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Applied Calculus Math 215

Karl Heinz Dovermann Professor of Mathematics University of Hawaii

July 5,

c Copyright 1999 by the author

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publication may be reproduced,

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the American Mathematical Society’s TEX macro system,

- and LATEX 2ε

This book is dedicated to my wife Emily (Eun Hee) and my sons Christopher and Alexander

Mathematica Version 2

Wolfram Research,

Champaign,

- 0 A Preview
- 1 Some Background Material 1

1 Lines

- 2 Parabolas and Higher Degree Polynomials
- 3 The Exponential and Logarithm Functions
- 4 Use of Graphing Utilities

2 The 2

Differentiability as a Local Property

Slopes of Secant Lines and Rates of Change Upper and Lower Parabolas

Exponential Growth and Decay

Being Close Versus Looking Like a Line

Rules of Differentiation

- 1 Linearity of the Derivative
- 2 Product and Quotient Rules
- 3 Chain Rule
- 4 Derivatives of Inverse Functions
- 12 Implicit Differentiation
- 13 Related Rates
- 14 Numerical Methods
- 7 7 13 20 31
- 35 42 47 48 57 61 69 70 76 81 83 84 85 87 91 101 111 115 118
- 1 Approximation by Differentials 2
- 2 Newton’s Method
- 3 Euler’s Method

15 Summary

- 3 Applications of the Derivative 3
- 1 Differentiability on Closed Intervals
- 2 Cauchy’s Mean Value Theorem
- 3 The First Derivative and Monotonicity 3
- 4 The Second and Higher Derivatives
- 5 The Second Derivative and Concavity 3
- 6 Local Extrema and Inflection Points
- 7 The First Derivative Test
- 8 The Second Derivative Test
- 9 Extrema of Functions
- 10 Detection of Inflection Points
- 11 Optimization Problems
- 12 Sketching Graphs
- 118 120 125 134
- 4 Integration 4
- 1 Upper and Lower Sums
- 2 Integrability and Areas
- 3 Some elementary observations
- 4 Integrable Functions
- 5 Anti-derivatives
- 6 The Fundamental Theorem of Calculus
- 7 Substitution
- 8 Areas between Graphs
- 9 Numerical Integration
- 10 Applications of the Integral
- 11 The Exponential and Logarithm Functions
- 5 Prerequisites from Precalculus 5
- 1 The Real Numbers
- 2 Inequalities and Absolute Value
- 3 Functions,

- 4 Graphing Equations
- 5 Trigonometric Functions
- 6 Inverse Functions
- 7 New Functions From Old Ones
- 263 263 266 268 274 276 286 292

Preface These notes are written for a one-semester calculus course which meets three times a week and is,

- preferably,
- supported by a computer lab

The course is designed for life science majors who have a precalculus back ground,

and whose primary interest lies in the applications of calculus

We try to focus on those topics which are of greatest importance to them and use life science examples to illustrate them

we try of stay mathematically coherent without becoming technical

we are willing to sacrifice generality

Instead,

more complex and demanding problems find their place in a computer lab

In this sense,

we are trying to adopt several ideas from calculus reform

We typically explore new ideas in examples before we give formal definitions

In one more way we depart radically from the traditional approach to calculus

We introduce differentiability as a local property without using limits

The philosophy behind this idea is that limits are the a big stumbling block for most students who see calculus for the first time,

and they take up a substantial part of the first semester

our approach to the derivative makes no use of limits,

allowing the students to get quickly and without unresolved problems to this concept

and fewer functions are differentiable in this manuscript than in a standard text

But the functions which we do not recognize as being differentiable are not particularly important for students who will take only one semester of calculus

in our opinion the underlying geometric idea of the derivative is at least as clear in our approach as it is in the one using limits

More technically speaking,

instead of the traditional notion of differentiability,

we use a notion modeled on a Lipschitz condition

function and the tangent line satisfies a Lipschitz condition2 of order 2 in x − x0 for all x in an open interval around x0 ,

instead of assuming that this difference is o(x − x0 )

which should be to easy to follow for anyone with a background in analysis,

has been used previously in teaching calculus

The au¨ thor learned about it when he was teaching assistant (Ubungsgruppenleiter) for a course taught by Dr

There this approach was taken for the same reason,

to find a less technical and efficient approach to the derivative

Schmidt followed suggestions which were promoted and carried out by Professor H

Professor Karcher had learned calculus this way from his teacher,

There are German language college level textbooks by K¨ utting and M¨ oller and a high school level book by M¨ uller which use this approach

Calculus was developed by Sir Isaac Newton (1642–1727) and Gottfried Wilhelm Leibnitz (1646–1716) in the 17th century

The emphasis was on differentiation and integration,

and these techniques were developed in the quest for solving real life problems

Among the great achievements are the explanation of Kepler’s laws,

the development of classical mechanics,

and the solutions of many important differential equations

Though very successful,

the treatment of calculus in those days is not rigorous by nowadays mathematical standards

In the 19th century a revolution took place in the development of calculus,

foremost through the work of Augustin-Louis Cauchy (1789–1857) and Karl Weierstrass (1815–1897),

when the modern idea of a function was introduced and the definitions of limits and continuous functions were developed

This elevated calculus to a mature,

- well rounded,

mathematically satisfying theory

mathematically challenging setup is required (limits) before one comes to the central ideas of differentiation and integration

A second revolution took place in the first half of the 20th century with the introduction of generalized functions (distributions)

This was stimulated by the development of quantum mechanics in the 1920ies and found is final mathematical form in the work of Laurent Schwartz in the 1950ies

? We want to introduce the concepts of differentiation and integration

- we do not 2 see page 42 of: A

Cambridge University Press,

reprinted with corrections and some additions 1968

v need the powerful machine developed in the 19th century

we like to be mathematically rigorous because this is the way mathematics is done nowadays

This is possible through the use of the slightly restrictive notion of differentiability which avoids the abstraction and the delicate,

technically demanding notions of the second period

Often times we accept computer generated graphics without having developed the background to deduce their correctness from mathematical principles

and sometimes the mathematical theory was

and it is also not our goal to teach them

PREFACE

Chapter 0

differentiation and integration

population of live yeast bacteria in a bun of pizza dough

time by t (say measured in hours) and the size of the population by P (say measured in millions of bacteria),

we denote this function by P (t)

- say at time t0 = 4
- • For a straight line,

the rate of change is its slope

although its graph is certainly not a straight line

What can we do

? Let us try to replace the function P (t) by a line L(t),

at least for values of t near t0

then we call L(t) the tangent line to the graph of P at t0

or the rate of change of P (t) at the time t = t0

Figure 3: Graph & tangent line

- on your graphing calculator,
- zoom in on the point (4,

P (4)) on the graph

This process works for the given example and most other functions treated in these notes

You see the zoom picture in Figure 2

- 3 scrutiny,

you detect that the graph is not a line,

- but still bent

let us ignore this bit of bending and pretend that the shown piece of graph is a line

Actual measurements in the picture let you suggest that the slope of that line should be about −70

This translates into the statement that the population of the live bacteria decreases at a rate of roughly 70 million per hour

which you should be able to carry out only after having studied a good part of this manuscript,

shows that the value of the slope of this line is about −67

that the geometric determination of the rate of change was quite accurate

To some extent,

it is up to us to decide the meaning of the requirement • |P (t) − L(t)| is small for all t near t0

One possible requirement1 ,

which it technically rather simple and which we will use,

is: • The exists a positive number A and an open interval (a,

- b) which contains t0 ,
- such that (2)

|P (t) − L(t)| ≤ A(t − t0 )2

- for all t in (a,

The inequality in (2) dictates how close we require the graph of P (t) to be to line L(t)

- or there may not,

exist an interval and a number A such that the inequality holds for an appropriate line

If the line,

- the interval,
- and A exist,
- then the line is unique

- it is denoted by P 0 (t0 ),

and we say that P (t) is differentiable at t0

we say • If P (t) is a function which is differentiable at t0 ,

- then P 0 (t0 ) is,
- by definition,

the rate at which P (t) changes when t = t0

which depends on the notion of limits,

- is imposed at this point

Our choice of requirement and our decision to avoid limits is based on the desire to keep the technicalities of the discussion at a minimum,

and to make these notes as accessible as possible

Different interpretations of the word ‘small’ lead to different ideas about differentiability

- if it exists,

is not effected by the choice of meaning for the word

the interpretation of the word ‘small’ has to imply the uniqueness of the derivative

CHAPTER 0

You noticed that we need the idea of a line

When you look at (2) and see the square of the variable you can imagine that we need parabolas

So we review and elaborate on lines and parabolas in Chapter 1

We also introduce the,

- possibly,

two most important functions in life science applications,

the exponential function and the logarithm function

Chapter 2 is devoted to the precise definition of the derivative and the exploration of related ideas

we calculate the derivative for some basic functions

which allow us to differentiate many more functions

Chapter 3 is devoted to applications

We investigate the ideas of monotonicity and concavity and discuss the 1st and 2nd derivative tests for finding extrema of functions

One finds a mathematical formulation for a problem which one encounters in some other context

One formulates the problem so that its solution corresponds to an extremum of its mathematical formulation

Having found the solution for the mathematically formulated problem one draws conclusions about the problem one started out with

- look at a drop of mercury

Physical principles dictate that the surface area be minimized

- given a fixed volume,
- is a ball

This is roughly what you see

for which the internal forced are not as strong as the ones in a drop of mercury

Often calculus is used to solve differential equations

for some simple population models the equation (Malthusian Law) P 0 (t) = aP (t) is asserted

The rate at which the population changes (P 0 (t)) is proportional to the size of the population (P (t))

analytical and numerical means

Suppose you need to take a certain medication

the relation may also involve the independent variable and higher derivatives

Figure 4: Constant Rate

us say that the rate at which the medication is absorbed through the skin is a function R(t),

where R stands for rate and t for time

that over some period of time R(t) is constant,

3 mg/hr

The situation is graphed in Figure 4

- 9 mg of the medication

We multiplied the rate at which the medication is absorbed with the length of time over which this happened

Assuming that you applied the patch at time t = 0,

the three hours would end at time t = 3

An interpretation of the total amount of medication which is absorbed between t = 0 and t = 3 is the area of the rectangle bounded by the line t = 0,

- the line t = 3,
- the x-axis,

and the graph of the function R(t) =

In Figure 5 you see the function A(t) =

- as a function of time,

how much medication has been absorbed

As the pill dissolves in the stomach,

it sets the medication free so that your body can absorb it

the medication is moved through your digestive system,

and decreasing amounts are available to being absorbed

We denote it once more by R(t)

Again you may want to find out how much medication has been absorbed within a given time,

say within the first 4 hours after swallowing the pill

Set the time at which you took the pill as time t = 0

CHAPTER 0

A PREVIEW

Figure 7: Amount absorbed

medication which has been absorbed between t = 0 and t = T is the area under the graph of R(t) between t = 0 and t = T

- we found the function A

one may want to find the area under the graph of a function f (x) between x = a and x = b

To make sense out of this we first need to clarify what we mean when we talk about the area of a region,

in particular if the region is not bounded by straight lines

finding the area between the graph of a non-negative function f and the x-axis between x = a and x = b means to integrate f from a to b

Both topics are addressed in the chapter on integration

If we differentiate the function shown in Figure 7 at some time t,

then we get the function in Figure 6 at t

In this section we also discuss the Fundamental Theorem of Calculus,

which is our principal tool to calculate integrals

The two basic ideas of the rate of change of a function and the area below the graph of a function will be developed into a substantial body of mathematical results that can be applied in many situations

You are expected to learn about them,

so you can understand other sciences where they are applied

Some Background Material Introduction In this chapter we review some basic functions such as lines and parabolas

- or typically intervals in,
- the real line

unless we really want to emphasize it

and they are fundamental for the understanding of almost everything which follows

- y(x) = 2x − 3
- drawn in Figure 1

More generally,

one may consider functions of the form (1

- y(x) = mx + b

where m and b are real numbers

Their graphs are straight lines with slope m and y-intercept (the point where the line intersects the y axis) b

we have the following definition

CHAPTER 1

SOME BACKGROUND MATERIAL

Definition 1

y) in the x − y-plane which satisfy the equation (1

- ax + by = c

for some given real numbers a,

- b and c,

where it is assumed that a and b are not both zero

If b = 0,

then we can write the equation in the form x = c/a,

and this means that the solutions of the equation form a vertical line

The value for x is fixed,

and there is no restriction on the value of y

then we can write the equation in the form y = c/b,

and this means that the solutions of the equation form a horizontal line,

- the value for y is fixed,

and there is no restriction on the value of x

If b 6= 0,

then ax + by = c'translates into y = − ab x + cb ,

and the equation describes a line with slope −a/b and y-intercept c/b

Figure 1

- 1: y(x) = 2x − 3

Figure 1

- 2: 2x = 3 & 2y = 5

we are often given the slope of a line and one of its points

Using functional notation,

the line is the graph of the function (1

- y(x) = m(x − x0 ) + y0

To see this,

- observe that y(x0 ) = y0 ,
- so that the point (x0 ,
- y0 ) does indeed lie on the graph

In addition,

you can rewrite the expression for the function in the form y(x) = mx + (−mx0 + b) to see that it describes a line with slope m

Its y-intercept is −mx0 + b

Example 1

The line with slope 3 through the point (1,

- 2) is given by the equation y = 3(x − 1) + 2

we want to find the equation of a line through two distinct,

- given points (x0 ,
- y0 ) and (x1 ,

otherwise the line is vertical

- y1 − y0 (x − x0 ) + y0
- x1 − x 0

This is slope formula for a line through the point (x0 ,

- y0 ) with h the point i y1 −y0 slope x1 −x0

You should check that y(x1 ) = y1

This means that (x1 ,

- y1 ) is also a point on the line

the equation of the line is: y1 − y0 y1 − y0 y(x) = x+ − x0 + y 0

x1 − x 0 x1 − x 0 Example 1

y0 ) = (1,

- −1) and (x1 ,

y1 ) = (3,

Putting the points into the equation of the line,

we find 4 − (−1) 5 7 y(x) = (x − 1) + (−1) = x −

♦ 3−1 2 2 The line is shown in Figure 1

CHAPTER 1

SOME BACKGROUND MATERIAL

Figure 1

- 3: Line through (1,
- −1) & (3,

we ended up with three different ways to write down the equation of a non-vertical line,

depending on the data which is given to us: • Intercept-Slope Formula: We are given the y-intercept b and slope m of the line

• Point-Slope Formula: We are given a point (x0 ,

- y0 ) on the line and its slope m

The equation of the line is y = m(x − x0 ) + y0

• Two-Point Formula: We are given two points (x0

- y0 ) and (x1 ,
- y1 ) with different x-coordinate on the line

- y1 − y0 (x − x0 ) + y0
- x1 − x 0

- 1) is a point on the line

Using the point (2,

write down the point slope formula for the line and convert it into the slope intercept formula

Find the x and y-intercept for the line and sketch it

Exercise 5

1) and (2,

What is the slope of the line

? Where does the line intersect the coordinate axes

- ? Sketch the line

Consider the lines l1 : ax + by = c'&

- l2 : Ax + By = C

They intersect in the point (x0 ,

- y0 ) if this point satisfies both equations

to find intersection points of two lines we have to solve two equations in two unknowns simultaneously

Example 1

Find the intersection points of the lines 2x + 5y = 7 &

- 3x + 2y = 5

both equations hold if we set x = 1 and y = 1

As an exercise you may verify that (1,

- 1) is the only intersection point for these two lines

♦ The lines ax + by = c'and Ax + By = C are parallel to each other if (1

Ab = aB,

and in this case they will be identical,

or they will have no intersection point

- 4x + 10y = 14
- are identical

observe that the second equation is just twice the first equation

y) will satisfy one equation if and only if it satisfies the other one

So the lines are identical

The lines 2x + 5y = 7

- & 4x + 10y = 15

are parallel and have no intersection point

observe that the first equation,

- multiplied with 2,
- is 4x + 10y = 14

There are no numbers x and y for which 4x + 10y = 14 and 4x + 10y = 15 at the same time

and the two lines do not intersect

♦ To be parallel also means to have the same slope

then the condition says that the slopes −a/b of the line l1 and −A/B of the line l2 are the same

If both lines are vertical,

then we have not assigned a slope to them

then the lines are not parallel to each other,

and one can show that they intersect in exactly one point

You saw an example above

If Aa = −bB,

then the lines intersect perpendicularly

Assuming that neither line is vertical (b 6= 0 and B 6= 0),

the equation may be written as a A × = −1

b B This means that the product of the slopes of the first line and −A/B the one of the of one line is the negative reciprocal of the the condition which you have probably seen perpendicularly

of the lines (−a/b is the slope second line) is −1

The slope slope of the other line

This is before for two lines intersecting

The lines 3x − y = 1 &

- x + 3y = 7
- have slopes 3 and −1/3,

and intersect perpendicularly in (x,

- y) = (1,
- ♦ Exercise 6

Find the intersection points of the lines l1 (x) = 3x + 4

- l2 (x) = 4x − 5

PARABOLAS AND HIGHER DEGREE POLYNOMIALS

Exercise 7

decide whether the lines are parallel,

- perpendicular,
- or neither

Find all intersection points for each pair of lines

l1 : 3x − 2y = 7 l2 : 6x + 4y = 6 l3 : 2x + 3y = 3 l4 : 6x − 4y = 5 Exercise 8

- 2) and intersects the line 3x − 4y = 5 perpendicularly

? Find its slope point formula (use (1,

- 2) as the point on the line) and its slope intercept formula

Sketch the line

- a function of the form (1
- y(x) = ax2 + bx + c
- where a,

and c'are real numbers and a 6= 0

Depending on whether a is positive or negative the parabola will be open up- or downwards

Abusing language slightly,

we say that y(x) is a parabola

and they will be of importance to us in one interpretation of the derivative

Typical examples of parabolas are the graphs of the functions p(x) = x2 − 2x + 3 and

- q(x) = −x2 − x + 1
- shown in Figures 1

4 and 1

- the second one downwards

The x-intercepts of the graph of p(x) = ax2 + bx + c'are also called roots or the zeros of p(x)

The solutions of this equation are found with the help of the quadratic formula i p 1 h p(x) = 0 if and only if x = (1

- 8) −b ± b2 − 4ac
- 2a The expression b2 − 4ac under the radical is referred to as the discriminant of the quadratic equation

CHAPTER 1

SOME BACKGROUND MATERIAL

6 1 5 4

Figure 1

- 4: y = x2 − 2x + 3

Figure 1

- 5: y = −x2 − x + 1

• p(x) has two distinct roots if the discriminant is positive

• p(x) has exactly one root if the discriminant is zero

• p(x) has no (real) root if the discriminant is negative

Example 1

Find the roots of the polynomial p(x) = 3x2 − 5x + 2

- 6 So the roots of p(x) are 1 and 2/3

- (1) p(x) = x2 − 5x + 2
- (3) r(x) = 2x2 − 12x + 18
- (2) q(x) = 2x2 + 3x − 5
- (4) s(x) = −x2 + 5x − 7

- say p(x) = a1 x2 + b1 x + c1
- and q(x) = a2 x2 + b2 x + c2

In other words,

we look for the roots of p(x) − q(x) = (a1 − a2 )x2 + (b1 − b2 )x + (c1 − c2 )

The highest power of x in this equation is at most 2 (this happens if (a1 − a2 ) 6= 0),

and this means that it has at most two solutions

- 150 125 100 75 50 25

- 6: Intersecting parabolas

Example 1

- and q(x) = 2x2 + 3x − 5

the solutions are √ √ 1 x = − [8 ± 64 + 28] = −4 ± 23

- 2 √ So the parabolas intersect at x = −4 ± 23

and you can check that our calculation is correct

- ♦ Exercise 10

Graph the pairs of parabolas and verify your calculation

We will study how parabolas intersect in more detail in Section 2

Right now we like to turn our attention to a different matter

- 1 we used the slope-intercept and the point-slope formula to write down the equation of a line

- y = mx + b = mx1 + bx0

SOME BACKGROUND MATERIAL

- expresses y in powers of x

- 9) we added some redundant notation to make this point clear

When we write down the point slope formula of a line with slope m through the point (x0 ,

y = m(x − x0 ) + y0 = m(x − x0 )1 + y0 (x − x0 )0 ,

then we expressed y in powers of (x − x0 )

- y(x) = x2 + 5x − 2
- in powers of (x − 2)

Our task is to find numbers A,

- such that (1

y(x) = A(x − 2)2 + B(x − 2) + C

- 11) and gathering terms according to their power of x we find y(x) = A(x2 − 4x + 4) + B(x − 2) + C = Ax2 + (−4A + B)x + (4A − 2B + C) Two polynomials are the same if and only if their coefficients are the same

comparing the coefficients of y in (1

- 10) with those in our last expression for it,
- we obtains equations for A,

and C: A = 1 −4A + B = 5 4A − 2B + C = −2 These equations can be solved consecutively,

- and C = 12

We expanded y(x) in powers of (x − 2)

we come up with the following formula: (1

y(x) = ax2 + bx + c'= A(x − x0 )2 + B(x − x0 ) + C

- where A=a (1

B = 2ax0 + b C = ax20 + bx0 + c'= y(x0 )

In fact,

given any polynomial p(x) and any x0 ,

one can expand p(x) in powers of (x − x0 )

- only it gets lengthier

On the computer you can do it in a jiffy

Expand y(x) = x2 − x + 5 in powers of (x − 1)

Exercise 12

What is the purpose of expanding a parabola in powers of (x − x0 )

? Let us look at an example and see what it does for us

Consider the parabola p(x) = 2x2 − 5x + 7 = 2(x − 2)2 + 3(x − 2) + 5

The last two terms in the expansion form a line: l(x) = 3(x − 2) + 5

- |p(x) − l(x)| = 2(x − 2)2
- and in particular,
- p(2) = l(2)

we found a line l(x) which is close to the graph of p(x) near x = 2

and the estimate holds for all x in (−∞,

- ∞) (or any interval)

For each of the following parabolas p(x) and points x0 ,

find a line l(x) and a constant A,

such that |p(x) − l(x)| ≤ A(x − x0 )2

p(x) = 3x2 + 5x − 18 and x0 = 1

p(x) = −x2 + 3x + 1 and x0 = 3

p(x) = x2 + 3x + 2 and x0 = −1

CHAPTER 1

Let p(x) = x4 − 2x3 + 5x2 − x + 3 and x0 = 2

Find a line l(x) and a constant A,

such that |p(x) − l(x)| ≤ A(x − x0 )2 for all x in the interval I = (1,

(Note that the open interval I contains the point x0 = 2

) Expanding p(x) in powers of (x − 2) we find p(x) = (x − 2)4 + 6(x − 2)3 + 17(x − 2)2 + 27(x − 2) + 21

Then |p(x) − l(x)| = (x − 2)4 + 6(x − 2)3 + 17(x − 2)2 = (x − 2)2 + 6(x − 2) + 17 (x − 2)2 ≤ (1 + 6 + 17)(x − 2)2 ≤ 24(x − 2)2

- 9) in Section 5
- 2 to get the first inequality

If x ∈ (1,

then |x − 2| < 1 and |x − 2|k < 1 for all k ≥ 1

This helps you to verify the second inequality

with A = 24 and l(x) = 27(x − 2) + 21,

we find that |p(x) − l(x)| ≤ A(x − x0 )2 for all x ∈ (1,

Exercise 16

Find a line l(x) and a constant A,

such that |p(x) − l(x)| ≤ A(x − x0 )2 for all x in the interval I = (4,

Consider a polynomial p(x) = cn xn + cn−1 xn−1 + · · · + c1 x + c0

Pick a point x0 ,

and expand p(x) in powers of x0 : p(x) = Cn (x − x0 )n + Cn−1 (x − x0 )n−1 + · · · + C1 (x − x0 ) + C0

- and we learned how to do this

Set l(x) = C1 (x − x0 ) + C0

PARABOLAS AND HIGHER DEGREE POLYNOMIALS

Then |p(x) − l(x)| = Cn (x − x0 )n−2 + · · · + C3 (x − x0 ) + C2 (x − x0 )2 ≤ |Cn (x − x0 )n−2 | + · · · + |C3 (x − x0 )| + |C2 | (x − x0 )2 ≤ (|Cn | + |Cn−1 | + · · · + |C2 |) (x − x0 )2 for all x ∈ I = (x0 − 1,

x0 + 1)

To get the equation,

we took |p(x) − l(x)| and factored out (x − x0 )2

- 9) in Section 5

The last inequality follows as (x − x0 )k < 1 if k ≥ 1

for l(x) = C1 (x − x0 ) + C0 and A = (|Cn | + · · · + |C2 |) we have seen that |p(x) − l(x)| ≤ A(x − x0 )2 for all x ∈ (x0 − 1,

x0 + 1)

In the sense of our preview,

and the upcoming discussion about derivatives,

this means • The rate of change of p(x) at the point (x0 ,

- p(x0 )) is C1 ,
- the slope of the line l(x)

find the rate of change of p(x) when x = x0

p(x) = x2 − 7x + 2 and x0 = 4

- p(x) = 2x3 + 3 and x0 = 1

p(x) = x4 − x3 + 3x2 − 8x + 4 and x0 = −1

that we began to omit labels on the axes of graphs

that we displayed more than one function in one graph,

and that means that there is no natural name for the variable associated to the vertical axis

that we use the horizontal axis for the independent variable and the vertical one for the dependent one1

- 1 If you like to review the terms independent and dependent variable,

then we suggest that you read Section 5

- 3 on page 268

CHAPTER 1

where a is a positive real number and x is a rational number

102 = 100,

101/2 =

10−1 =

if x = n/m and n and m are natural numbers,

then ax is obtained by taking the n-th power of a and then the m-root of the result

- one sets a−x = 1/ax

Find exact values for −2 1 43/2 2

3−1/2

25−3/2

Use your calculator to find approximate values for 34

Until now you may not have learned about irrational (i

not rational) √ √ π 2 exponents as in expressions like 10 or 10

We like to give a meaning to the expression ax for any positive number a and any real number x

A new idea is required which does not only rely on arithmetic

- recall what we have

If a > 1 (resp

- 0 < a < 1) and x1 and x2 are two rational numbers such that x1 < x2 ,
- then ax1 < ax2 (resp
- ax1 > ax2 )

So far,

this function is defined only for rational arguments (values of x)

The function is monotonic

it is increasing if a > 1 and decreasing if 0 < a < 1

called the exponential function with base a and denoted by expa (x),

which is defined for all real numbers x such that expa (x) = ax whenever x is a rational number

- ax > 0 for all x,
- and so we use (0,

∞) as the range2 of the exponential function expa (x)

- 2 You may want to review the notion of the range of a function in Section 5
- 3 on page 268

This will be quite easy once we have more tools available

the exponential function is of great importance and has many applications,

so that we do not want to postpone its introduction

It is common,

and we will follow this convention,

to use the notation ax for expa (x) also if x is not rational

7 and 1

In another graph,

- see Figure 1

you see the graph of an exponential function with a base a smaller than one

We can allowed a = 1 as the base for an exponential function,

- but 1x = 1 for all x,

and we do not get a very interesting function

5 600 2

400 300

1 200 0

- 7: 2x for x ∈ [−1,

Figure 1

- 8: 2x for x ∈ [−1,

Suppose you like to find 2π

14 and 3

Saying that exp2 (x) is increasing just means that 23

- 14 < 2π < 23

- places 2π between 8

81 and 8

In fact,

if r1 and r2 are any two rational numbers,

- r1 < π < r2 ,

then due to the monotonicity of the exponential function,

- 2r1 < 2π < 2r2

CHAPTER 1

SOME BACKGROUND MATERIAL

The theorem asserts that there is at least one real number 2π which satisfies these inequalities,

and the uniqueness part asserts that there is only one number with this property,

- making 2π unique

9: (1/2)x

also called the exponential laws,

are collected in our next theorem

The theorem just says that the exponential laws,

which you previously learned for rational exponents,

also hold in the generality of our current discussion

You will derive the exponential laws from the logarithm laws later on in this section as an exercise

Theorem 1

- 13 (Exponential Laws)

For any positive real number a and all real numbers x and y a0 = 1 a1 = a ax ay = ax+y ax /ay = ax−y (ax )y = axy Some of the exponential laws can be obtained easily from the other ones

Assuming the third one,

one may deduce the first and third one

You are invited to carry out these deductions in the following exercises

THE EXPONENTIAL AND LOGARITHM FUNCTIONS

Exercise 20

Show: If a 6= 0,

- the a0 = 1

Although we did not consider an exponential function with base 0,

- it is common to set 00 = 1

If x 6= 0,

- then 0x = 0

Show ax /ay = ax−y

the proof of which we also postpone for a while (see Section 4

Theorem 1

There exists a unique (i

exactly one) real number x such that ax = b

If ax = ay ,

- then x = y,
- or equivalently,
- if x 6= y,
- then ax 6= ay

- as an equation in x

For a given a and b we want to (and the theorem says that we can) find a number x,

- so that the equation holds
- if a=2 a=4 a = 1/2 √ a= 2
- and and and and
- b = π,
- then then then then
- x = 1/2
- x = −1

The value for x in the last example was obtained from a calculator and is rounded off

(1) (a,

- b) = (10,

(3) (a,

- b) = (2,

(5) (a,

- b) = (2,

(2) (a,

- b) = (1000,

(4) (a,

- b) = (4,

(6) (a,

- b) = (100,

- 16) by loga (b)

- a 6= 1,

then loga (b) is the unique number,

- such that aloga (b) = b
- expa (loga (b)) = b

- log 2 (1/8) = −3
- log2 16 = 4

2 = 1/2

and for the base 10: log10 1 = 0

- log10 100 = 2
- log10 (1/10) = −1

Exercise 23

- (3) log10 π
- (5) log10 25
- (2) log10 100
- (4) log10 (1/4)
- (6) log10 1

Mathematically speaking,

- we just defined a function

called the logarithm function with base a

It is defined for all positive numbers,

and its range is the set of real numbers

Part of the graph of log2 (x) is shown in Figure 1

In Figure 1

- 11 you see the graph of a logarithm function with base a less than 1

- loga (ay ) = y
- log a (expa (y)) = y

Setting b = ay in (1

- 17) we have that y)
- aloga (a

- 15) says that loga (ay ) = y

Taken together,

17) and (1

- 18) say that for every a > 0,
- a 6= 1,
- we have aloga (y) = y x

loga (a ) = x This just means that

for all y > 0 and for all x ∈ (−∞,

THE EXPONENTIAL AND LOGARITHM FUNCTIONS

- 10: log2 (x)

Figure 1

- 11: log(1/2) (x)

The exponential function expa (x) = ax and the logarithm function loga (y) are inverses3 of each other

we obtain the graph of the logarithm function by reflecting the one of the exponential function at the diagonal in the Cartesian plane

This is the general principle by which the graph of a function and its inverse are related

The role of the independent and dependent variables,

and with this the coordinate axes,

- are interchanged

- see Figure 1
- 10 is a reflection of the one in Figure 1

you need to take into account that the parts of the function shown are not quite the same and that there is a difference in scale

Once you make these adjustments you will see the relation

If loga (u) = loga (v),

- then u = v,
- and equivalently,
- if u 6= v,
- then loga (u) 6= loga (v)

It is a general fact,

that the inverse of an increasing function is increasing,

and the inverse of a decreasing function is decreasing (see Proposition 5

- 25 on page 291)

So the monotonicity statements for the logarithm functions follow from the monotonicity properties of the exponential functions (see Theorem 1

- 14) because these functions are inverses of each other
- 3 A quick review of the idea of inverse functions is given in Section 5
- 6 on page 286,

and you are encouraged to read it in case you forgot about this concept

SOME BACKGROUND MATERIAL

Furthermore,

loga (u) = loga (v) implies that u = aloga (u) = aloga (v) = v

This verifies the remaining claim in the theorem

Corresponding to the exponential laws in Theorem 1

- 13 on page 22 we have the laws of logarithms

The other parts are assigned as exercises below

- 19 (Laws of Logarithms)

for all positive real numbers x and y,

and any real number z loga (1) = 0 loga (a) = 1 loga (xy) = loga (x) + loga (y) loga (x/y) = loga (x) − loga (y) loga (xz ) = z loga (x) Because the exponential and logarithm functions are inverses of each other,

- their rules are equivalent

Assume the exponential laws and deduce the laws of logarithms

Exercise 25

To show you how to solve this kind of problem,

we deduce one of the exponential laws from the laws of logarithms

and the remaining two equations hold because of the way the logarithm function is defined

- we deduce from Theorem 1
- 18 the third exponential law: ax ay = ax+y

loga (xy) = loga (x) + loga (y)

Show that loga (x/y) = loga (x) − loga (y)

The Euler number e as base You may think that f (x) = 10x is the easiest exponential function,

at least you have no problems to find 10n if n is an integer (a whole number)

- the number e,
- named after L

so the decimal expansion does not have a repeating block

- 71828182845904523536028747135266249775724709369996

- 12 on page 52 where it is stated that this function is its own derivative

the logarithm function for the base e,

is called the natural logarithm function

- see Theorem 2
- 13 on page 52

let us state the definitions formally

We graph these two functions on some reasonable intervals to make sure that you have the right picture in mind when we talk about them,

- see Figure 1
- 12 and Figure 1

- 12: ex for x ∈ [−2,

- 13: ln x for x ∈ [

Definition 1

Leonard Euler (1707–1783),

one of the great mathematicians of the 18th century

CHAPTER 1

Example 1

- 21 (Exponential Growth)

It is not so apparent from the graph how fast the exponential function grows

You may remember the tale of the ancient king who,

as payment for a lost game of chess,

was willing to put 1 grain of wheat on the first square on the chess board,

- 2 on the second,
- 4 on the third,
- 8 on the forth,

doubling the number of grains with each square

and that commits him to 263 grains on the 64th square for a total of 264 − 1 = 18,

615 grains

In mathematical notation,

you say that he puts f (n) = 2n−1 grains on the n-th square of the chess board

let us graph the function f (x) = 2x for 0 ≤ x ≤ 63,

- see Figure 1

even an already enormous number like 254 ,

cannot be distinguished from 0

10 18 6

10 18 4

10 18 2

- 14: Graph of f (x) = 2x

The amount of grain which the king has to put on the chess board suffices to feed the current world population (of about 6 billion people) for thousands of years

- 15: Compare 2x and x6

- 16: Compare 2x and x6

Example 1

- 22 (Comparison with Polynomials)

A different way of illustrating the growth of an exponential function is to compare it with the growth of a polynomial

15 and 1

- 16 you see the graphs of an exponential function (f (x) = 2x ) and a polynomial (p(x) = x6 ) over two different intervals,
- 23] and [0,

In each figure,

the graph of f is shown as a solid line,

and the one of p as a dashed line

- on the given interval,

the polynomial p is substantially larger than the exponential function f

In the second figure you see how the exponential function has overtaken the polynomial and begins to grow a lot faster

Other Bases Finally,

let us relate the exponential and logarithm functions for different bases

for any positive number a (a 6= 1),

Theorem 1

- ln a Proof

- loga x =
- ax = (eln a )x = ex ln a

use eln x = x = aloga x = (eln a )loga x = eln a loga x

- or loga x =
- ln x ln a ,
- as claimed

and with this the function f (t) itself,

can be determined if we give the value of f at two points

We call a the growth rate5

Example 1

- f (0) = 3,
- and f (5) = 7

- its relative growth rate a,

and the time t0 for which f (t0 ) = 10

Solution: By assumption,

the function is of the form f (t) = Ceat

Substituting t = 0,

we find 3 = f (0) = Cea·0 = Ce0 = C

we substitute t = 5 into the expression of f (t): 7 = f (5) = 3e5a

using arithmetic and the fact that the natural logarithm function is the inverse of the exponential function,

- that e5a = 7/3
- ln(7/3) =

the growth rate of the function is (approximately)

- and f (t) = 3e

- t0 is determined by the equation 3e
- 16946t0 = 10

- 16946t0 = 10/3 The value for t0 is rounded off
- ln(10/3) = 7

others the relative growth rate

the rate of change of f (t) at time t0 is af (t0 ),

so that the name relative growth rate (i

relative to the value to f (t)) is quite appropriate

- in the long run,

you may get tired of having to say relative all the time,

and with the exact meaning understood,

you are quite willing to drop this adjective

Exercise 27

- f (1) = 3,
- and f (4) = 7

Find the function f ,

- its relative growth rate a,

and the time t0 for which f (t0 ) = 10

Exercise 28

Suppose the function f (t) grows exponentially,

- and f (T ) = 2f (0)

What is the relative growth rate for the population

? How long did it take for the population to double

- ? Remark 4

We still have justify our characterization of the exponential function in Theorem 1

- 19: loga (xy) = loga (x) + loga (y) and
- loga (xz ) = z loga (x),

and we have to define the Euler number e

All of this will be done in Sections 4

Use of Graphing Utilities

A word of caution is advised

- in our case Mathematica,
- to draw graphs of functions

We use these graphs to illustrate the ideas and concepts under discussion

no graphing utility is perfect and we cannot uncritically accept their output

Given any computer and any software,

- no matter how good they are,

with some effort you can produce erroneous graphs

That is not their mistake,

it only says that their abilities are limited

17 and 1

- 18 you see two graphs of the function p(x) = (x + 1)6 = x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1

and then we asked it to use the expanded expression

The outcome is remarkably different

? The program makes substantial round-off errors in the calculation

Which one is the correct graph

- ? Calcul
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