FUNCTIONS, LIMITS
AND CONTINUITY
Prof. Janette C. Lagos
Chapter Outline
1.1
1.2
1.3
1.4
Introduction
Concepts on Functions
Concepts on Limits
Concepts on Continuity of a Function at a
Given Point
Learning Objectives:
1.
2.
3.
4.
5.
6.
7.
1
0
Define a function
Evaluate functions
Perform operations on functions
Define different types of functions
Graphing functions
Determining the domain and range of functions
Define the limit of a function, one sided limit, limit at
infinity and infinite limit
8. Familiarize with the theorems on limits
9. Apply the theorems on limits in evaluating the limit of any
function
10. Define continuity of a function
11. Determine whether the function is continuous at a given
point or not.
12. Determine permissible values of the independent variable
at which the function be continuous
FUNCTIONS AND LIMITS P a g e | 2
1.1 Introduction
This chapter deals with three fundamental concepts of Calculus: functions, limits and continuity.
Function shows the dependency of one or more variables to another variable. Function also describes
the relationship that exists among the interplaying variables. The concept of limits allows us to
investigate thoroughly the movement of a function around a given point even when the function is
not defined at that point. Continuity tells us that there are no breaks or jumps at a given point of a
defined function.
Since calculus is a study of continuous change, these three concepts are embodied throughout the
discussion of all calculus courses. These three are also the primary considerations in developing
technologies. The function defines how the technology would probably work. The concepts of limit
and continuity help the developer of the technology to determine the optimum efficiency of the
technology and to determine factors, which may be represented by points along the function, at
which the technology would work effectively or not.
1.2 Concepts on Functions
Definition 1.2.1 A function is a set of ordered pairs of numbers (π₯, π¦) in which no two distinct
ordered pairs have the same first number. The set of all admissible values of π₯is called the domain
of the function, and the set of all resulting values of π¦ is called the range of the function.
If π is a function, then the graph of π is the set of all points (π₯, π¦) in π
2 for which (π₯, π¦) is an
ordered pair in π.
The graph of a function can be intersected by a vertical line in at most one point.
Definition 1.2.2
Given the two functions π and π:
a. their sum, denoted by π + π, is the function defined by
(π + π)(π₯) = π(π₯) + π(π₯)
b. their difference, denoted by π − π, is the function defined by
(π − π)(π₯) = π(π₯) − π(π₯)
c. their product, denoted by π ⋅ π, is the function defined by
(π ⋅ π)(π₯) = π(π₯) ⋅ π(π₯)
f
d. their quotient, denoted by g , is the function defined by
ο¦ f οΆ (x) = f (x) where π(π₯) ≠ 0
g (x)
ο¨gοΈ
Definition 1.2.3 Given the two functions π and π, the composite function, denoted by π β π, is
defined by (π β π)(π₯) = π(π(π₯)) and the domain of π β π is the set of all numbers π₯ in the domain
of π such that π(π₯) is in the domain of π.
FUNCTIONS AND LIMITS P a g e | 3
Definition 1.2.4
A function π is said to be an even function if for every π₯ in the domain of π,
(π)(−π₯) = π(π₯).
Definition 1.2.5 A function π is said to be an odd function if for every π₯ in the domain of π,
(π)(−π₯) = −π(π₯).
Definition 1.2.6
The linear function is a polynomial function of degree 1.
Definition 1.2.7
The linear function defined by π(π₯) = π₯ is called the identity function.
Definition 1.2.8
The quadratic function is a polynomial function of degree 2.
Definition 1.2.9
The cubic function is a polynomial function of degree 3.
Definition 1.2.10 If a function can be expressed as the quotient of two polynomial functions, it is
called a rational function.
Definition 1.2.11 An algebraic function is one formed by a finite number of algebraic operations on
the identity function and a constant function.
Definition 1.2.12 Transcendental functions include trigonometric functions, logarithmic functions,
exponential functions and, hyperbolic functions.
Definition 1.2.13 A piecewise defined function is a function defined by at least two equations,
each of which applies to a different part of the domain. It can take on a variety of forms. The
equations may be all linear, or a combination of functional forms such as constant, linear, quadratic,
cubic, radical and transcendental.
Definition 1.2.14 The absolute value function is defined by π(π₯) = |π₯|
Definition 1.2.15 The greatest integer function is defined by β¦π₯β§, which is used to denote the
greatest integer less than or equal to π₯ that is β¦π₯β§ = π ππ π ≤ π₯ < π + 1, π€βπππ π ππ ππ πππ‘ππππ
1.3 Concepts of Limits
Definition 1.3.1 Limit of a Function
Let π be a function that is defined at every number in some open interval containing π, except
possibly at the number π itself. The limit of π(π₯)as π₯ approaches π is πΏ, written as πππ π(π₯) = πΏ if
π₯→π
the following statement is true:
Given π > 0, however small, there exists a πΏ > 0 such that if 0 < |π₯ − π| < πΏ then |π(π₯) − πΏ| < π.
FUNCTIONS AND LIMITS P a g e | 4
Theorems on Limits of Functions
a.
If π and π are any constants, πππ (ππ₯ + π) = ππ + π
π₯→π
b. If π is a constant, then for any number π, πππ π = π
π₯→π
c. πππ π₯ = π
π₯→π
d. If πππ π(π₯) = πΏ and πππ π(π₯) = π, then πππ [π(π₯) ± π(π₯)] = πΏ ± π
π₯→π
π₯→π
π₯→π
e. If πππ π1 (π₯) = πΏ1 , πππ π2 (π₯) = πΏ2 , . . . , πππ πππ ππ (π₯) = πΏπ , then
π₯→π
π₯→π
π₯→π
πππ [π1 ± π2 (π₯)±. . . ±ππ (π₯)] = πΏ1 ± πΏ2 ± . . . ±πΏπ
π₯→π
f.
If πππ π(π₯) = πΏ and πππ π(π₯) = π, then πππ [π(π₯) ⋅ π(π₯)] = πΏ ⋅ π
π₯→π
π₯→π
π₯→π
g. If πππ π1 (π₯) = πΏ1 , πππ π2 (π₯) = πΏ2 , . . . , πππ πππ ππ (π₯) = πΏπ , then
π₯→π
π₯→π
π₯→π
πππ [π1 ⋅ π2 (π₯) ⋅. . .⋅ ππ (π₯)] = πΏ1 ⋅ πΏ2 ⋅ . . .⋅ πΏπ
π₯→π
h. If πππ π(π₯) = πΏ and π is any positive integer, then πππ π(π₯)π = πΏπ
π₯→π
π₯→π
π(π₯)
πΏ
=
π(π₯)
π
π₯→π
i.
If πππ π(π₯) = πΏ and πππ π(π₯) = π, then πππ
j.
If π is a positive integer and πππ π(π₯) = πΏ, then πππ √π(π₯) = √πΏ with the
π₯→π
π₯→π
ππ π ≠ 0
π
π₯→π
π
π₯→π
restriction that if π is even, πΏ > 0
k. πππ π(π₯) = πΏ if and only if πππ [π(π₯) − πΏ] = 0
π₯→π
l.
π₯→π
πππ π(π₯) = πΏ if and only if πππ[π(π‘ + π) − πΏ] = 0
π₯→π
π‘→0
Definition 1.3.2 Let π be a function that is defined at every number in some open interval (π, π).
Then the limit of π(π₯), as π₯ approaches a from the right, is πΏ, written πππ+π(π₯) = πΏ if for any π > 0,
π₯→π
however small, there exists a πΏ > 0 such that if 0 < |π₯ − π| < πΏ then |π(π₯) − πΏ| < π.
Definition 1.3.3 Let π be a function that is defined at every number in some open interval (π, π).
Then the limit of π(π₯), as π₯approaches a from the left, is πΏ, written πππ−π(π₯) = πΏ if for any π > 0,
π₯→π
however small, there exists a πΏ > 0 such that if 0 < |π₯ − π| < πΏ then |π(π₯) − πΏ| < π.
Definition 1.3.4 The πππ π(π₯) exists and is equal to πΏ if and only if πππ+π(π₯) and πππ−π(π₯) both
π₯→π
π₯→π
π₯→π
exists and both are equal to πΏ.
Definition 1.3.5 Let π be a function that is defined at every number in some open interval
πΌcontaining π, except possibly at the number π itself. As π₯ approaches π, π(π₯) increases without
FUNCTIONS AND LIMITS P a g e | 5
bound, which is written πππ π(π₯) = +∞ if for any number π > 0there exists a πΏ > 0 such that if 0 <
π₯→π
|π₯ − π| < πΏ then π(π₯) > π.
Definition 1.3.6 Let π be a function that is defined at every number in some open interval
πΌcontaining π, except possibly at the number π itself. As π₯ approaches π, π(π₯) decreases without
bound, which is written πππ π(π₯) = −∞ if for any number π > 0there exists a πΏ > 0 such that if 0 <
π₯→π
|π₯ − π| < πΏ then π(π₯) < π.
Theorems on Infinite Limits
a. If π is any positive integer, then
1
a.1 πππ+ π₯ π = +∞
π₯→π
1
a.2 πππ− π₯ π = {
π₯→π
−∞ ππ π ππ πππ
+∞ ππ π ππ ππ£ππ
b. If π is any real number, and if πππ π(π₯) = 0 and πππ π(π₯) = π, where π is a
π₯→π
π₯→π
constant not equal to 0 then
b.1 if π > 0 and if π(π₯) → 0 through positive values of π(π₯),
πππ
π(π₯)
π₯→π + π(π₯)
= +∞
b.2 if π > 0 and if π(π₯) → 0 through negative values of π(π₯),
πππ
π(π₯)
π₯→π − π(π₯)
= −∞
b.3 if π < 0 and if π(π₯) → 0 through positive values of π(π₯),
πππ
π(π₯)
π₯→π + π(π₯)
= −∞
b.4 if π < 0 and if π(π₯) → 0 through negative values of π(π₯),
πππ
π(π₯)
π₯→π − π(π₯)
= +∞
The theorem is also valid if “π₯ → π” is replaced by “π₯ → π+ ” or “π₯ → π− ”
c. If πππ π(π₯) = +∞, and πππ π(π₯) = π, where π is any constant, then
π₯→π
π₯→π
πππ [π(π₯) + π(π₯)] = +∞. The theorem is valid if “π₯ → π” is replaced by “π₯ →
π₯→π
+
π ” or “π₯ → π− ” .
FUNCTIONS AND LIMITS P a g e | 6
Theorems on Infinite Limits (continued)
d. If πππ π(π₯) = −∞, and πππ π(π₯) = π, where π is any constant, then
π₯→π
π₯→π
πππ [π(π₯) + π(π₯)] = −∞. The theorem is valid if “π₯ → π” is replaced by “π₯ → π+ ” or
π₯→π
“π₯ → π− ”
d. If πππ π(π₯) = +∞, and πππ π(π₯) = π, where π is any constant except 0, then
π₯→π
π₯→π
e.1 if π > 0, πππ [π(π₯) ⋅ π(π₯)] = +∞
π₯→π
e.2 if π < 0, πππ [π(π₯) ⋅ π(π₯)] = −∞
π₯→π
The theorem is valid if “π₯ → π” is replaced by “π₯ → π+ ” or “π₯ → π− ”
e. If πππ π(π₯) = −∞, and πππ π(π₯) = π, where π is any constant except 0, then
π₯→π
π₯→π
f.1 if π > 0, πππ [π(π₯) ⋅ π(π₯)] = −∞
π₯→π
f.2 if π < 0, πππ [π(π₯) ⋅ π(π₯)] = +∞
π₯→π
The theorem is valid if “π₯ → π” is replaced by “π₯ → π+ ” or “π₯ → π− ”
Definition 1.3.7 The line π₯ = π is said to be a vertical asymptote of the graph of the function π if
at least one of the following statements is true:
a. πππ+π(π₯) = +∞
π₯→π
b. πππ+π(π₯) = −∞
π₯→π
c. πππ−π(π₯) = +∞
π₯→π
d. πππ−π(π₯) = −∞
π₯→π
Definition 1.3.8 Let π be a function that is defined at every number in some interval (π, +∞). The
limit of π(π), as π increases without bound , is π³, written πππ π(π₯) = πΏ if for any π > 0, however
π₯→+∞
small, there exists a number π > 0 such that if π₯ > π then |π(π₯) − πΏ| < π
Definition 1.3.9 Let π be a function that is defined at every number in some interval (−∞, π). The
limit of π(π), as π decreases without bound , is π³, written πππ π(π₯) = πΏ if for any π > 0, however
π₯→−∞
small, there exists a number π < 0 such that if π₯ < π then |π(π₯) − πΏ| < π
FUNCTIONS AND LIMITS P a g e | 7
Definition 1.3.10 If π is any positive integer, then
a.
b.
πππ
1
=0
π₯→+∞ π₯ π
1
πππ
=0
π₯→−∞ π₯ π
Definition 1.3.10
The πΎ = π is said to be a horizontal asymptote of the graph of the function
π if at least one of the following statements is true:
a. πππ π(π₯) = π, and for some number π, if π₯ > π, then π(π₯) ≠ π;
π₯→+∞
b. πππ π(π₯) = π, and for some number π, if π₯ < π, then π(π₯) ≠ π;
π₯→−∞
1.4 Concepts on Continuity of a Function at a Number
Definition 1.4.1 The function π is said to be continuous at a number π if and only if the following
three conditions are satisfied:
a. π(π) exists
b. πππ π(π₯) exists
π₯→π
c. πππ π(π₯) = π(π)
π₯→π
If one or more of these three conditions fails to hold for π, the function π is said to be
discontinuous at π.
Theorems on Continuity
a. If π and π are two functions that are continuous at the number π, then
a.1 π + π is continuous at π;
a.2 π − π is continuous at π;
a.3 π ⋅ π is continuous at π;
π
a.4 is continuous at π, provided that πΌ(π) ≠ 0.
π
b. A polynomial function is continuous at every number.
c. A rational function is continuous at every number in its domain.
π
d. If π is a positive integer and π(π₯) = √π₯ then
d.1 if π is odd, π is continuous at every number;
d.2 if π is even, π is continuous at every positive number;
FUNCTIONS AND LIMITS P a g e | 8
Watch the following videos for further explanation and examples:
FUNCTIONS AND LIMITS P a g e | 9
Exercise 1.1
Functions
Name: ____________________________________________________
Course-Block: _________________ Schedule: ____________________
Professor: _________________________________________________
A. Evaluate the following:
2
3
3
1. π(0), π ( ), π(2β), if π(π₯) = √π₯ 3 − 4
1
2. π(−2), π (− 2), π(β + 3), if π(π₯) = 5π₯+2
π
3π
3. π ( 3 ), π (− 2 ), π(2ππ), if π(π) = πππ‘ 2 π = πππ (2π)
π₯+2
4. π(−3), π(√2), π(β2 ), if π(π₯) = ππ|7π₯ 2 + 2π₯ + 3| − π₯−2
Score:
FUNCTIONS AND LIMITS P a g e | 10
1
2
β
4
5. π(−2), π ( ), π ( ), if π(π₯) = π₯ π₯
2
x-2
B. Given the functions f(x) = x , g(x) = x2+1 and h(x) = x2 -4 , perform the indicated
operations
1. 2π − 3β
h
2. g + 4f4
3.
g
h
4.
g(2+k)-h(2)
k
5. π β β − π β β
FUNCTIONS AND LIMITS P a g e | 11
C.
Find the domain of the variable π₯ for which the following equations determine π¦ as a real
function of π₯.
1. 3π¦ − π₯π¦ = 5 + π₯
2. π¦ = √36 − π₯ 2
3. π¦ 3 = 4 + π₯ 2
4. π¦ = ππ π π₯
5. π¦ = 7π₯
D. Graph the following piecewise, absolute value and greatest integer functions
1.
2π₯ + 1 ππ π₯ ≠ 2
π(π₯) = { 2
π₯ − 4 ππ π₯ < 3
FUNCTIONS AND LIMITS P a g e | 12
2. π(π₯) = 6 + |π₯ + 5|
3. π(π₯) = {
π₯+5
ππ π₯ ≤ −5
√25 − π₯ 2
ππ − 5 < π₯ < 5
4. π(π₯) = β¦π₯ − 9β§
5. π(π₯) =
β¦π₯+2β§
|π₯|
FUNCTIONS AND LIMITS P a g e | 13
Exercise 1.2
Limits
Name: ____________________________________________________
Course-Block: _________________ Schedule: ____________________
Professor: _________________________________________________
A. Find the limit of the following:
1.
πππ (3π₯ 2 + 6π₯ − 5)
π₯→−4
2. πππ(π₯ 3 − 64)
π₯→5
3.
4.
2π₯ 3 +6
π₯→−2 5π₯−1
πππ
πππ
8π₯+1
π₯→−1 π₯ 2 +3π₯+4
3
2π₯ 2 −π₯−1
5. πππ √ π₯ 3 +1
π₯→1
Score:
FUNCTIONS AND LIMITS P a g e | 14
3π₯ 2 −8π₯−16
6. πππ 2π₯ 2 −9π₯+4
π₯→4
7.
8π₯ 3 +1
πππ1 2π₯+1
π₯→−
2
√π₯−1
π₯→1 π₯−1
8. πππ
9. πππ
π₯→0
√π₯+3−√3
π₯
√π₯+5−7
π₯→−1 π₯+1
10. πππ
B. Find the indicated limit if it exists
1. πππ+π(π₯) ππ π(π₯) = {
π₯→3
2π₯ + 1 ππ π₯ < 3
10 − π₯ ππ 3 ≤ π₯
FUNCTIONS AND LIMITS P a g e | 15
2. ππππ(π₯) ππ π(π₯) = 5 + |2π₯ − 4|
π₯→2
|π₯|
3. πππ−π(π₯) ππ π(π₯) = π₯
π₯→0
π₯ + 1 ππ π₯ < −1
4. πππ+π(π₯) ππ π(π₯) = {π₯ 2 ππ − 1 ≤ π₯ ≤ 1
π₯→−1
2 − π₯ + 1 ππ 1 < π₯
√π₯ 2 − 9 ππ π₯ ≤ −3
5. ππππ(π₯) ππ π(π₯) = {√9 − π₯ 2 ππ − 3 < π₯ < 3
π₯→3
√π₯ 2 − 9 ππ 3 ≤ π₯
C. Find the indicated limit if it exists
1.
x+3
πππ− x2-9
π₯→3
√5+π₯ 2
π₯→0 π₯
2. πππ
FUNCTIONS AND LIMITS P a g e | 16
3.
4.
5.
6.
7.
8.
5
3
πππ+ (π₯−7 + π₯ 2 −49)
π₯→7
6π₯ 2 +π₯−3
πππ + 2π₯2 +3π₯−2
π₯→−2
πππ
π₯+3
π₯→3− π₯ 2 −9
5−15π₯ 3
2 +7π₯ 3
6π₯
π₯→0
πππ+
πππ
π₯→8−
√64−π₯ 2
π₯−8
π ππ π₯
πππ+ π ππ 2π₯
π
π₯→
2
π‘ππ 2π₯
9. πππ π‘ππ π₯
π₯→0
FUNCTIONS AND LIMITS P a g e | 17
πππ π₯
10. πππ+ ππ π₯
π₯→0
D. Find the indicated limit if it exists
1.
2π₯+1
πππ 3π₯+1
π₯→−∞
1+5π₯
2. πππ 4−5π₯
π₯→∞
π₯ 2 +π₯
3. πππ π₯+2
π₯→∞
4.
2π₯
π₯
πππ (
− π₯+1)
π₯→−∞ π₯ 2 −1
5.
πππ √ 8π₯3 +π₯+2
3
4π₯ 3 +2π₯ 2 −5
π₯→−∞
2
π₯→∞ π₯
6. πππ ( 3 − 4π₯)
FUNCTIONS AND LIMITS P a g e | 18
√π₯ 2 +9
π₯→∞ π₯+9
7. πππ
8. πππ (√π₯ 2 + π₯) − π₯
π₯→∞
9.
√π₯ 2 −2π₯+3
π₯+7
π₯→−∞
πππ
3
3
10. πππ √π₯ 3 + 3 − √π₯ 3 + 1
π₯→−∞
FUNCTIONS AND LIMITS P a g e | 19
Exercise 1.3
Continuity
Name: ____________________________________________________
Course-Block: _________________ Schedule: ____________________
Professor: _________________________________________________
A. Determine whether the function is continuous at the given point, π = π
1. π(π₯) =
3π₯ 2 −10π₯+3
; ππ‘ π₯ = 0
π₯ 2 −3π₯
π₯ 3 −8
2. π(π₯) = π₯+2 ; ππ‘ π₯ = −2
3. π(π₯) =
4. π(π₯) =
π₯ 4 −81
; ππ‘ π₯ = 3
π₯ 2 −9
|π₯+5|
π₯
; ππ‘ π₯ = 0
5. π(π₯) = √π₯ 3 − 1; ππ‘ π₯ = −1
Score:
FUNCTIONS AND LIMITS P a g e | 20
π₯+6
6. π(π₯) = √π₯−6 ; ππ‘ π₯ = 6
7. π(π₯) = {
4 − π₯2
2π₯ + 3
ππ π₯ < 1
; ππ‘ π₯ = 1
ππ 1 ≤ π₯
ππ π₯ ≤ 0
√−π₯
8. π(π₯) = { 3
; ππ‘ π₯ = −1
√π₯ + 1 ππ π₯ ≠ 0
5
9. π(π₯) = π π₯ −8 ; ππ‘ π₯ = 0
10. π(π₯) = π ππ(π₯ − π); ππ‘ π₯ = 0
B. Using the functions in A, determine the values of π at which each function is continuous.
1. π(π₯) =
3π₯ 2 −10π₯+3
π₯ 2 −3π₯
π₯ 3 −8
2. π(π₯) = π₯+2
FUNCTIONS AND LIMITS P a g e | 21
π₯ 4 −81
3. π(π₯) = π₯ 2 −9
4. π(π₯) =
|π₯+5|
π₯
5. π(π₯) = √π₯ 3 − 1
π₯+6
π₯−6
6. π(π₯) = √
7. π(π₯) = {
4 − π₯2
2π₯ + 3
ππ π₯ < 1
ππ 1 ≤ π₯
ππ π₯ ≤ 0
√−π₯
8. π(π₯) = { 3
√π₯ + 1 ππ π₯ ≠ 0
5
9. π(π₯) = π π₯ −8
10. π(π₯) = π ππ(π₯ − π)
FUNCTIONS AND LIMITS P a g e | 22
Chapter
Review Exercise
Name: ____________________________________________________
Score:
Course-Block: _________________ Schedule: ____________________
Professor: _________________________________________________
Answer the following problems by showing the complete solution
π(2)−π(5)
3
if π(π₯) = 2π₯ + 3, π(π₯) = π₯ 2 − π₯ πππ β(π₯) = √π₯
π(−3)
πβπ
1
Evaluate 2π if π(π₯) = π₯ 3 πππ π(π₯) = π₯ 2
π₯ 3 +3π₯ 2 −4π₯−12
Determine the domain and the range of π(π₯) =
π₯ 2 +π₯−6
1. Evaluate
2.
3.
π₯+3
ππ π₯ < −2
2
4. Determine the domain and the range of π(π₯) = {4 − π₯
ππ − 2 ≤ π₯ ≤ 2
3−π₯
ππ 2 < π₯
√4π₯+3
5. Find the πππ
π₯→1 5−π₯−π₯ 2
6. Find the πππ
π₯→0
π₯ 2 − 9 ππ π₯ ≠ −3
7. Find the πππ π(π₯) if π(π₯) = {
π₯→−3
4
ππ π₯ = −3
8. Find the πππ −π(π₯) if π(π₯) = 5 + |3π₯ − 2|
π₯→−2
3
√π₯
9. Find the πππ +π(π₯) if π(π₯) = {
π₯→−0
√π₯
2
10. Find the πππ (π₯ + √π₯ + 2π₯)
ππ π₯ < 0
ππ 0 ≤ π₯
π₯→∞
π₯ 2 +4π₯+3
11. Find the πππ π₯−1
π₯→1
8π₯ 2 +14π₯+3
12. Find the πππ3 2
π₯→− 4π₯ +12π₯+9
2
13. Find the πππ (√π₯ + √π₯ + √π₯)
π₯→∞
14. Find the πππ π ππ π₯
π₯→−∞
πππ 2 π₯+1
π₯
π₯→−∞
π₯+2
16. Find the πππ (π₯−1)3
π₯→1
15. Find the πππ
17. Find the πππ
π‘ππ π₯
3π−
π₯→
2
18. Determine if the function π(π₯) = √4 − π₯ 2 is continuous at π₯ = −3
π
π
19. Determine if the function π(π₯) = πππ (π₯ − 2 ) is continuous at π₯ = 2
π
20. At what values of π₯ will the function π(π₯) π −π₯ −1be continuous
FUNCTIONS AND LIMITS P a g e | 23
References:
https://mathbitsnotebook.com/Algebra1/FunctionGraphs/FNGTypePiecewise.html
The Calculus with Analytic Geometry, 6th Edition. Leithold, Louis, 1990
Calculus with Analytic Geometry. Peterson, T.S., 1964
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