1. International Honor Program, School of Business, Kainan University
微積分
Calculus
ㄔㄣˊ ㄆㄨˇ ㄏㄨㄞˊ
陳 浦 淮
CHEN, Pu Huai
International Honor Program,
School of Business, Kainan University
商學院IHP課程授課內容
中華民國107年10月14日修訂
Revised on 14th, Oct., 2018.

2. About Dr. Pu-Huai Chen
Dr. Pu-Huai Chen was born in Kaohsiung, Taiwan, R.O.C., in 1965.
He received the B.Sc. degree from the Chung Cheng Institute of Technology (CCIT), Taiwan, in 1987, and the M.Sc. degree and the Ph.D. degree from University College London (UCL), London, U.K., in 1993 and 2000, respectively.
He was a Teaching Assistant from 1987 to 1992, and was appointed Lecturer in 1994 and Associate Professor in 2000 at CCIT.
In 2005, he moved to Kainan University. He is currently with the Department of Business and Entrepreneurial Management.

3. Syllabus (授課大綱) 1. Introduction 2. Algebra Reference
3. Linear Functions
4. Nonlinear Functions
5. The Derivative
6. Calculating the Derivative
7. Graphs and the Derivative
8. Applications of the Derivative
9. Integration

4. Syllabus (--- continued)
10. Applications of Integration
11. Multivariable Calculus

5. Textbook, Dictionary, and Handbook─
Lial, Greenwell and Ritchey, Calculus with Applications, 11th ed.. Pearson Education Limited, England. ISBN:
James and James, Mathematics Dictionary, 5th ed. Van Nostrand Reinhold, New York. ISBN:
Bronshtein and Semendyayev, Handbook of Mathematics (English translation edited by Hirsch), reprint of the 3rd ed. Springer-Verlag, Berlin. ISBN x

6. Chapter 4. Nonlinear Functions
Properties of Functions
Function.
A function is a rule that assigns to each element from one set exactly one element from another set.
The “rule” is expressed as an equation.
Each “set” will ordinarily be the real numbers or some subset of the reals.
When an equation is given for a function, we say that the equation defines the function.

7. 4.1 Properties of Functions
Functions arise in numerous applications. and an understanding of them is critical for understanding calculus.
Domain and Range
The set of all possible values of the independent variable in a function is called the domain of the function
The resulting set of possible values of the dependent variable is called the range.

8. 4.1 Properties of Functions
Domain and Range (-continued)
Several different values of the independent variable can have the same value for the dependent variable.
One value of the independent variable cannot lead to several different values of the dependent variable.
Decide whether an equation or a graph represents a function.
Examples

9. 4.1 Properties of Functions
VII-3
Agreement on Domains
Unless otherwise stated, assume that the domain of all functions defined by an equation is the largest subset of real numbers that are meaningful replacement for the independent variable.
Suppose y = −𝟐𝒙 𝟑𝒙−𝟏 .
The domain of this function is the set of all real numbers except 1/3, which we denote {x | x ≠ 1/3}, or (-∞, 1/3)∪(1/3, ∞).

10. 4.1 Properties of Functions
Caution! When finding the domain of a function, there are two operations to avoid:
Dividing by zero;
Taking the square root (or any even root) of a negative number.
Some functions, such as logarithms, require further restrictions on the domain.
Find the domain and range for functions.
Examples

11. 4.1 Properties of Functions
Domain and Range
Find the domain and range for each function defined as follows.
Example
f(x) = x2
Solution:
Any number may be squared, so the domain is the set of all real numbers, written (－∞, ∞). Since x2 ≥ 0 for every value of x, the range is
[0, ∞). ■

12. 4.1 Properties of Functions
Example
y = x2, with the domain specified as {－2, －1, 0, 1, 2}.
Solution:
With the domain specified, the range is the set of values found by applying the function to the domain.
Since f(0) = 0 , f(－1) = f(1) = 1, and
f(－2) = f(2) = 4, the range is {0, 1, 4}. ■

13. 4.1 Properties of Functions
Example
y = 𝟔−𝒙
Solution:
For y to be a real number, 6 – x must be nonnegative. This happens only when 6 – x ≥ 0, or 6 ≥ x, making the domain (－∞, 6].
The range is [ 0, ∞) because 𝟔−𝒙 is always nonnegative. ■

14. 4.1 Properties of Functions
Example
y = 𝟐𝒙𝟐+𝟓𝒙−𝟏𝟐
Solution:
The domain includes only those values of x satisfying 2x2 + 5x – 12 ≥ 0. Using the methods for solving a quadratic inequality produces the domain
(－∞, 4]∪[3/2, ∞).
The range is [0, ∞) because
y = 𝟐𝒙𝟐+𝟓𝒙−𝟏𝟐 is always nonnegative. ■

15. 4.1 Properties of Functions
Example
y = 𝟐 𝒙𝟐−𝟗
Solution:
Since the denominator cannot be zero, x ≠ 3 and
x ≠ －3. The domain is
(－ ∞, － 3)∪(－ 3, 3)∪(3, ∞).
Because the numerator can never be zero, y ≠ 0. The denominator can take on any real number except for 0, allowing y to take on any value except for 0, so the range is (－ ∞, 0)∪(0, ∞).

16. 4.1 Properties of Functions
Understanding how a function works
Think of a function f as a machine, such as a computer---that takes an input x from the domain and use it to produce an output f(x) (representing the y-value) within the range.
Domain X
Range Y x f
f(x)
Fig How a function works

17. 4.1 Properties of Functions
Evaluating functions.
Examples
Caution! Notice that g(x＋h) is not the same as g(x)＋h.
Vertical Line Test
If a vertical line intersects a graph in more than one point, the graph is not the graph of a function.
A graph represents a function if and only if every vertical line intersects the graph in no more than one point.

18. 4.1 Properties of Functions
Even function. f(－x) = f(x)
The graph is symmetric about the y-axis.
Odd function. f(－x) = － f(x)
The graph is symmetric about the origin.
Step function
Examples

19. 4.2 Quadratic Functions
Quadratic Functions: Translation and Reflection
Quadratic Function
A quadratic function is defined by
f(x) = ax2 ＋ bx ＋ c
where a, b, and c are real numbers (a≠0)
In a quadratic function the independent variable is squared.
A quadratic function is an especially good model for many situations with a maximum or a minimum function value.

20. 4.2 Quadratic Functions Parabola
Every quadratic function has a parabola as its graph.
The lowest (or highest) point on a parabola is the vertex of the parabola.
The graph of a quadratic function is symmetric with respect to a vertical line through the vertex; this line is the axis of symmetry of the parabola.
A projectile thrown in the air follows a parabolic path.

21. Fig. 4.2.1 The simplest quadratic functionx y
y = x2
is a parabola.
The dependent variable y is proportional to the square of the dependent variable x.
The y-axis
is the axis of symmetry of the parabola.
If the graph were folded in half along the y-axis,
the two halves of the parabola would match exactly.
the vertex of the parabola is (0, 0)
Fig The simplest quadratic function

22. 4.2 Quadratic Functions Vertical Translation
The effect of c in ax2 ＋ bx ＋ c is to lower the graph if c is negative and to raise the graph if c is positive.
The movement up or down is referred to as a vertical translation of the function.

23. Fig. 4.2.2 Vertical translationx y
y = x2＋2
vertical translation
y = x2－14
Fig Vertical translation

24. 4.2 Quadratic Functions Vertical Reflection
The sign of a in ax2 ＋ bx ＋ c determines whether the parabola opens upward or downward.
Multiplying f(x) by a negative number flips the graph of f upside down. This is called a vertical reflection of the graph.
The magnitude of a determines how steeply the graph increases or decreases.

25. Fig. 4.2.3 Graphing quadratic functionsx y
y = 0.2 x2
vertical reflection
y = －0.2x2
y = －0.5x2
y = －x2
y = －2x2
Fig Graphing quadratic functions

26. 4.2 Quadratic Functions Horizontal Translation
If a quadratic equation is given in the form as ax2 ＋ bx ＋ c, we can identify the translation and any vertical reflection by rewriting it in the form
y = a(x－h)2 ＋ k,
where the vertex is given as (h, k).
In this form, h determines the left (if h is positive) or (if h is negative) right shift of the graph of the quadratic equation.

27. horizontal translationx y
y = (x＋10)2
y = x2
y = (x－10)2
horizontal translation
Fig Horizontal translations

28. 4.2 Quadratic Functions A Formula for the Vertex
By the quadratic formula, if ax2 ＋ bx ＋ c, where a≠0, then
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
= −𝑏 2𝑎 ± 𝑏 2 −4𝑎𝑐 2𝑎 = −𝑏 2𝑎 ±𝑸,
where Q = 𝑏 2 −4𝑎𝑐 /(2a).
Since a parabola is symmetric with respect to its axis, the x-coordinate of the vertex is halfway between its two roots, or x=－b/(2a).

29. 4.2 Quadratic Functions The Graph of the Quadratic Function
f(x) = ax2 ＋ bx ＋ c, where a≠0, is a parabola with the following characteristics:
The parabola opens upward if a＞0 and downward if a＜0;
The y-intercept is at (0, c);
The x-intercepts, if they exist, occur where f(x) = 0;
The vertex is at (－b/(2a), f(－b/(2a)));
The axis of symmetry is x =－b/(2a).

30. 4.2 Quadratic Functions Examples Graphing a Quadratic Function
y = x2＋4x＋6.
Solution:
Since a = 1＞0, the parabola opens upward.
The y-intercept is at (0, 6).
We try the quadratic formula. x = −𝟒± −𝟖 𝟐 There are no real number solution.
Therefore, there are no x-intercept.
The vertex is at (－2, 2).
The axis of symmetry is the line x =－2.

31. Fig. 4.2.5 The graph of y = x2＋4x＋6 (-5, 11) (1, 11) (-4, 6)x y
(-5, 11)
(1, 11)
(-4, 6)
(0, 6) y-intercept
y = x2＋4x＋6
vertex (-2, 2)
The axis of symmetry is the line x =－2.
There are no real number solution.
Therefore, there are no x-intercept.
Fig The graph of y = x2＋4x＋6

32. 4.2 Quadratic Functions Examples Management Science Profit
Elderly in the U.S. Labor Force

33. 4.2 Quadratic Functions Examples
When Power and Money, Inc. , charges $600 for a seminar on management techniques, it attracts 1000 people. For each $20 decrease in the fee, an additional 100peoplewill attend the seminar. The managers are wondering how much to charge for the seminar to maximize their revenue.
Solution:
Let x be the number of $20 decreases in the price. Then the price charged per person will be
Price per person = 600 – 20x,
and the number of people in the seminar will be
Number of people = x.

34. 4.2 Quadratic Functions
Solution: (cont.) The total revenue, R(x), is given by the product of the price and the number of people attending, or R(x) = (600 – 20x)(1000 – 100x) =600, ,000x – 2000x2. We see by the negative in the x2 – term that this defines a parabola opening downward, so the maximum revenue is at the vertex. The x-coordinate of the vertex is x = −𝒃 𝟐𝒂 = −𝟒𝟎,𝟎𝟎𝟎 𝟐(−𝟐𝟎𝟎𝟎) =𝟏𝟎. The y-coordinate is then y = 600, ,000(10) – 2000(102) = 800,000. Therefore, the maximum revenue is $800,000, which is achieved by charging 600 – 20x = 600 – 20(10) = $400 per person.

35. 4.2 Quadratic Functions Examples
A deli owner has found that his revenue from producing x pounds of vegetable cream cheese is given by R(x) = -x2 + 30x, while the cost in dollars is given by C(x) = 5x
Find the minimum break-even quantity.
Solution:
Notice from the graph in Figure 21 that the revenue function is a parabola opening downward and the cost function is a linear function that crosses the revenue function at two points. To find the minimum break-even quantity, we find where the two functions are equal.
R(x) = C(x)
-x2 + 30x = 5x + 100
0 = x2 – 25x + 100
= (x - 5)(x - 20)
The two graphs cross when x = 5 and x = 20. The minimum break-even point is at x = 5. The deli owner must sell at least 5 lb of cream cheese to break even.

36. 4.2 Quadratic Functions Examples Find the maximum revenue. Solution:
Since the revenue function is a quadratic with a negative x2-term, the maximum is at the vertex. The vertex has a value of x = -b/(2a) = -30/(-2) = 15. The maximum revenue is R(15) = (15) = 225, or $225.

37. 4.2 Quadratic Functions Examples Find the maximum profit. Solution:
The profit is the difference between the revenue and the cost, or
P(x) = R(x) – C(x)
= (-x2 + 30x) – (5x + 100)
= -x2 + 25x – 100.
The value of x at the vertex is x = -b/(2a) = -25/(-2) = The value of the function here is P(12.5) = (12.5) – 100 = It is clear that this is a maximum, not only from Figure 21, but also because the profit function is a quadratic with a negative x2-term. A maximum profit of $56.25 is achieved by selling 12.5 lb of cream cheese.

38. 4.3 Polynomial and Rational Functions
Polynomial Functions
A polynomial function of degree n, where n is a nonnegative integer, is defined by
f(x) = anxn＋an-1xn-1 ＋……＋ a1x ＋ a0
where an, an-1, ……, a1, a0 are real numbers,
, called coefficients, with a≠0. The number
an is called the leading coefficient.
For n = 1, a polynomial function takes the form f(x) = a1x ＋ a0, a linear function. or a polynomial function of degree 1.

39. 4.3 Polynomial and Rational Functions
Polynomial Functions (-continued)
A polynomial function of degree 0 is the constant function.
A polynomial function of degree 2 is a quadratic function.
A polynomial function of degree 3 is known as a cubic function.
A polynomial function of degree 4 is known as a quartic function.
The graphs of all polynomials are smooth curves.

40. 4.3 Polynomial and Rational Functions
Polynomial Functions (-continued)
The simplest polynomial functions of higher degree are those of the form f(x) = xn, or a power function.

41. Fig. 4.3.1 Graphing odd functionsx y
The graphs of the odd functions are symmetric about the origin.
f(x) = 0.05x3
f(x) = 0.05x5
Fig Graphing odd functions

42. Fig. 4.3.2 Graphing even functionsx y
f(x) = 0.1x4
f(x) = 0.1x2
The graphs of the even functions have symmetry about the y-axis.
Fig Graphing even functions

43. 4.3 Polynomial and Rational Functions
Properties of Polynomial Functions
A polynomial function of degree n can have at most n－1 turning points. Conversely, if the graph of a polynomial function has n turning points, it has degree at least n＋1.
In the graph of a polynomial function of even degree, both ends go up or both ends go down. For a polynomial function of odd degree, one end goes up and one end goes down.

44. 4.3 Polynomial and Rational Functions
Properties of Polynomial Functions (-contd.)
If the graph goes up as x becomes a large positive number, the leading coefficient must be positive. If the graph goes down as x becomes a large positive number, the leading coefficient is negative.
Examples

45. Fig. 4.3.3 To be compared with Fig. 4.3.1.x y
The graphs of the odd functions with negative leading coefficients.
f(x)=－0.05x3
f(x) =－0.05x5
Fig To be compared with Fig

46. 4.3 Polynomial and Rational Functions
A rational function is defined by
f(x) = 𝒑(𝒙) 𝒒(𝒙) ,
where p(x) and q(x) are polynomial functions
and q(x) ≠ 0.
Since any value of x such that q(x) = 0 are excluded from the domain, a rational function often has a graph with one or more breaks.
Examples

47. Fig. 4.3.4 Graphing a rational functionx y
The line x = 0 (the y-axis) is a horizontal asymptote.
The graphs of the function approaches the vertical line x = 0 ( the y-axis) without ever touching it.
y = 𝟏𝟎 𝒙
The graphs of the function approaches the horizontal line y = 0 (the x-axis) without ever touching it.
The line y = 0 (the x-axis) is a horizontal asymptote.
Fig Graphing a rational function

48. 4.3 Polynomial and Rational Functions
A rational function is defined by
f(x) = 𝒑(𝒙) 𝒒(𝒙) ,
where p(x) and q(x) are polynomial functions
and q(x) ≠ 0.
Since any value of x such that q(x) = 0 are excluded from the domain, a rational function often has a graph with one or more breaks.
Examples

49. 4.3 Polynomial and Rational Functions
Asymptotes
If a function gets larger and larger in magnitude without bound as x approaches the number k, then the line x = k is a vertical asymptote.
If the value of y approach a number k as |x| gets larger and larger, the line y = k is a horizontal asymptote.
If a number k makes the denominator 0 but does not make the numerator 0, then the line x = k is a vertical asymptote. See Fig, 4.3.4

50. 4.3 Polynomial and Rational Functions
Examples
Graphing a Rational Function, p.102.

51. 4.3 Polynomial and Rational Functions
Examples
y = 𝟑𝒙+𝟐 𝟐𝒙+𝟒 .
Solution:
The value x=-2 makes the denominator 0, but not the numerator, so the line x=-2 is a vertical asymptote. To find a horizontal asymptote, let x get larger and larger, so that 3x+2 ≈ 3x because the 2 is very small compared with 3x. Similarly, for x very large, 2x+4 ≈ 2x, Therefore, y = (3x+2)/(2x+4) ≈ (3x)(2x) =3/2. This means that the line y = 3/2 is a horizontal asymptote. (A more precise way of approaching this idea will be seen in the next chapter when limits at infinity are discussed.)

52. 4.3 Polynomial and Rational Functions
Solution: (cont.) The intercepts should also be noted. Setting x=0, we find that y-intercept is y=2/4=1/2. To find the x-intercept( also called the root), we set y=0. Note that to make a fraction zero, the number must be zero. Solving 3x+2=0, we get the x-intercept x=-2/3. We can also use these values to determine where the function is positive and where it is negative on(-2, -2/3) and positive on(-∞, -2)U(-2/3,∞). With this information, the two asymptotes to guide us, and the fact that there are only two intercepts, we suspect the graph is as shown in Figure 47. A graphing calculator can support this.

53. 4.3 Polynomial and Rational Functions
Solution: (cont.) Rational functions occur often in practical applications. In many situations involving environmental pollution, much of the pollutant can be removed from the air or water at a fairly reasonable cost, but the small part of the pollutant can be very expensive to remove. Cost as a function of the percentage of pollutant removed from the environment can be calculated for various percentages of removal, with a curve fitted through the resulting data points. This curve then leads to a mathematical model of the situation. Rational functions are often a good choice for these cost-benefit models because they rise rapidly as they approach a vertical asymptote.

54. 4.3 Polynomial and Rational Functions
Examples
Suppose a cost-benefit model is given by
y = 𝟏𝟖𝒙 𝟏𝟎𝟔−𝒙 ,
where y is the cost(in thousands of dollars) of removing x percent of a certain pollutant. The domain of x is the set of all number from 0 to 100 inclusive; any amount of pollutant from 0% to 100% can be removed. Find the cost to remove following amounts of the pollutant: 100%,95%,90%, and80%. Graph the function.
Solution:
Removal of 100% of the pollutant would cost
𝒚= 𝟏𝟖(𝟏𝟎𝟎) 𝟏𝟎𝟔−𝟏𝟎𝟎 =𝟑𝟎𝟎,
or$300,000. Check that 95% of the pollutant can be removed for $155,000,90% for $101,000, and 80%for $55,000. Using these points, as well as others obtained from the function ,gives the graph shown in Figure48.

55. 4.3 Polynomial and Rational Functions
If a cost function has the function has the from C(x)=mx + b, where x is the number of items produced, m is the marginal cost per item and b is the fixed cost, then the average cost per item is given by 𝑪 (x)= 𝑪(𝒄) 𝒙 = 𝒎𝒙+𝒃 𝒙 . Notice that this is a rational function with a vertical asymptote at x=0 and a horizontal asymptote at y=m. The vertical asymptote reflects the fact that, as the number of items produced approaches 0, the average cost per item becomes infinitely large, because the fixed costs are spread over fewer and fewer items. The horizontal asymptote shows that, as the number of items becomes large, the fixed costs are spread over more and more items, so most of the average cost per item is the marginal cost to produce each item. This is another example of how asymptotes give important information in real applications.

56. 4.4 Exponential Functions
An exponential function with base a is defined as
f(x) = ax, where a ＞ 0 and a ≠ 1.
Exponential functions are used to describe growth and decay, which are important ideas in management, social science, and biology.
The domain is the set of all real numbers and the range is the set of all positive number.

57. Fig. 4.4.1 Graphing an exponential functionx y
y = 2x
The line y = 0 (the x-axis) is a horizontal asymptote.
Fig Graphing an exponential function

58. 4.4 Exponential Functions
Exponential Equation
An equation with a variable in the exponent is called an exponential equation.
Exponential equation can be solved using the following property.
If a ＞ 0, a ≠ 1, and ax = ay, then x = y.
Examples

59. 4.4 Exponential Functions
Compound Interest
The cost of borrowing money or the return on an investment is called interest.
The amount borrowed or invested is the principal, P. The rate of interest r is given as a percent per year, and t is the time, measured in years.
Simple Interest
The product of the principal P, rate r, and the time t gives simple interest, I:
I = Prt

60. 4.4 Exponential Functions
Compound Interest
With compound interest, interest is charged (or paid) on interest as well as on the principal.
Suppose that the principal, P dollars, is deposited at a rate of interest r per year.
The interest earned during the first year is found using the formula for simple interest, or P．r．1 = Pr.
At the end of the first year, the amount on deposit will be P＋Pr = P(1＋r).

61. 4.4 Exponential Functions
If the deposit earns compound interest, the interest earned during the second year is [P(1＋r)]．r．1 = P(1＋r)r.
At the end of the second year, the total amount on deposit will be
P(1＋r)＋P(1＋r)r
= P(1＋r)(1＋r) = P(1＋r)2. ……
After t years, The total amount on deposit, called the compound amount, is P(1＋r)t.

62. 4.4 Exponential Functions
Compound Amount
When interest is compounded more than once a year, the compound amount formula is adjusted.
If P dollars is invested at a yearly rate of interest r per year, compounded m times per year for t years, the compound amount is
A = P 𝟏+ 𝒓 𝒎 𝒕𝒎 dollars.
Examples

63. 4.4 Exponential Functions
Examples
Morgan Presley invests a bonus of $9000 at 6% annual interest compounded semiannually for 4 years. How much interest will she earn?
Solution:
Use the formula for compound amount with P = 9000, r = 0.06 , m = 2(since interest is compounded semiannually), and t =4.
A = 𝑷 𝟏+ 𝒓 𝒎 𝒕𝒎
= 9000 𝟏+ 𝟎.𝟎𝟔 𝟐 4(2)
= 9000 (𝟏.𝟎𝟑) 𝟖
≈ 11,400.93
The compound amount (investment plus the interest) is $11, The interest is $11, $9000=$

64. 4.4 Exponential Functions
The Number e
Definition of e, an irrational number
As m becomes larger and larger, 𝟏+ 𝟏 𝒎 𝒎 becomes closer and closer to the number e , whose approximate value to 15 decimal places is …
The number e is often used as the base in an exponential function because it provides a good model for many natural, as well as economic, phenomena.

65. 4.4 Exponential Functions
Continuous Compounding
Let us rearrange the formula for compound amount to take advantage of this fact.
A=P 𝟏+ 𝒓 𝒎 𝒕𝒎 =𝑷 𝟏+ 𝟏 (𝒎/𝒓) 𝒕𝒎
= P 𝟏+ 𝟏 (𝒎/𝒓) 𝒕𝒎 𝒓𝒕
This last expression becomes closer and closer to Pert as m becomes larger and larger, which describes what happens when interest is compounded continuously.

66. 4.4 Exponential Functions
Continuous Compounding
If a deposit of P dollars is invested at a rate of interest r compounded continuously for t years, the compound amount is
A = Pert dollars.
Examples
Continuous Compound Interest
Oxygen Consumption
Food Surplus

67. 4.4 Exponential Functions
Examples
If $600 is invested in an account earning 2.75% compounded continuously, how much would be in the account after 5 years?
Solution:
In the formula for continuous compounding, let P=600, t = 5, and r = to get
A= 600e5(0.0275) ≈ ,
or$

68. 4.4 Exponential Functions
Examples
Biologists studying salmon have found that the oxygen consumption of yearling salmon(in appropriate units) increases exponentially with the speed of swimming according to the function defined by
f(x) = 100 𝑒 𝟎.𝟔𝒙 ,
where x is the speed in feet per second. Find the following.
The oxygen consumption when the fish are still
Solution:
When the fish are still, their speed is 0. Substitute 0 for x:
f(0) = 100e(0.6)(0) =100e0
= 100▪1 = 100.
When the fish are still, their oxygen consumption is 100 units.

69. 4.4 Exponential Functions
Examples
Biologists studying salmon have found that the oxygen consumption of yearling salmon(in appropriate units) increases exponentially with the speed of swimming according to the function defined by
f(x) = 100 𝑒 𝟎.𝟔𝒙 ,
where x is the speed in feet per second. Find the following.
The oxygen consumption at a speed of 2ft per second
Solution:
Find f(2) as follows.
f(2) = 100e(0.6)(0) =100e1.2 ≈ 332

70. 4.5 Logarithmic Functions
For a > 0, a ≠ 1, and x > 0
y = loga x means ay =x
(Read y = loga x as “y is the logarithm of x to the base a.)
A logarithm is an exponent: logax is the exponent used with the base a to get x.

71. 4.5 Logarithmic Functions
Examples
Doubling time
For $1 to double (become $2) in t years, assuming 5% annual compounding, means that
A=P 𝟏+ 𝒓 𝒎 𝒎𝒕 becomes 2=1 𝟏+ 𝟎.𝟎𝟓 𝟏 𝟏(𝒕)
or 2 = (1.05)t .
(1.05)t = 2 can be rewritten as t = log1.05 2
Evaluating Logarithms

72. 4.5 Logarithmic Functions
For a given positive value of x, the definition of logarithm leads to exactly one value of y, so y = loga x defines the logarithmic function of base a (the base a must be positive, with a≠1)
Logarithmic Function
If a > 0 and a ≠ 1,then the logarithmic function of base a is defined by
f(x) = 𝐥𝐨𝐠 𝒂 𝒙
for x > 0.

73. 4.5 Logarithmic Functions
Graphing Logarithmic Functions
For any number m, if f(m) = p, then g(p)= m. Function related in this way are called inverse functions of each other.
The graphs also show that the domain of the exponential function (the set of real numbers) is the range of the logarithmic function.
Also, the range of the exponential function(the set of positive real numbers) is the domain of the logarithmic function.

74. Fig. 4.5.1 Inverse functions (3, 8) (2, 4) f(x) = 2x (0, 1) (8, 3)x y
(3, 8)
(2, 4)
f(x) = 2x
(0, 1)
(8, 3)
(4, 2)
(1, 0)
g(x) = log2 x
Fig Inverse functions

75. 4.5 Logarithmic Functions
Every logarithmic function is the inverse of some exponential function. This means that we can graph logarithmic functions by rewriting them as exponential functions using the definition of logarithm.
The graphs in Fig. show a characteristic of a pair of inverse functions: Their graphs are mirror images about the line y = x.
Caution! The domain of loga x consists of all x > 0. In other words, you cannot take the logarithm of zero or a negative number.

76. 4.5 Logarithmic Functions
Properties of Logarithms
Let x and y be any positive real numbers and r be any real number. Let a be a positive real number, a≠1. Then
log 𝒂 𝒙𝒚= log 𝒂 𝒙+ log 𝒂 𝒚
log 𝒂 𝒙 𝒚 = log 𝒂 𝒙 － log 𝒂 𝒚
log 𝒂 𝒙 𝒓 = r log 𝒂 𝒙
log 𝒂 𝒂 = 1
log 𝒂 𝟏=𝟎
log 𝒂 𝒂 𝒓 = r.

77. 4.5 Logarithmic Functions
Properties of Logarithms
To prove property (a), let m = loga x and n= loga y. Then, by the definition of logarithm,
am = x and an = y.
Hence,
aman = xy.
By a property of exponents, aman = 𝒂 𝒎+𝒏 , so
𝒂 𝒎+𝒏 =𝒙𝒚
Now use the definition of logarithm to write
log 𝒂 𝒙𝒚=𝒎+𝒏.
Since m = log 𝒂 𝒙𝒚 = log 𝒂 𝒙 + log 𝒂 𝒚.

78. 4.5 Logarithmic Functions
Base 10 logarithms were called common logarithms.
𝐥𝐨𝐠 𝟏𝟎 𝒙 is abbreviated log x.
Logarithms to base e are called natural logarithms.
𝐥𝐨𝐠 𝒆 𝒙 is abbreviated ln x.
Change-of-Base Theorem for Logarithms
If x is any positive number and if a and b are positive real number, a ≠ 1, b ≠ 1, then
𝐥𝐨𝐠 𝒂 𝒙= 𝐥𝐨𝐠 𝒃 𝒙 𝐥𝐨𝐠 𝒃 𝒂 .

79. 4.5 Logarithmic Functions
Using ln x for loge x gives the special case of the theorem using natural logarithms.
For any positive numbers a and x, a≠1,
𝐥𝐨𝐠 𝒂 𝒙= 𝐈𝐧 𝒙 𝐈𝐧 𝒂 .

80. 4.5 Logarithmic Functions
Logarithmic Equation
Equations involving logarithms are often solved by using the fact that exponential functions and logarithmic functions are inverses, so a logarithmic equation can be rewritten (with the definition of logarithm) as an exponential equation. In other cases, the properties of logarithm may be useful in simplifying a logarithmic equation.
Solving Logarithmic Equations
Solving Exponential Equations

81. 4.5 Logarithmic Functions
Change-of-Base Theorem for Exponentials
For every positive real number a,
𝒂 𝒙 = e(In a)x.
Examples

82. 4.5 Logarithmic Functions
Examples
log 𝒙 + log 𝒙−𝟑 =𝟏
Solution:
Since log 𝒂 𝒙 + log 𝒂 𝒚 = log 𝒂 𝒙𝒚, we have
log 𝒙 + log (𝒙−𝟑) = log 𝒙 𝒙−𝟑 =𝟏.
Since the logarithm base is 10, this means that
𝒙 𝒙−𝟑 =𝟏𝟎
𝒙 𝟐 −𝟑𝒙−𝟏𝟎=𝟎
(x-5)(x+2) = 0.
This leads to two solutions: x = 5 and x = -2. But notice that -2 is not a valid value for x in the original, since the logarithm of a negative number is undefined. Therefore, the only solution is x = 5.

83. 4.5 Logarithmic Functions
Examples
Solve each equation.
3x = 5
Solution:
Taking natural logarithms (logarithms to any base could be used) on both sides gives
ln3x = ln5
x ln3= ln5
x = 𝒍𝒏𝟓 𝒍𝒏𝟑 ≈𝟏.𝟒𝟔𝟓.

84. 4.5 Logarithmic Functions
Examples
With an inflation rate averaging 5% per year, how long will it take for prices to double?
Solution:
Recall that if prices will double after t years at an inflation rate of 5%, compounded annually, t is given by the equation A = P(1 + r)1
2 = (1.05)1.
We solve this equation by first taking natural logarithms on both sides. ln2 = ln(1.05)t
ln2 = t ln 1.05
t = 𝒍𝒏𝟐 𝒍𝒏𝟏.𝟎𝟓 ≈𝟏𝟒.𝟐
It will take about 14 years for prices to double.