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McGraw-Hill Ryerson, MATHEMATICS 11
CONTENTS
Using
McGraw-Hill Ryerson Calculus & Advanced Functions, Solutions
v
CHAPTER 1 Functions and Models
1
CHAPTER 2 Polynomials
23
CHAPTER 3 Limits
99
CHAPTER 4 Derivatives
168
CHAPTER 5 The Chain Rule and Its Applications
263
CHAPTER 6 Extreme Values: Curve Sketching and Optimization Problems
311
CHAPTER 7 Exponential and Logarithmic Functions
437
CHAPTER 8 Trigonometric Functions and Their Derivatives
513
Using
McGraw-Hill Ryerson Calculus & Advanced Functions
,
Solutions
McGraw-Hill Ryerson Calculus & Advanced Functions
,
Solutions
provides complete model
solutions to the following:
for each numbered section of
McGraw-Hill Ryerson Calculus & Advanced Functions
,
- every odd numbered question in the
Practise
- all questions in the
Apply, Solve, Communicate
Solutions are also included for all questions in these sections:
- Review
- Chapter Check
- Problem Solving: Using the Strategies
Note that solutions to the Achievement Check questions are provided in
McGraw-Hill Ryerson
Calculus & Advanced Functions
, Teacher’s Resource.
Teachers will find the completeness of the
McGraw-Hill Ryerson Calculus & Advanced
Functions
,
Solutions
helpful in planning students’ assignments. Depending on their level of
ability, the time available, and local curriculum constraints, students will probably only be able to
work on a selection of the questions presented in the
McGraw-Hill Ryerson Calculus & Advanced
Functions
student text. A review of the solutions provides a valuable tool in deciding which
problems might be better choices in any particular situation. The solutions will also be helpful in
deciding which questions might be suitable for extra practice of a particular skill.
In mathematics, for all but the most routine of practice questions, multiple solutions exist. The
methods used in
McGraw-Hill Ryerson Calculus & Advanced Functions
,
Solutions
are generally
modelled after the examples presented in the student text. Although only one solution is presented
per question, teachers and students are encouraged to develop as many different solutions as
possible. An example of such a question is Page 30, Question 7, parts b) and c). The approximate
values can be found by substitution as shown or by using the Value operation on the graphing
calculator. Discussion and comparison of different methods can be highly rewarding. It leads to a
deeper understanding of the concepts involved in solving the problem and to a greater
appreciation of the many connections among topics.
Occasionally different approaches are used. This is done deliberately to enrich and extend the
reader’s insight or to emphasize a particular concept. In such cases, the foundations of the
approach are supplied. Also, in a few situations, a symbol that might be new to the students is
introduced. For example in Chapter 3 the dot symbol is used for multiplication. When a graphing
calculator is used, there are often multiple ways of obtaining the required solution. The solutions
provided here sometimes use different operations than the one shown in the book. This will help
to broaden students’ skills with the calculator.
There are numerous complex numerical expressions that are evaluated in a single step. The
solutions are developed with the understanding that the reader has access to a scientific
calculator, and one has been used to achieve the result. Despite access to calculators, numerous
problems offer irresistible challenges to develop their solution in a manner that avoids the need
for one, through the order in which algebraic simplifications are performed. Such challenges
should be encouraged.
iv
There are a number of situations, particularly in the solutions to Practise questions, where the
reader may sense a repetition in the style of presentation. The solutions were developed with an
understanding that a solution may, from time to time, be viewed in isolation and as such might
require the full treatment.
The entire body of
McGraw-Hill Ryerson Calculus & Advanced Functions
,
Solutions
was created
on a home computer in Te
x
tures. Graphics for the solutions were created with the help of a variety
of graphing software, spreadsheets, and graphing calculator output captured to the computer.
Some of the traditional elements of the accompanying graphic support are missing in favour of
the rapid capabilities provided by the electronic tools. Since many students will be working with
such tools in their future careers, discussion of the features and interpretation of these various
graphs and tables is encouraged and will provide a very worthwhile learning experience. Some
solutions include a reference to a web site from which data was obtained. Due to the dynamic
nature of the Internet, it cannot be guaranteed that these sites are still operational.
CHAPTER1FunctionsandModels
1.1FunctionsandTheirUseinModelling
Practise
Section1.1Page18Question1
a)
x
2[ 2
;
2]
-2
-1
0
1
2
b)
x
2[4
;
13]
-8
-4
0
4
8
12
13
16
c)
x
2( 4
;
1)
-5
-4
-3
-2
-1
0
d)
x
2(0
;
4)
0
1
2
3
4
e)
x
2( 1
;
2)
-2
-1
0
1
2
f)
x
2( 1
;
1)
-1
0
1
2
3
g)
x
2( 1
;
1]
-5
-4
-3
-2
-1
h)
x
2[0
;
1)
0
1
2
3
4
Section1.1Page18Question3
f
(
x
) = 2(
x
)
3
b)
g
(
x
) = (
x
)
3
4
c)
h
(
x
) = 1 (
x
)
2
= 2
x
3
=
f
(
x
)
=
x
3
4
6=
g
(
x
) or
g
(
x
)
= 1
x
2
=
h
(
x
)
f
(
x
) = 2
x
3
is o
d
d.
g
(
x
) =
x
3
4 is n
ei
ther even nor odd.
h
(
x
) = 1
x
2
is e
v
en.
8
8
8
6
6
6
y
4
y
4
y
4
2
2
2
–4 –3 –2
0
1 2 3 4
x
–4 –3 –2
0
1 2 3 4
x
–4 –3 –2
0
1 2 3 4
x
–2
–2
–2
–4
–4
–4
–6
–6
–6
–8
–8
–8
1.1 Functions and Their Use in Modelling MHR
1
a)
Section1.1Page19Question5
¬
¬
¬
¬
a)
f
(
x
) = j
x
j
b)
g
(
x
) =j
x
j+ 1
c)
h
(
x
) =
¬
¬
2(
x
)
3
¬
¬
= j
x
j
=
f
(
x
)
=j
x
j+ 1
=
g
(
x
)
=
2
x
3
=
h
(
x
)
The function
f
(
x
) = j
x
jis even.
The function
g
(
x
) =j
x
j+ 1 is even.
The function
h
(
x
) =
¬
¬
2
x
3
¬
¬
is even.
4
6
6
3
5
5
y
2
4
4
1
y
3
y
3
–4 –3 –2 –1
0
1 2 3 4
x
2
2
–1
1
1
–2
–4 –3 –2 –1
0
1 2 3 4
x
–4 –3 –2 –1
0
1 2 3 4
x
–3
–1
–1
–4
–2
–2
Section1.1Page19Question7
a)
Use
f
(
x
) =
1
x
2
.
f
(3) =
²
³
1
3
2
1
( 3)
2
1
3
1
i)
ii)
f
( 3) =
iii)
f
=
¡
2
1
3
1
9
1
9
=
=
1
1
9
= 9
²
1
4
³
1
²
1
k
³
1
²
k
1 +
k
³
1
v)
f
=
iv)
f
=
¡
2
vi)
f
=
¡
¡
1
4
2
1
k
k
1+
k
2
1
1
k
2
=
k
2
1
1
16
= 16
1
k
2
(1+
k
)
2
=
=
=
(1 +
k
)
2
k
2
=
b)
Use
f
(
x
) =
x
1
x
.
²
³
3
1 ( 3)
1
3
1
3
1
3
3
1 3
i)
f
(3) =
ii)
f
( 3) =
iii)
f
=
=
3
2
=
3
4
=
1
3
2
3
1
2
=
2
MHR Chapter 1
=
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