene for

1

1

f  f  a 

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1

DO NOT RECOPY

2

INTRO CALCULUS

Antiderivative Formulas:

 K dx  Kx  C n n  x dx  n x  C 1

1

1

1   

x

dx  ln x  C

x x  e dx  e  C

,

so

ax ax  e dx  a e  C

e.g.

x x  e dx  e  C

1

1

5

5

5

x  a dx 

 

ax C ln a

x dx   cos x  C ,

sin

cos



sec



csc



sec



csc

x dx  sin x  C ,

so

 ax dx   a

cos

e.g.

  x  dx  

1

so



cos

e.g.



cos 7

1

sin

sin 3

3

ax  C

ax dx  1 sin ax  C a

 x  dx  71 sin  7x   C

x dx  tan x  C ,

since

tan x   sec x

x dx   cot x  C ,

since

cot x    csc x

x tan x dx  sec x  C ,

since

sec x   sec x tan x

x cot x dx   csc x  C ,

since

csc x    csc x cot x

2

2

 x  C

cos 3

2

2

Trigonometric Values to Memorize 0

sin

cos

0

1

6 1 2

3 2

4

2 2 2 2

3

2

3

1

2 1

0

2

1 tan

© 1997-2011 Grant Skene for

0

3

1

3

undefined

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INTRO CALCULUS

3

Required Theorem Proofs to Memorize (See pages 152 to 155 for hints to help understand these proofs.) Theorem 1: (a)

Prove:

(Differentiable Functions are Continuous) If a function

f is differentiable at point a, then it is

Theorem 4:

(The Product Rule)

a. Is it true that if f is continuous at a it is also differentiable at a? Justify your answer.

f x  g x    f  x  g  x   f x g x 

continuous at

(b)

f and g are differentiable functions, prove:

Given that

Proof: Proof: (a)

If

f is differentiable at a, then f   a  f  a  h  f  a  . h f  a  h  lim f  a  h 

f  a   lim

h 0

lim

h 0

h 0



lim

h

0

 f a  h   f a   f a  

 f a  h  f a   h  f a   h   f a  h   f a   lim  lim h  lim f a  h h0 h0 h0  f  a  0  f a  

lim

h

0

 f  a We have proven

continuous at

(b)

lim

h 0

f  x  g  x    h 

lim

exists where

0

f  x  h  g  x  h   f  x  g x  h

f x  h  g x  h   g x  h  f x   g x  h  f  x   f  x  g  x 



lim



lim

 f  x  h  g  x  h   g  x  h  f  x  g  x  h f  x   f  x  g  x      h h  



lim

g  x  h  g  x   f x  h  f x   g  x  h  f  x     h h



lim

h0

h0

h0

h

f x  h   f x  h

h0



lim

h0

 f  x  g  x   f  x  g   x  (Note that

lim

h 0

g x  h  

f x  

lim

g x  h  g x  h

h0

proven

g x  h   g  x 

f  a  h  f  a

meaning

f is

a but not be f x   x is continuous at 0

Theorem 5:

since

(The Derivative of sin

Prove:

sin x   cos x

(i.e.

FALSE. A function can be continuous at

a.

lim

h0

g is continuous;

differentiable functions are continuous.)

a.

differentiable at

.

For example,

but it is not differentiable at 0.

Theorem 2: (The Constant Multiple Rule) Given that c is a constant and f is a differentiable function, prove:

c  f  x    c  f  x 

x.)

d sin x  cos x dx

).

Proof: sin

sin x   hlim

x  h  sin x

h x cos h  cos x sin h  sin x lim h h  sin x cos h  sin x  cos x sin h lim h  h sin x cos h  sin x cos x sin h   lim   h   h h  cos h  1 sin h   cos x  lim sin x  h h  h   cosh  1 lim sin x  lim  lim cos x  lim h h  h  h  h  sin x  0  cos x  1 cos x proven 0



sin

0

.



0

Proof:

c  f  x  h  c  f  x  h f  x  h  f  x   c  lim h h  c  f  x  proven

c  f  x    h 

lim



0

0



0

0

Theorem 3:



0

 (The Sum Rule)

Given that

f and g are differentiable functions, prove:

 f  x   g  x    f  x   g x  Proof:

 f  x   g  x    h 

lim

 

0

lim

h 0 lim

h 0



0

0

0

sin

h

h

.

f  x  h   g x  h   f x   g x   h f  x  h   f x   g x h   g x  h  f  x  h  f  x  g  x  h   g  x      h h  

f  x  h   f x  g x h   g x   lim h h h   f x   g x  proven 

lim

h 0

© 1997-2011 Grant Skene for

0

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(Intro Calculus) LESSON 2: LIMITS

31

Lesson 2: Limits Memorize these two trig limits:

sin

lim

h

0

h

h

1

lim

and

cos

h 0

h 1 0 h

Lecture Problems: (Each of the questions below will be discussed and solved in the lecture that follows.)

For each of questions 1 to 14, find the value of the limit, if it exists. If it does not exist, is it infinity, negative infinity, or neither? Justify your answers.

x x2 x 1

x 2

2

1.

3.

lim

x  1

lim

x3

2.

x 7 4 x  5x  21 4



lim

4.

x  3x  4 5x  9 x  4

lim

x

2

2

5.

lim

x  1

6.

2

x 9 x  x2

lim

x5

3

9.

11.

lim

x  2

8.

2

x  7 x  18

x  10x  25

 x  2

x 3x 2 x   3 x  5 x  1 lim

15.

lim

lim

x

12.

2

2 x  3   x

2

 6x  5



x  x 6

x

4

lim

x  

14.

3

x x 2 1 3 x

2

3

4

2

13.

x 6

lim

x 6

10.

2

lim

x

2

2

7.

2

x 2 x2

2

2

x4 

x  2

lim

x  

3

3

x  21 4 x  3x  7 5

2

1  3 x  4 x 5 x  4 

2

5

2

Which limits in questions 1 to 14 above indicate the existence of a Vertical or Horizontal Asymptote?

16.

Find the following limits.

(a)

17.

 x  7 x

sin 2 lim

x

x

0

(b)

3

 x sin  5 x 

sin 2 lim

x

0

(c)

 x

tan 4 lim

x

0

x

2

Use the Squeeze Theorem to solve the following limits.

(a)

lim

x

0

x

© 1997-2011 Grant Skene for

4

2

sin

1 x  

(b)

lim

x

0

x

5

cos

3    x  3

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32

© 1997-2011 Grant Skene for

LESSON 2: LIMITS (Intro Calculus)

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(Intro Calculus) LESSON 2: LIMITS

© 1997-2011 Grant Skene for

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33

DO NOT RECOPY

34

© 1997-2011 Grant Skene for

LESSON 2: LIMITS (Intro Calculus)

Grant’s Tutoring (text or call (204) 489-2884)

DO NOT RECOPY

(Intro Calculus) LESSON 2: LIMITS

© 1997-2011 Grant Skene for

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35

DO NOT RECOPY

(Intro Calculus) LESSON 3: CONTINUITY

79

Lesson 3: Continuity Memorize the Definition of Continuity:

f x 

is continuous at

x=a if and only if

lim

xa

f  x   f a  .

Lecture Problems: (Each of the questions below will be discussed and solved in the lecture that follows.)

1.

For the function shown below, determine for which

x it is continuous.

Justify your

answer.

  x2  x 4 f  x    x 2   4 x 

if

x  2

if

2 x 

2

2.

Is

g  x

below continuous at

x=1?

if

1

x 1

2

For

f  x

below, find

2

4.

Show that

f  x   x  3x  1

5.

Show that

f  x   x  x  7x  4

3

© 1997-2011 Grant Skene for

x1

a and b which will make the function continuous everywhere. x  4   f  x    ax  b   6x 

3

x

(Justify your answer.)

x 1    x 32 g x    3  3.

1

2

x0

if if

0

 x3

if

has a zero between

x3

x0

and

has at least three zeros on

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x  1.

4, 4  . DO NOT RECOPY

96

LESSON 4: DEFINITION OF DERIVATIVE (Intro Calculus)

Lesson 4: The Definition of Derivative Memorize the Definition of Derivative:

f

  x   lim h

f

 x  h   f x 

0

h

Lecture Problems: (Each of the questions below will be discussed and solved in the lecture that follows.)

1.

2.

 x   2x

2

 3x  1 :

For

f

(a)

Find the average rate of change of

(b)

Use the definition of derivative to find the instantaneous rate of change of

(c)

Find the equation of the tangent line to

For the functions below, find

(a)

f

(b)

f

 x 

 x 

f

 x 

f

 x  for the interval 0  x  2.

y



f

x

at x=1 in

y

 mx  b

f

 x .

form.

using only the definition of derivative.

2x 9



x

2

1 x

1

© 1997-2011 Grant Skene for

Grant’s Tutoring (text or call (204) 489-2884)

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110

LESSON 5: DIFFERENTIATION RULES (Intro Calculus)

LESSON 5: Differentiation Rules All of the formulas on this page must be MEMORIZED

 x n   nx n

The Power Rule:

 f  g

The Product Rule:

The Quotient Rule:

1

 f  g  f g

 T  T B  T B    B  B 2

 f  u   f   u  u

The Chain Rule:

 un   nun

1

The Chain Rule Version of Power Rule:

 u

Derivatives of Trigonometric Functions:

sin u   cos u  u

tan u 

 sec u  u

sec u 

cos u 

cot u 

 csc u  u

csc u    csc u cot u  u

 sin u u 

2

2

 sec u tan u  u

Derivatives of Exponential and Logarithmic Functions:

 e u   e u  u

© 1997-2011 Grant Skene for

ln u  

u u

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(Intro Calculus) LESSON 5: DIFFERENTIATION RULES

111

Lecture Problems: (Each of the questions below will be discussed and solved in the lecture that follows.)

1.

Find the indicated derivatives for the following. You need not simplify your answer. 2

4

x

dy dx

9 1 y  x  x  , find y  3x  2

(d)

y   x  1

(e)

y   4  6x

(f)

y  2x


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