COURSE OVERVIEW

Why learn Calculus?

What we will talk about in this class, a lot. 

• Graphs
• Cryptocurrency Graphs
• Black Holes, limits and infinity
• Graphs of functions
• change
• Graphs
• Applications in business administration, finance, engineering and industrial design.

 Why learn Mathematics?

1. You don’t have to learn Mathematics — most of the time —.

You do, however, need to learn how to deal with failure. Calculus is designed to teach you that. Why?

2. Struggling is important. Failing is important. This class will be your best practice to fail. Do not underestimate failure.

3 If you are obsessed with a grade, then you should start looking for counseling since you will struggle and you will arrive at incorrect answers MANY times. It is just normal. Don't panic. I will guide you through the whole process, and will give you enough resources and quality time to make you succeed if you WANT to succeed.

4. One of the most important things in life is knowing how to recover from failure. Life is full of failures.

5. Because it is a profound personal experience that also stimulates your brain.

Recover

How you recover is one the most important things in life as well.
 

Do not confuse failure and failing grades. The more you fail in my class the more you will learn  and get you a higher grade.
 

Don’t stigmatize mistakes. Mistakes are key in mathematics. 

Algebra is the reason #1 students struggle

The usual trouble that people have with ‘calculus’ is with algebra

Calculus itself is quite easy

I personally think that I make this class extremely easy for everyone to get a 100, but you have to take Calculus seriously.

It is not like any other course you have taken in the past.

Learn from the past 

• Students who complain usually fail or do very bad in the class. Students who commit and work hard usually do very good.
• Students who copy homework fail the class. Always. 100% of the time. Last semester 15 failed due to continued plagiarism during the semester. 
•  Students who ask questions and participate in class usually score the best grades in exams. 
• Students who revisit assignments never fail the class. 
• Dishonest students never get a point from me. I will never encourage dishonesty in such way.
• Students who miss class, sleep during class or use their phone during class, ALWAYS struggle. 

Keys to success in my class

• Take notes in class. Smart notes, not compulsive notes.
• Start you homework ahead of time and workout problems on your own.
• Learn when to ask for help. Do not wait until the end.
• Make many mistakes and fix them. This is key.
• Make yourself notice. Let me see that you are engaged with the class and that you are genuinely trying. 
• Cheating does not work in Math, period. Don’t copy. You will fail.
• But the master key is to learn how to deal with failure. 

Is time-travelling possible?

Title Text

Calculus will help you understand this 

WebAssign

cufm.mx 1125 1555

WeBWorK 

Take responsibility 

for your learning 

for your life 

for your relationships with people

for your actions and decisions

for your mistakes

Grading Criteria

In-Class Assignments - 15%

Homework - 15% 

Laboratory - 10%

Quizzes - 10% 

Demonstration - 10%

Exam 1 - 40%

Functions

 Quick Review

Limits

Rational Functions

Rational Functions

Continuity

A gateway to limits

Limits

Evaluating Limits:

Graphical Approach

Try on your own until you get it 100% right 

$$\lim_{x\to3^{-}} = $$

$$\lim_{x\to3^{+}} = $$

$$\lim_{x\to3} = $$

$$f(3) = $$

$$ \text{Intervals where }f(x) \text{ increases} = $$ 

$$ \text{Intervals where }f(x) \text{ decreases} = $$ 

$$ \text{absolute maxima} = $$ 

$$ \text{local maxima} = $$ 

$$x$$

Continuity and Direct Substitution

A limit is the point where the function would converge (or where both sides of the function would meet) as we make \(x\) get closer and closer to some value. To remember this, think of it as a question and ask yourself:

Where is \(f(x)\) approaching as \(x\) approaches \( a\)?

We don't care what the value of \(f(x)\) is at \( a\), we just care about the infinitesimal points before and after \(a\). The substitution rule is a method of finding limits, by simply substituting  with \(x\) in the function \(f(x)\). This method can only be used if the function \(f(x)\) is continuous at \(a\).

You can look at it as evaluating any other function, except \(f(a)\) may not exist whereas \( \lim_{x\to a} f(x) \) may. However, \(f(x)\) must be continuous in order to accurately employ the substitution method.  In other words, a function is continuous if :

\[\lim _{x \to a} f(x) = f(a)\]

\(\text{ Where is } f(x) \text{ approaching as } x \rightarrow 2 ? \)

Limit of piecewise functions

Evaluating Limits 

$$\text{The }\dfrac{0}{0} \text{ case} $$

You  might stumble find functions that will evaluate to be undefined upon direct substitution. This is known as the                  However, you should not automatically declare that the function is discontinuous. The 0/0 case is usually an indicator of further algebraic manipulation.

\[\dfrac{0}{0} \text{ case.}\]

Nonetheless, remember that we are interested in the values that are really close to \(a\), not \(a\) itself. So let's take a look at some examples to illustrate this situation. 

Computing Limits

Example 1

$$\lim_{x\to-2} 3x^2+5x-9 $$

Example 2

$$\Rightarrow 12-19= -7$$

$$\lim_{x\to0} {(x+e^{x})}^{-1}$$

$$= \lim_{x\to0}\dfrac{1}{(x+e^{x})}$$

$$= \lim_{x\to-2} 3(-2)^2+5(-2)-9 $$

$$= \lim_{x\to-2} 12-10-9 $$

$$\lim_{x\to-2} 3x^2+5x-9 = -7 $$

$$= \lim_{x\to0}\dfrac{1}{(0+e^{0})}$$

$$\Rightarrow \lim_{x\to0}\dfrac{1}{1}=1$$

$$\lim_{x\to0} {(x+e^{x})}^{-1} = 1$$

Example 3

$$\lim_{x\to2} \dfrac{x^3-4x}{x^2-4} $$

Example 4

$$=\lim_{x\to2} \dfrac{x(x^2-4)}{x^2-4} $$

$$=\lim_{x\to2} {x} =2$$

$$=\lim_{x\to2} {x} $$

$$\lim_{x\to6} \dfrac{x^2+x-42}{-12+2x} $$

$$= \lim_{x\to2} \dfrac{(x+7)(x-6)}{2(x-6)} $$

$$= \lim_{x\to2} \dfrac{(x+7)}{2} $$

$$ \Rightarrow \lim_{x\to2} \dfrac{(2+7)}{2} =\dfrac{9}{2} $$

$$\lim_{x\to6} \dfrac{x^2+x-42}{-12+2x} = \dfrac{9}{2} $$

$$\lim_{x\to2} \dfrac{x^3-4x}{x^2-4} = 2 $$

Example 5

$$\lim_{x\to-4} x^2+\sqrt{-4x} $$

Example 6

$$\Rightarrow 16+ 4= 20$$

$$\lim_{x\to0} {(x+1)^2}\cos{(\dfrac{x}{1-x})}$$

$$= \lim_{x\to-4} (-4)^2+\sqrt{-4(-4)} $$

$$= \lim_{x\to-4} 16+\sqrt{16} $$

$$\lim_{x\to-4} x^2+\sqrt{-4x} = 20$$

$$= \lim_{x\to0} {((0)+1)^2}\cos{(\dfrac{0}{1-0})}$$

$$\Rightarrow \lim_{x\to0} {(1)^2}\cos{(0)}$$

$$\Rightarrow \lim_{x\to0} {(1)}{(1)}=1$$

$$\lim_{x\to0} {(x+1)^2}\cos{(\dfrac{x}{1-x})}=1$$

Limits at Infinity

Limits Involving Infinity \(\infty\)

Before jumping into limits involving infinity, let's recall what indeterminate forms are and how to "evaluate" expressions that oscillate near these values. For instance, let's take a look at the expression \(\dfrac{1}{x}\). 

 Let's start with \(x=1\), plotting each time a greater value for \(x\)

\(\dfrac{1}{10000000}= 0.0000001\)

\(\dfrac{1}{1}=1\)

\(\dfrac{1}{100}=0.01\)

\(\dfrac{1}{10000}=0.00001\)

Do you see where the value of \(\dfrac{1}{x}\) is going?  What would \(\dfrac{1}{x}\) be if \(x=1,000,000,000\)?

The expression \(\dfrac{1}{x}\) would get closer and closer to zero as \(x\) gets bigger and bigger.  We can represent this mathematical behavior by assigning this huge number an infinite value: \(\infty\).

\(\text{ Any number divided by } \pm\infty \text{ approaches } 0\)

\[\lim_{x\to\infty}\dfrac{a}{x}=0 \]

\(\dfrac{a}{\infty}\ne0 \)

but!

'Intuitive' Forms:

\[ \dfrac{a}{\infty} = 0\] 

\[ \dfrac{\infty}{a} = \infty\]

\[ a^{-1}=\dfrac{1}{a}\]

\[a^{\infty}=\infty\]

\[a^{-\infty}=\dfrac{1}{a^{\infty}}\]

\[a*\infty=\infty, a\ne 0\]

\[ \infty-\infty\]

\[a^{0}=1\]

\[\infty^{\infty}\]

\[\infty^{0}\]

\[0^{0}\]

\[1^{\infty}\]

\[\dfrac{0}{0}\]

\[0*\infty\]

Indeterminate Forms:

\[ \dfrac{a}{0^{+}} = \infty\] 

\[ \dfrac{a}{0} = \text{und. }\] 

\[ \dfrac{a}{0^{-}} = -\infty\] 

\(e^{\pm\infty}\)

\[\lim _{x\to\infty} e^x = \infty \]

Trick:   \(e^{-\infty} = \dfrac{1}{e^{\infty}}\) and any number \(a^{\infty} = \infty\)

\[\lim _{x\to\infty} e^{-x} = 0 \]

\[\lim _{x\to-\infty} e^{x} = 0 \]

\[\lim _{x\to-\infty} e^{-x} = \infty \]

\[\lim_{x\to\infty} e^{-x}\]

\[\lim_{x\to-\infty} e^{x}=\]

\[\lim_{x\to\infty} e^{-x}=0\]

\[\lim_{x\to -\infty} e^{x}=0\]

Intermediate Value Theorem

 For any function \(f(x)\) that is continuous on \([a, b]\), and \(f(a)\) and \(f(b)\) exist, (at each end of the interval), then some value \(c\) must exist between \(a\) and \(b\) such that \(f(b)>f(c)>f(a)\)

\(f(c)\)

\(f(a)\)

\(f(b)\)

Intermediate Value Theorem

Domain:

$$[a,b]$$

$$\text{abs. minimum }$$

$$(x_{max}, f(x_{max}))$$

$$\text{abs. maximum }$$

Extreme Value Theorem

There is at least one point \(c\) in the interval \([a,b]\) such that \(f(c) \leq f(x) \) for all \(x\) in \([a,b]\) and there is at least one point \(d\) in the interval \([a,b]\) such that \(f(d) \geq f(x)\) for all \(x\) in  \([a,b]\).

\(f(x)\)

Extreme Value Theorem

There is at least one point \(c\) in the interval \([a,b]\) such that \(f(c) \leq f(x) \) for all \(x\) in \([a,b]\) and there is at least one point \(d\) in the interval \([a,b]\) such that \(f(d) \geq f(x)\) for all \(x\) in  \([a,b]\).

Consider the function \(f(x)\). Assume it is continuous on \([a,b]\). According to  The Extreme Value Theorem, the function \(f\) will have a maximum value and also a minimum value within [a,b]

\(f(x)\)

Chain Rule 

Differentiation 

The Derivative

What the hell is a derivative, anyways?

are all the derivative of \(f(x)\):

\[f'(x) = \lim _{h\to0}\dfrac{f(x+h)-f(x)}{h}\]

The slope of the tangent line

\[m_{tan} = \dfrac{\Delta y}{\Delta x}\]

and the instantaneous rate of change

 \[\lim_{\Delta x\to0}\dfrac{f(x+\Delta x) - f(x)}{\Delta x}\]

...yup, we are bound to have some fun! :)

Examples in Spanish 

Differentiation Rules

(beta, in construction)

CRITICAL VALUES

We say that $x=c$ is a critical point of the function $f\left(x\right)$ if $f\left(c\right)$

exists and if either of the following are true:

$$f^{}\text{'}(c)=0\text{OR}{f}^{}\mathrm{\text{'}(c})\text{doesn't exist}$$

FIND ALL CRITICAL VALUES

MINS AND MAXS

DEFINITION OF CONCAVITY

Let f(x) be a differentiable function on an open interval I.

The graph of f(x) is:

1. concave upward on I if f '(x) is increasing on the interval

2. concave downward on I if f '(x) is decreasing on the interval

CONCAVITY TEST

Let f(x) be a function whose second derivative exists on an open interval I.

   1. If f ''(x) > 0 for all x in I , then the graph of f (x) is concave upward on I .

    2. If f ''(x) < 0 for all x in I , then the graph of f (x) is concave downward on I .

INFLECTION POINT

If the graph of a continuous function possesses a tangent line at a point where its concavity changes from upward to downward (or vice versa), then the point is called a point of inflection.

SECOND - DERIVATIVE TEST FOR RELATIVE EXTREMA

Let f ''(x) exists on some open interval containing c and f '(c) = 0.

1. If f ''(x) > 0, then f(c) is a relative minimum.

2. If f ''(x) < 0, then f(c) is a relative maximum.

3. If f ''(x) = 0, then nothing can be decided

Examaple 1

Function f(x) = x^2 - 1 is concave upward on the set of all real numbers because its second derivative
f''(x) = 2 is positive for all x.

Find any local extrema  using the Second Derivative Test.

FIND THE CRITICAL VALUES

L'Hospital's Rule

It is possible to find the limit of some function whose limit evaluates to \( \dfrac{0}{0} \) or any indeterminate form. To reviews indeterminate forms, click here.

L'Hospital's Rule states that whenever

\[ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0} \]
 

we can find the limit by taking the derivative of the numerator and the denominator:

\[ \lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{f'(x)}{g'(x)} \]

The Integral

(beta, still in construction)

Position  s(t)

Velocity  v(t)

Acceleration a(t)

Initial Value Problems

The Area Problem

Imagine you want to find the area below the curve shown at the right.

Of course, there is no formula that can nearly be fixed for every area under every curve. However, we can approximate, to an infinitesimal point, the total area if we break it down into rectangles.

Let's see what I mean by this.

The Sum of the Areas

As observed in both curves, the more rectangles (\(n)\) we draw, the more accurate will be our result.

\(n=4\)

\(n=4\)

\(n=8\)

\(n=14\)

Limits in Integration

As we anticipated earlier, limits are also a fundamental concept behind integration. As a matter of fact, the limit is the most important concept in all calculus.  Let's see the role that the limit plays in integration.

What is the base of each rectangle that is drawn between 0 and 1?

If we draw 5 rectangles \(n=5\), then the base of each rectangle will be \(\dfrac{1}{5}\) or \( \dfrac{1}{n}\).  The height will vary, but the base will remain constant for every triangle.
Furthermore, we can find the area of each rectangle and add them up to come up with the exact area. The more rectangles we draw, the smaller will be the base of each triangle, but the more accurate we will get.

Solving the area problem

In the make as of July 31st, 2018

Title Text

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Techniques of Integration

Integration by Substitution

Integration by Parts