Chapter 1

1.1 Lines

1.2 Functions and Graphs

1.3 Exponential Functions

1.5 Functions and Logarithms

1.6 Trigonometric Functions

Chapter 2

2.1 Rates of Change and Limits

2.2 Limits involving Infinity

2.3 Continuity

2.4 Rates of Change and Tangent Lines

Chapter 3

3.1 Derivative of a Function

3.2 Differentiability

3.3 Rules for Differentiation

3.4 Velocity and Other Rates of Change

3.5 Derivatives of Trigonometric Functions

3.6 Chain Rule

3.7 Implicit Differentiation

3.8 Derivatives of Inverse Trigonometric Functions

3.9 Derivatives of Exponential and Logarithmic Functions

Chapter 4

4.1 Extreme Values of Functions

4.2 Mean Value Theorem

4.3 Relationship between f ' and f " with f

4.4 Modeling and Optimization

4.5 Linearization

4.6 Related Rates

Chapter 5

5.1 Estimating with Finite Sums

5.2 Definite Integrals

5.3 Definite Integrals and Antiderivatives

5.4 Fundamental Theorem of Calculus

5.5 Trapezoidal Rule

6.1 Anti derivatives

6.2 Integration by Substitution


Introduction

Introducing, Connie Lu's Calculus AB study guide. In this guide, you will find everything you need to review or study for the AP Calculus AB exam. It begins with the basics that are learned in pre-calculus and proceeds through the basics of calculus including limits, derivatives, and integrals. I will try to make the study guide as simple and easy to follow as possible without using too many complicated Calculus terms and symbols. It will also include applicable examples to help explain the concepts.

Chapter 1 - Pre-calculus Review

Introduction

In this chapter, you will find a review of all the pre-calculus skills necessary to learn calculus. This includes the basics of a line, a function, finding the slope, logs and trigonometry. If you already are familiar with this, you can skip this section because it is pretty basic.

Lines

Calculus is important because it helps us analyze the changing quality of things, whether it be a business model or reproducing bacteria. Often we can find a pattern in the change and calculus is used to analyze these patterns.

A line is used to show an increment between two coordinates (x1, y1) and (x2, y2). The change in x and y is called "delta x" and "delta y" respectively and are shown as Dx and Dy. Dx and Dy can be positive, negative or zero.

Special cases include:

Slope is the change of x and y in a line. Slope is denotaed by "m".

m = rise/ run = Dy / Dx = y2- y1 / x2 - x1

Special cases include:

The equation of a line follows the general form y = mx + b where b is the y intercept.

We find the equation for a line one of two ways.

Functions and Graphs

A function is important in scientific applications of math. A function is used to describe the relationship between the two variables. It is often seen in the form f(x) = y which means "y is the function of x". A function is like a translation in which one enters a value "x" and gets a result "y". For every value x there is only one y result and that result will be the same everytime the function is executed.

The domain of a function are all the possible x values or inputs.

For example, in the equation f(x) = 1/ x, the domain would be all real numbers except 0, because 0 is not a valid input value.

The range of a function are all the possible y values or results.

For example in the equation f(x) = x^2, the range would be f(x) or y ˜ 0 because for any value of x, the result can is between 0 and infinity

Symmetry of a Function

The names come from the powers of x, because an odd power (x, x^3, x^5, etc..) is an odd function and even powers (x^2, x^4, x^6, etc...) are even functions.

Draw diagrams??

Special types of functions

To solve a piecewise function, just handle each section seperately, and then combine the result. If an x value is given and you need to find f(x), first chose the section in which the value fits and solve for f(x) using that sections definition of the function. (x^2) / 4, x > 4

f (-1) = -1

f (3) = 1

f (5) = 25 / 4

Logarithms

Even if you're not an idiot, inverse functions can be difficult to comprehend. Logarithms make it easier to deal with inverse functions.

The equation y = loga x is the inverse of y = ax

Special logarithms

** Properties of Logarithms

1) base a: a^ (log ax) = x, a > 1, x > 0

2) base e: e ln x = x, x > 0

These are very useful for simplifying equations.

Review:

Log Rules

For any real numbers x > 0 and y > 0

Product Rule

loga xy = loga x + loga y

Quotient Rule

loga (x/y) = loga x - loga y

Power Rule

loga xy = y loga x

Trigonometry Review

Trigonometry is important for calculus because it gives some patterns of change by using angles.

Radians vs. Degrees

Radians and Degrees are two ways of measuring an angle.

To convert radians to degrees you multiply times 180/pi

For example: 2pi = 2 pi * 180/pi = 360 degrees

To convert degrees to radians you multiply times pi/180

For example: 90 degrees = 90 * pi /180 = pi/2 radians

Basic Trigonemtetric Functions

draw unit circle pg 41

r is the radius, h is the angle with it's vertex on the origin, and x and y are the coordinates on the unit circle.

sine : sin h = y / r

cosine : cos h = x/r

tangent : tan h = y / x

secant : csc h = r / y

secant : sec h = r / x

cotangent : cot h = x/y

How to remember:

These formulas can be remembered using the phrase "SOH CAH TOA". The first letter stands for Sin, Cos and Tan. O stands for Opposite, A for Adjacent, and H for Hypotenuse. Thus, to find the Sin of an angle, you would divide Opposite over Hypotenuse.

** Important trig relationships

tan x = sin x / cos x = 1 / cot x

cot x = cos x / sin x = 1/ tan x

sec x = 1 / cos x

csc x = 1 / sin x

tan2x + 1 = sec2x

sin2x + cos2x = 1

1 + cot2x = csc 2x

sin 2x = 2 sinx cosx

cos 2x = cos 2x - sin2x

Inverse Trigonometric functions

an inverse function only works with one-to-one functions. Since none of the trig functions are one-to-one, we must limit their domains to one cycle to find the inverse function of each trig function.

The purpose of the inverse function is to find an angle measure when you know what the sides are to a triangle.

Example : sin o/2 = 0 , sin-1 (0) =o/2

cos o/2 = 1, cos-1 (1) =o/2

Inverse Trig Identities

cos-1 x = o/2 - sin-1 x

cot-1x = o/2 - tan-1 x

csc-1x = o/2 - sec-1 x

Chapter 2 - Limits and Continuity

If you are a beginner at calculus and you are an idiot, this will be the most frustrating chapter of all. It is difficult to understand, but the concepts are essential to learning calculus. The idea of limits is what seperates calculus from algebra and trigonometry.

Rates of Change and Limits

As we said before, calculus is used to study the change in some property. It can also be used to study the change another change. For example, we can study the change in position, and we can study velocity which is the change in the change in position. Or we can study acceleration which is the change in the change of velocity. In other words, while velocity might be an analysis of some change, we can also study its change.

Average and Instantaneous Speed

Average speed is found over a period of time. It is found by dividing the time interval by the distance covered

Average Velocity = < x / < t

Instantaneous speed is the speed at a given moment in time

Limits *take your dunce cap off and put on your thinking cap

Definition: Let c and L be real numbers. The function f has limit L as x approaches c if, given an y positive number e there is a positive number d such that for all x 0 < | x- c| <d -> | f (x) - L | < e

Written as: lim f(x) = L

x -> c

If this confuses you, here are some ways of looking at limits:

What that really means: Is that for a function, the limit as the x variable gets closer to the constant "c" it will just about make it to L. It does not go exactly to L but we say it does because for human purposes it gets so close to L that it does not really matter that it does not actually get there.

Example: Say I am 40 ft from a door, and with each move I make, I will move exactly half the distance between me and the door. My first move takes me to 20 ft from the door. Then 10 ft, then 5 ft, then 2.5 ft, then 1.25 ft, and so on. Each time I divide my distance to the door by two. Eventually I will be within millimeters to the door. However, if I keep dividing by two, I will never actually be at the door. But for human purposes it is good enough to say that if I am 1 nanometer from the door, I am at the door. This is an example of limits. The function is my distance to the door. The constant c is infinity because I can take an infinite number of moves. We say that the door is the limit because as far as L because my limit is the door because if I take infinity number of steps, I will never get to the exact door, but I will be so close to it that I wouldn't even be able to tell and just say I am at the door.

How to actually solve limits:

Properties of Limits

Limits may seem impossible to solve at times. It is important to remember these properties.

We can find the limit of all polynomal and rational functions using these properties

Theorem 1

If L, M, c and k are real numbers and

lim f(x) = L and lim g(x) = M

x-> c x -> c

Rule

Description

Formula

Sum Rule

The limit when you ADDtwo functions is the sum of their limits

lim (f(x) + g(x)) = L + M

x-> c

Difference Rule

The limit when you SUBTRACT two functions is the difference of their limits.

lim (f(x) - g(x)) = L - M

x-> c

Product Rule

The limit when you MULTIPLY two functions is the product of their limits

lim (f(x) * g(x)) = L * M

Constant Multiple Rule

The limit of a constant multiplied by a CONSTANT is the limit of just the function.

lim (k * f(x)) = k * L

x-> c

Quotient Rule

The limit when you DIVIDE two limits

lim (f(x) / g(x)) = L/ M

x -> c

Power Rule

The limit of a function raised to a rational power is the limit of the function raised to that power, provided the power is a real number.

lim (f(x))r/s = Lr/s

x-> c

Use theorem 1 when you need to find the limit of one or two functions for which you need to do some operation, ie. added, subtracted, multiplied, divided together with anothe function or raised to a power.

Theorem 2

Finding the limits of Polynomial and Rational functions

Technical definition

1) If f(x) = an xn + a n-1 xn-1 + ... + a0 is any polynomail function and c is any real number then

lim f(x) = f(c) = an cn + a n-1 cn-1 + ... + a0

x->c

2) If f (x) and g(x) are polynomials and c is any real number, then

lim f(x) / g(x) = f(c) / g(c) as long as g(c) g0

x-> c

What it really means : To find the limit of a function f(x), or f(x) / g(x) as x approaches c, just plug in the value of c. This is also known as SUBSTITUTION

BEWARE! - If it were this easy, everyone would be taking calculus and getting 100s. However, this does not always work because often plugging in c for x will give you a fraction with 0 in the denominator, which of course doesn't work. Fortunately, there are ways around this. Of course, they are not easy. In this case we have to use SOME OTHER METHOD. This often requires creativity.

Here are some examples of SOME OTHER METHOD

1) factor and hope things cancel out

2) rationalize

3) mulitpy

4) divide

Example: lim tan x / x

x -> 0

Why it's a problem: Plugging in 0 for x would give you tan 0 / 0, which is bad because 0 on the bottom is a no-no. Thus we have to get around this by first converting tan x to sin x / cos x

The problem now becomes:

lim (sin x / x) (1/ cos x)

x -> 0

Using the property in theorem 1 with the product rule we can say that:

lim (sin x / x) * lim (1/ cos x)

x -> 0 x-> 0

The lim sin x/ x is known to be 1

x-> 0

Thus, 1 * lim (1/ cos x) = 1 * (1/ cos 0) = 1

x-> 0

Also be aware that not all functions have a limits.

We can find out if a limit actually exists by breaking the limit down.

One-sided and Two Sided Limits

looks like: lim f(x)

x -> c+

What it means: Take the limit as you approach x = c from the right. Imagine the x axis and evaluate the function coming from the right side or bigger x values to the smaller x values until you get to x = c.

looks like : lim f(x)

x -> c-

What it means: Take the limit as you approach x = c from the left. If you imagine the x axis you evaluate the function as you move from smaller x values to the bigger ones until you get to x = c.

Theorem 3

A function f(x) has a limit as x approaches c if and only if the right and left hand limits at c exist and are equal.

If you evaluate both the right hand and left hand limit, and they are equal, you have a two sided limit.

looks like: lim f(x) means that lim f(x) must equal = lim f(x)

x -> c x -> c+ x -> c-

Even with all these theorems and stuff, you still might not be able to determine the limit. So we introduce the Sandwich Theorem.

Theorem 4

Formal definition

Sandwich Theorem : If g(x) f(x) h(x) for all x g c sin some interval about c, and

and lim g(x) = lim h(x) = L

x->c x -> c

then lim f(x) = L

x -> c

What it really means: Beats me!

Limits with Infinity

Limits involving infinity tell us where the function will be if we keep adding or keep subtracting to the x value. In other words, it tells us where the function will be if we go as far to the right on the x axis as is possible for the limit of infinity and as far to the left as possible for the limit of negative infinity.

Theorem 5 - Properties of Limits as x approaches ! º

If L,M, and k are real numbers and

lim f(x) = L and lim g(x) = M, then

x d ! º x d ! º

  1. Sum Rule:
  2. Difference Rule:
  3. Product Rule:
  4. Constant Multiple Rule:
  5. Quotient Rule:
  6. Power Rule: If r and s are integers, s ! 0, then:
  7. provided that L^(r/s) is a real number.

lim (f(x) + g(x)) = L + M

x d ! º

lim (f(x) - g(x)) = L - M

x d ! º

lim (f(x) · g(x)) = L · M

x d ! º

lim (k · f(x)) = k · L

x d ! º

lim f(x) = L , M ! 0

x d ! º g(x) M

lim (f(x))^(r/s) = L^(r/s)

x d ! º

Asymptotes - a line which the function will keep aproaching but never intersect.

A horizontal line fitting (y = b, where b is a constant) such that for function y = f(x) either

lim f(x) = b or lim f(x) = b

x -> º x -> º

Solve a function by plugging in º or - º for x.

It takes a little thought to do this. For example, if you have 1/x and plug in 1/º, you have to imagine if the numerator keeps getting bigger and bigger, the number gets closer and closer to 0. Thus the limit must be 0.

A vertical line fitting (x = a, where b is a constant) such that for function y = f(x) either

lim f(x) = !º or lim f(x) = !º

x -> a+ x -> a-

Find a value "a" for which the limit as the function approaches that value is either º or -º.

More simply, you simplify the function and find when the denominator is equal to 0.

The idea of infinity is difficult to visualize. It is essentially the greatest number which really doe not actually exist. Thus, it must be equally difficult to imagine what a function is doing at this non-existant point. We must use end behavior models to help us think about what a function is doing at infinity.

The basic idea of end behavior models is to look at the degree of the powers of x. The end behavior model assumes that the function will follow the greatest degree of the power of x in the function.

Ex) In the function f(x) = 3x5 + 2x4 - 3x2 + 9x + 20 we ignore everything except 3x5 because in comparison, all the rest is extremely small compared to the x5.

What it really means: If we try to analyze a function at either end, it might not be the same for each end. Therefore, we can say that the limit as x approaches either º or- º of the original function divided by the function of a behavior model is 1, meaning their limits are the same.

Continuity

Continuity is best explained to a dummy by saying that you can draw the function without picking up your pencil. If you have to pick up your pencil to draw it, the function is not continuous.

Unfortunately, we have to use complicated calculus to actually prove this and further explain continutity

Rules for a Function Being Continuous

For all values "c" ...

1) The lim f(x) must exist

x -> c

2) The function must be equal to the limit at c

3) The function must be defined at c

What it really means: All points on the function must exist for all values of x without any gaps.

Note: All polynomial functoins will be continuous, all compositions will be continuous, as long as there is no division by zero.

Intermediate Value Theorem For Continuous Functions

A function y = f(x) that is continuous on a closed interval [a, b] takes on every value between f(a0 and f(b).

What it really means: It may sound complicated, but it really just means that if a function is continuous in some interval, then the result of the function must consist of at least all the points result from the entering the beginning and end of the interval. If the function is continuous between 4 and 7, the results must contain all possible values between f(4) and f(7). If the result of f(4) = 2 and the result of f(7) = 10, then f(x) must result in all values between 2 and 10 because it must go from 2 to 10 without missing any values. If it did, there would be a gap and then it wouldn't be considered continuous.

Tangent Lines and Rates of Change

A normal line is the line perpendicular to the tangent of a curve, or any line for that matter. The slope of a normal line is the negative reciprocal of the tangent to the line. In other words it is 1 over the negative value of the tangent. So change the sign and flip it over!


Chapter 3 - Derivatives

What is a derivative?

If you think this looks complicated, with the limits and all, you are right. Thankfully, there is an easier way to do this which we will discuss later.

* If you know two points on the function use the following formula to find the derivative labeled

f '(a)

The derivative at x = a :

f '(a) = lim f (x) - f (a)

x-> a x - a

How it works: At first it may look complicated but just think of two points. One is at x and one is at a. The coordinates of the two points will be (x, f(x)) and (a, f(a)). So if you just use the formula for finding slope : (y2-y1) / (x2-x1) you get the slope of a secant between the two lines. Then to find the tangent you just take the limit as you approach the exact point at which you want to find the derivative.

Diferentiability

A function is differentiable if there is a derivative at every point of its domain. In other words, every point on the curve must have a slope, and therefore must be continuous and have a limit at each point.

More simply: The curve should be smooth, and have no gaps to be differentiable.

Taking the derivative to one side

Often, the slope of a line at a point may be different depending which side you come from. If this is true, it is not a differentiable function. It does not have a derivative at this point. However, we may be able to find the derivative coming from each side at this point.

The derivative from the right is not the same as the derivative to the left at point (0,0).

One-sided derivative

As you may suspect, to find the derivative from one said, you just use your formula for finding the derivative, but find the limit as it approaches the point from one side, the positive side for the right hand derivative and the negative side for the left hand derivative.

Right-hand:

Left-Hand

Cases when the function is not differentiable

It is easy to think of ways that a function is not differentiable. At one point, the derivative taken from one side is not equal to the derivative taken from the other side.

Although the slope might be the same, it is not continuous, and can not have a derivative.

* Because a differentiable function must be continuous, we can also assume that if a function is differentiable, then it is continuous.

Intermediate Value Theorem for Derivatives

Similar to the intermediate value theorem for continuity

If a and b are any two points in an interval on which f is differentiable, then f' takes on every value between f '(a) and f '(b).

What it means: If you analyze the curve, the slope between an interval takes on every value between the slope at the beginning and the slope at the end. This is clear because in order to get from one slope to the next, it must take on all slope values in between.

** Rules for Differentiation

Rule

When to use it

Decription

Formula

Derivative of a Constant

Taking the derivative of a constant

The derivative of a constant is 0. So the derivative of 1 is zero, the derivative of 100 is zero, the derivative of 439242 is zero.

df / dx = d/dx (c) = 0

Power Rule for Positive Integer Powers of x

Taking the derivative of a polynomial or function with an x term.

For a variable taken to a power, multiply times the power, THEN subtract one from the power.

d/dx (xn) = nxn-1

So d/dx (x2) = 2x1 = 2x

Constant Multiple Rule

When there ia constant times a differntiable function.

When there is a constant times a differentiable function, take the derivative of the function, then muliply times the multiple.

for constant "c" and differentiable function "u" :

d/dx (cu) = c du / dx

So d/dx 5x2 = 5 (2x) = 10 x

Sum and Difference Rule

When taking the derivative of the sum or difference of two differentiable functions.

The derivative of the sum is equal to the derivative of each one taken seperately, then added together.

The derivative of the difference is the derivative of one subtracted from the other.

With differentiable funcions u and v:

d/dx (u!v) = du/dx ! dv/dx

Product Rule

To find the derivative of two differentiable functions multiplied together.

The derivative of the product of two functions is the first times the derivative of the second, plus the second times the derivative of the first.

d/dv (uv) = u dv/dx + v du/dx

d/dv (x(3+x)) = x(1) + (3+x) (1) = 2x + 3

Quotient Rule

To find the derivative when one function is divided by the other.

Find the derivative using the following rhyme: "lo" is the denominator, "hi" is the numerator, and "di" means "the derivative of". Lo, di hi, minus hi, di lo, over lo lo. Thus remembering this will help you remember that the derivative is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, over the denominator squared.

u and v are differentiable functions.

d/dx (u/v) = [v du/dx - u dv/dx ] / v2

d/dx (x2 / (x3 +2))=

(x3 +2)(2x)-(x2(3x2))

= 2x4 + 4x -3x4

= -x4 + 4x

Velocity and Other Rates of Change

Using these rules, we should be able to find the derivative of many functions. The question is what to do next.

One common use of derivatives is to study the position, velocity and acceleration of an object over time.

Derivatives of Trigonometric Functions

The derivative rules above are fine and dandy with polynomials and powers. However, trig functions are an important part of math and science to determine patterns. Thus we need to figure out how to find the derivative of trig functions at any point on the function.

Trig function

derivative

d/dx sin x

cos x

d/dx cos x

- sin x

d/dx tan x

sec2 x

d/dx cot x

- csc2 x

d/dx sec x

sec x tan x

d/dx csc x

- csc x cot x

*When taking the derivative of functions that begin with "co", ie. cos, cot, co-secant, you must remember to have a negative

The Chain Rule

AKA the "outside inside rule" or "thumb print rule".

When to use it: When you have a composite of two functions

ie. y = (3x2 + 1)2, so f(x) = x2 and g(x) = 3x2 + 1. Thus it can be called f(g(x)).

How do we solve it?

If f is differentiable at the point u = g(x), and g is differentiable at x, then the composite function

(f Û g)(x) = f(g(x)) is dfferentiable at x, and (f Û g)'(x) = f'(g(x)) g'(x)

In idiot's terms, you take the derivative of the outside using the inside alone. Then multiply times the derivative of the inside.

For example d/dx = sin (x3 +3x)

1) Take the derivative of the outside, sin (inside stuff) and leave the stuff inside alone.

d/dx sin (inside stuff) = cos (inside stuff) = cos (x3 +3x)

2) multiply times the derivative of the stuff inside (x3 + 3x)

d/dx (inside stuf) = d/dx (x3 + 3x) = 3x + 3

multiply and get a final answer of cos (x3 +3x) (3x+3)

Power Chain Rule

Basically, we comine the power rule and the chain rule.

When do we use it?

Power Chain Rule is used when you have a differentiable function is raised to a power.

ie. (sin x + 5x)5

What to do?

Use the formula : d/dx un = nun-1du/dx

Apply the same principles as the chain rule. Take the derative of the outside leaving the inside as it is. The mulitply times the derivative of the inside.

ie. (sin x + 5x)5

1) We find the derivative of the outside, (inside stuff)5 using the power rule

5 (inside stuff)4 = 5 (sin x + 5x)4

2) Then we multiply times the derivative of the inside stuff.

d/dx (inside) = d/dx (sin x + 5x) = (cos x + 5)

Multiply and we get a final answer of 5 (sin x + 5x)4 (cos x + 5)

Implicit Differentiation

When to use it:

Used to find the derivative when the function has two variables.

ie. f(x) = 4xy - y = 10

What to do:

1) Separate the two variables, y on one side and x on the other side.

2)Then take the derivative with respect to x on both sides.

3) Factor out dy/dx and solve for dy/dx

ie. Take the first derivative of sin y - 7y - 8x2= 0

sin y - 7y - 8x2 = 0

sin y - 7y = 8x2

dy/dx (siny - 7y) = dy/dx (8x2)

cos y dy/dx - 7 dy/dx = 16x

dy/dx (cos y - 7) = 16 x

dy/dx = 16 x / (cos y - 7)

Now to take the second derivative use the product rule

dy/dx = 16 x / (cos y - 7)

d2y/dx2 = (cos y -7) (16) - (16x)(-siny)dy/dx

(cos y -7)2

Substitute dy/dx from the original equation

d2y/dx2 = (cos y -7) (16) - (16x)(-siny)[16 x / (cos y - 7)]

(cos y -7)2

simplify

d2y/dx2 = (cos y -7) (16) - (16x)(-siny)[16 x / (cos y - 7)] * (cos y - 7)

(cos y -7)2 (cos y - 7)

d2y/dx2 = (cos y -7) (16)(cos y -7) - (16x)(-siny)(16 x)

(cos y - 7)3

d2y/dx2 = 16 [ (cos y - 7)2 + 16x2 sin y ]

(cos y - 7)3

Derivative of Inverse Trigonometric Functions

function

derivative

d/dx sin-1 u

1 / (sqrt (1-u2)) du/dx, |u| < 1

d/dx tan-1 u

1 / (1 + u2) du/dx

d/dx sec-1 u

1 / (|u| sqrt (u2-1)) du/dx

Derivatives of Exponential and Logarithmic Functions

There are a few more special cases in solving derivatives

type

function

derivative

Natural number "e" raised to some differentiable function

d/dx eu

eu du/dx

A constant base "a" raised to a differential function

d/dx au

au ln a du/dx

Natural log of a differentiable function

d/dx ln u

1/u du/dx

Base "a" logarithm of a differentiable function "u"

d/dx loga u

1/ (u ln a) du/dx

Don't forget to multiply times du/dx, which is the derivative of the differentiable function "u".

Chapter 4 - Applications of Derivatives

Now that we know how to find the derivative of a function, we can apply it to analyze graphs and functions.

Extreme Value of Functions

Extreme Values - values which represent the high or low points of a function.

Extreme Value Theorem

If f is continuous on a closed interval [a,b], then f has both a maxium value and a minimum value on the interval.

A function will have a max and a min as long as it is continuous. Sometimes the max will equal the min.

Critical Values

Critical values indicate possible max or minimum values.

A point in the domain of a function f at which f ' = 0 or f' does not exist

What it means: We can tell where a critical point is because the derivative at that point is zero pr does not exist.

This indicates it is changing directions so it might be either a maximum or minimum point.

Mean Value Theorem

If y = f(x) is continuous at every point of the closed interval [a, b] and differentiable at every point of its interior (a,b) then there is at least one point c in (a, b) at which

f ' (c) = f (b) - f (a)

b - a

What it really means: For a function in a given interval, there exists a point between the beginning and the end for which the slope at that point equals the slope of a line connecting the first point in the interval to the last point.

Increasing and Decreasing Functions

The derivative of a function can be used to tell whether it is increasing or decreasing at a point.

1) if f ' > 0 at a point of (a, b) then f increases on [a, b]

2) if f ' < 0 at a point of (a, b) then f increases on [a, b].

More clearly, if the derivative is negative, then the function is decreasing. If the derivative is positive, the function is increasing. If it is zero, it is stable and not changing and therefore can be considered a critical point.

Analyzing a Function

For a function f(x) :

1) Solve for f (x) = 0. These indicates the points at which the function crosses the x axis.

2) Solve for f ' (x) = 0 or is undefined. These indicate possible critical values. Each critical value marks the boundaries of the intervals. When f ' (x) = 0, the slope 0 and the tangent is a horizontal line.

3) Test each interval by plugging in a value in the interval in the f ' (x) equation.

4) Solve for f " (x) = 0. When f " (x) = 0, it is an inflection point and each x value are the boundaries of the intervals.

5) Test each interval by plugging in a value of x that is within the interval.

Modeling and Optimization

Analyzing the function helps us optimize by finding which variables give the best result. The best example of this is with max/min problems.

Here is the way to solve max/min problems.

1) Draw a picture and label with variables. This may seem dumb, but it makes everything much easier when you can see. Also it helps you remember everything so you don't leave anything out.

2) Write the formula for what you are trying to maximize or minimize. ie. Perimiter of Square = P = 4x or Area = A = or2

3) Use the given information to solve one variable in terms of the other.

4) Substitute one variable so that you have an equation for what you are trying to maximize or minimize, now all with one variable.

5) Make domain restrictions based on this equation.

6) Find the derivative and set equal to zero. Then solve for when the variable makes the derivative equal to zero.

7) In your restricted domain, create intervals whose boundaries are made from the values which make the derivative of the function equal to zero.

8) Plug in values between each interval and find out which intervals are positive or negative.

9) Depending on what you are looking for, plug in the possible places where the max or min may occur into the original function. Then compare the results and chose the one that is the min or max. Then solve for the values which the question asks for.

Example :

The village idiot needs to build a cage for his hamster. It does not need a top or a bottom and he has 8 ft of wire fensing which he hopes to attach to a wood bottom. He wants it to be a rectangular shape but and wants it to have the maximum area. Help him make the best cage for his hamster.

1) draw the picture

w = width

l = length

P = perimeter = 8

2) Write an equation for what you are trying to solve:

a = bh

3) Write an equation to solve one variable in terms of another.

2l + 2w = 8

2 (l + w) = 8

l+w = 4

l = 4 - w

4) Substitute to make an equatoin with only one variable.

a = (4-w) (w)

5) Restrict the domain. w > 0 and w < 4 because the area can not be zero and the width or length can not be zero or negative.

6) Set the derivative equal to zero and solve for the variable.

a ' = (4-w) (1) + w (-1) = (4-w) - w = 4 - 2w

4 - 2w = 0

w = 2

7) Make intervals using these critical values

a ' = 4 - 2 h

a ' = 4 - 2 (1) = 2

2 is positive

a' = 4 - 2 (3)

a ' = 4 - 6 = -2

-2 is negative

We are looking for a max which can only occur at w = 2

Thus l = 4 - w

l = 4 - 2 = 2

a = lw = 2(2) = 4

l = 2, w = 2 will give you a maximum area of 4

Linearization

A tangent line to a function can be approximated with linearization.

If f is differentiable at a point x = a, then we can find a function of the tangent line to be approximated by L (x) where:

L(x) = f(a) + f '(a) (x-a)

Related Rates

Similar to max/min problems, we can use calculus to optimize conditions based on two functions. This often requires a bit of ingenuity.

How to solve relates rates problems

1) draw a picture and label the variables and constants.

2) write formulas including all the information that is availble.

3) Write a formula for what you are supposed to solve

4) Write an equation that relates the variables

5) Differentiate with respect to t

6) Use the values that you know to find the rate

Example

The idiot forgot to fasten the base of the ladder to the ground to reach his window 5 ft in the air. The ladder is 13 ft long. So when he climbs it, it slips backwards along the ground at a rate of 5 ft/sec. How fast is he going be dropping down?

Call the ladder "z"

dx/dt = - 5 ft/sec

dz/dt = 0 ft/sec (the length of the ladder does not change)

x2 + y2 = z2

x2 + 52 = 132

x2 + 25 = 169

x = 12

2x dx/dt + 2y dy/dt = 2z dz/dt

2(12) (-5) + 2 (5) (dy/dt) = 2(13) (0)

10 dy/dt = 120

dy/dt = 12 ft/sec

He falls 12 ft per second


Chapter 5 - Integrals

To find the area under a curve, we use Integrals. An integral is anti-differentiation.

Approximation methods of finding the area under a curve

RAM - rectangular approximation method - find the area of a series of rectangles are created based on the curve from a function.

How to do it:

If the curve goes under the x-axis, you have to subtract those rectangles.

Trapezoidal Approximation Method

Since a curve is round, it would seem that maybe a rectangle isn't the perfect shape to use to calculate the area inside the curve. It certainly is easy to do, but we can do a little more work and perhaps find a way to approximate the area more closely. One idea is the trapezoidal appoximation method.

How we do it:

Just as we did before with the rectangular methods, we need to separate the function into sections which serve as the base of each trapezoid. We can think of a trapezoid basically as a rectangle with a triangle put on top. Thus we can find the area of each section using the right side and left side as the two heights. The shorter height makes the triangle and the longer one minus the shorter one is the height of the triangle on top. The formula for a trapezoid is h (y0 + y1) / 2. Thus we just multiply the height divided by two times the sum of the right and left sides of each section. This will gives us all the trapezoids which fit under the curve.

With a little math, we can also find that this is the same as adding the RRAM and LRAM, then dividing by two.

Simpson's Rule

To approximate the integral over a given interval, we can use simpson's rule:

S = h/3 (yo + 4 y1 + 2y2 + 4 y3+ ... 2 y n-2 + 4 y n-1 + yn)

where [a,b] is partitioned into an even number n of subintervals of equal length as described with RAM.


What it really means: To approximate using Simpson's Rule, you just divide the length of the base (length of each section = (b-a) / n ), here called "h", by three and then multiply times the sum of the height at b and a, the beginning and end, then starting with the second to the second last add them multiplying by four, then 2, then four and alternating between 4 and 2, until you reach the scond last division.

Reimann Sums and Definate Integrals

By now, you've probably had enough with approximating. You want to find the exact area under a curve. After all, you're doing all this work, and only getting something that ABOUT the right answer, but not exactly. What's even better is that the anti-differentiation method is much easier, and gives an exact value.

The concept works by taking the samllest possible subintervals by taking the limit as the size of the subinterval approaches zero. This gives you the exact curve when you add up each tiny subinterval.

Theorem 1

All continuous functions are integrable. That is, if a function f is continous on an interval [a, b]., then its definate integral over [a,b] exists.

So basically, if a function is continuous you can find the area under it to an exact measure.

Theorem 2

The integral of a constant

If f(c) = c, where c is a constant, on the interval [a,b], it is written as

°(b a) f (x) dx

= equals °(b a)c dx = c (b-a)

What it really means: If you have a function that is of the form f (x) = c, that means it's a horizontal line, ie, f (x) = 3, means that it is a horizontal line y = 3. Thus the area under the curve is just height times base, because it forms a rectangle. The height is c and the base is (b - a), which is c (b-a).

Rules For Definite Integrals

  1. Order of Integration:
  2. Zero:
  3. Constant Multiple:
  4. Sum and Difference:
  5. Additivity:
  6. Max-Min Inequality: If max f and min f are the maximum and minimum values of f on [a,b], then:
  7. Domination:

° f(x)dx = - ° f(x)dx

° f(x)dx = 0

° kf(x)dx = k ° f(x)dx

° - f(x)dx = - ° f(x)dx

° (f(x) ! g(x))dx = ° f(x)dx ! ° g(x)dx

° f(x)dx + ° f(x)dx = ° f(x)dx

min f · (b-a) [ ° f(x)dx [ max f · (b-a)

f(x) m g(x) on [a,b] u ° f(x)dx m ° g(x)dx

f(x) m 0 on [a,b] u ° f(x)dx m 0

g = 0

Average (mean) value

The average value of a function can be found using integrals

Average of f (x) = 1/ (b-a) ° (b a) f (x) dx

How it works: You can find the average value by dividing the area under the curve by the size of the range. When you take the average of something you add all the values, then divide by how many values you have. This is the same idea. You take the sum of all the function values by finding the area. Then divide by how many function values you have, all the x values, which is b-a.

Theorem 3

Mean Value Theorem for Integrals

If function f is continuous on [a,b] then at some point c which is between a and b,

f (c) = the average of f = 1/ (b-a) ° (b a) f (x) dx

How it works: Since the function is continous, it contains all the values of a function between some interval. Since the mean, or average is the middle value of all the values it must be somewhere between the greatest and the least, and thus must be inside the range.

** Theorem 4

The Fundamental Theorem of Calculus

Part 1

If f is continous on [a,b] , then the function

F (x) =° (x a) f (t) dt has a derivative at every point x in [a, b], and dF/dx = d/dx ° (x a) f (t) dt =

f (x)

What it means: "F (x)" means the integral of f (x).

The intergral of x has a derivative at every point x that is between a and b, the interval.

Also, the derivative of the intergral of f (x) is f(x). It may seem awkward at first, but since an interval is an anti-derivative, the derivative of an anti-derivative will reverse the process and you are left with the original function.

Part 2

If f is continuous at every point of [a,b] and if F is any anti-derivative of f on [a,b] then

° (b a) f(x) dx = F (b) - F (a).

What it means: This is the way to evaluate an integral. You find a function "F (x)" which is the integral of the original function "f (x)". Then subtract the value when you plug the upper bounds into F (x) from the value when you plug in the lower bounds to F (x).

Now, that we have all this theory, we need an easier way to solve integrals.

Here are all the integral formulas (k and n are constants)

Function

The integral (anti-derivative)

Proof (find the derivative)

° xn dx

xn+1 + C, n g 1

n+1

d = (xn+1) = (x + 1) (xn+1-1) = xn, n g 1

dx x +1 (x +1)

°(1/x) dx

ln |x| + C

d ln |x| = 1

dx x

° ekx dx

ekx + C

k

d ekx = ekx

dx k

° sin kx dx

- cos kx + C

k

d - cos kx = sin kx

dx k

° cos kx dx

sin kx + C

k

d sin kx = cos kx

dx k

° sec2 x dx

tan x + C

d tan x = sec2 x

dx

° csc2 x dx

- cot x + C

d (-cot x) = csc2 x

dx

° sec x tan x dx

sec x + C

d sec x = sec x tan x

dx

° cscx cot x dx

- csc x + C

d (-csc x) = csc x cot x

dx

What's with the + C business?

When you take the integral, you always add + C because the derivative of a constant is 0. Therefore, there might have been a constant which when you take the derivative, becomes zero. When taking the anti-derivative, we must take this into account.

Substitution

An important skill or tool in integrating is substitution. Basically, we have to get functions into the form that we can use one of the rules from above to solve. Thus we use substitution to try and achieve this.

Set a variable "u" equal to part of the function. Then plug in u to represent this. Then take the intergral. Finally, after you have an integral function, you can substitute back the original part for u.

Example:

° ( x + 5)3 dx

Set u = x + 5:

du = dx

° (u)3 du

Integrate:

° (u)3 = u4 / 4 + C

Substitute back:

(x + 5)4 / 4 + C

Substitute u = g(x) and du = g ' (x) dx, and integrate with respect to u from u = g(a) to u = g (b).

° (b a ) f (g(x)) * g ' (x) dx = ° (g (b) g(a)) f (u) du

What it means: When you use substitution, you have to change the upper and lower bounds to the function that you are substituting.

So if you make u = (x+5) and the upper and lower bounds are 2 and 1 respectively, then when you substitute, you can change the upper and lower bounds to (2 + 5) and (1 + 5) respectively. then leave the formula of the intergral in f (u) form and plug in your new values for upper and lower bounds.