Calculus I - Math 1431 - UH - class 18431 - section 50 by Adrian Radillo

Factoring a quadratic expression

adrian_radillo — 2016-08-21 01:45:40 UTC

http://www.mathsisfun.com/algebra/factoring-quadratics.html
Apart from the very good link above, I would like to add two techniques below.
Recall that the aim is to rewrite a polynomial that is initially expressed in the form \(P(x)=\alpha x^2+\beta x + \gamma\) into its factored form \(P(x)=(ax+b)(cx+d)\), where \(\alpha, \beta, \gamma, a,b,c,d \in \mathbb{Z}\).

One can recognize an identity of the form \(a^2\pm2ab+b^2\) or \(a^2-b^2\) (\(a\) and \(b\) here are not the same as in the previous paragraph). An example would be \(x^2-6x+9\). If you know your identities by heart, you quickly find \((x-3)^2\).
One can guess a root of the polynomial and use the fact that if \(x^*\) is a root of the polynomial \(P(x)\), then \(x-x^*\) will appear in the factorization of \(P(x)\). As an example consider the polynomial \(P(x)=x^2+2x-3\). It is not too hard to substitute in the values \(0,1,-1,2,-2\) for \(x\) and to stop as soon as one gives a result of zero. For instance here, we notice that the value \(x=1\) makes the expression vanish (that is to say, \(P(1)=0\), or equivalently, \(1\) is a root of \(P\)). We can then start writing out our polynomial as: \(P(x)=(x-1)(\ldots)\). But note that we still have to fill in those three dots \(\dots\) We can do this because we know that \(P(x)=x^2+2x-3\). TRY TO FIGURE IT OUT BY YOURSELF. Trust me, in many cases, the polynomials that you are given have an integer root between \(-2\) and \(2\) and this checking procedure can yield the factorization very quickly. The answer for the example is: \(P(x)=(x-1)(x+3)\).

CRUCIAL IDENTITIES FOR QUADRATICS!

adrian_radillo — 2016-08-21 01:51:08 UTC

\((a+b)^2=a^2+2ab+b^2\)
\((a-b)^2=a^2-2ab+b^2\)
\((a+b)(a-b)=a^2-b^2\)
LEARN THEM, PLEASE LEARN THEM. THEY ARE SO USEFUL!!!!!

Polynomials in one variable (or just polynomials)

adrian_radillo — 2016-08-21 02:14:04 UTC

For the purpose of this lab, a polynomial in one variable is a function,
\(P: \mathbb{R} \to \mathbb{R}\)
\( x \mapsto a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0 \)
where the fixed natural number \(n \geq 0\) is called the degree of \(P\) (we must assume \(a_n \neq 0\)) and the coefficients \(a_i \in \mathbb{Z}\) \((0\leq i \leq n)\) are all integers.

Note that I had already written with symbols that the coefficients were all integers. I added the text for you guys, but it is not necessary. When you feel comfortable enough with symbol notation, you can drop the English words with me. But make sure you use the symbols correctly!!!

Any \(x\in \mathbb{R}\) meeting the condition \(P(x)=0\) is called a root of \(P\).

Logic and Math

adrian_radillo — 2016-08-21 02:20:30 UTC

Knowing the rudiments of propositional logic helps a lot to do rigorous mathematics. If you spend some time learning the following, it will pay you back for the rest of your life!!!!!!

Great introductory text to \(\LaTeX\)

adrian_radillo — 2016-08-21 02:24:32 UTC

I learned almost everything I know about \(\LaTeX\) from this text.

What is this padlet???

adrian_radillo — 2016-08-21 03:13:09 UTC

Just double-click on the green area and you will be able to create a new post. Give it a title, and express yourself about anything that is Cal-1related! If your post ends up overlapping with pre-existing posts, don't worry, I will rearrange the layout regularly.

Usually you will appear as an anonymous author on your post, but feel free to sign it by writing your name at the bottom - as I did on this one - or to create an account on this website and sign in, if you want us to know who you are. Of course, it is all fine if you remain anonymous.

One of the moderators, which include myself and some of the instructors, will have to approve your post before it appears visible to everyone. It shouldn't take more than a day or two in general.

I also create posts that I deem important for your course every now and then, so feel free to have a look around to see what's new!
Don't forget to scroll all the way down and to the right not to miss out any posts.

Adrian :-D

Your TA

adrian_radillo — 2016-08-21 03:22:43 UTC

My full name is Adrian Ernesto Radillo. My office is in
PGH 347, but I am often somewhere else, so it is best to drop me an email at adrian@math.uh.edu to convene for a time to meet.

I have a personal webpage, but I will concentrate everything related to our lab on this padlet.

A few more important remarks:
I am French, so I apologize in advance for my English mistakes, and if I make you repeat your question in class.

Tambien hablo espanol, asi que si se les hace mas facil, les puedo ayudar en espanol.

I learned math in France. Therefore, I am not familiar with most of the techniques that you learned in high school. If your goal is to learn the concepts behind the techniques, this will not be a problem. After all, I am currently doing my PhD in mathematics at UH.

How to write mathematical formulae in your posts?

adrian_radillo — 2016-08-21 03:27:56 UTC

Check this out for an introduction.
Also, if you want to know how I typeset one of my post, just click on it and a window with the source code of my post should pop-up. You may copy portions of it and paste them in your own post. This will be a great way for you to learn the \(\TeX\) language.

CASA website (also called CourseWare)

adrian_radillo — 2016-08-21 03:33:16 UTC

www.casa.uh.edu
This website is very important! On it, you can check out the calendar for your course that contains lecture slides and deadlines. You can also access your textbook, your grades, and many other things... Check it out and ask your instructors about it. Note that you can also access this website via the university cougarnet login.

Discussion board
Once you are logged in to the casa website, a link named "Discussion Board" should appear next to the TEXTBOOK link. Clicking on it will give you access to a CAL 1 forum on which you can post questions. This is an alternative to this padlet. Use either one. Don't be shy.

Some other UH links

adrian_radillo — 2016-08-21 03:41:35 UTC

FAQ about the campus-carry law at UH.
Feeling unwell? Contact the student counseling service!

Your instructor's website

adrian_radillo — 2016-08-21 20:11:59 UTC

https://www.math.uh.edu/~beatrice/Math1431.htm

Homeworks info, VERY IMPORTANT:
https://www.math.uh.edu/~beatrice/1431AssignedHomework.htm

Trig identities

adrian_radillo — 2016-08-21 20:24:03 UTC

Learn them. BUT FIRST, LEARN THE MEANING OF THE COSINE, SINE AND TANGENT FUNCTIONS!

\(cos^2(x)+sin^2(x)=1\)
\(cos(a+b)=cos(a)cos(b)-sin(a)sin(b)\)
Deduce \(cos(2x)=cos^2(x)-sin^2(x)\) from it!
and further find \(cos(2x)=2cos^2(x)-1\)
and equivalently \(cos^2(x)=\frac{cos(2x)+1}{2}\)
\(sin(a+b)=sin(a)cos(b)+sin(b)cos(a)\)
Deduce \(sin(2x)=2sin(x)cos(x)\)

Cal I - Being correct, being quick... and understanding!

adrian_radillo — 2016-08-21 20:31:21 UTC

In this course you will find that in order to get the best grades, you will have to both be correct and quick in answering the tests questions.
Of course you could train your memory to remember formulae and strategies for every single problem you are presented. THIS IS DEFINITELY NOT WHAT I RECOMMEND YOU!

I encourage you to try first to understand, then to be correct, and only then to gain speed. I can teach you tricks to be quick in your tests. But I will NOT do it without making sure first that you understand the concepts that you are manipulating.

Understanding the concepts will help you immensely in at least two ways. First you will be able to think outside the box and tackle new problems by yourself. Secondly, these concepts will remain with you for the rest of your life. Sure, when you are 50 years old you might not remember the formula for the slope of a line, but you might remember the meaning of it! And believe me, many of the concepts that you will deal with during this semester are part of our everyday life. For instance, if I hear in the news that the amount of rain in a city is inversely proportional to its distance to the sea (I am making this up), the notion of slope is involved!!!

So please, be curious, try to drop your fears and bad feelings about maths, or about specific topics in math, and just ask questions about what's unclear. This padlet is one of a million ways in which you can learn something that will be useful to you!

One last thing, as you make progress in mathematics, you may find your head overloaded with formulae. Our heads are not dictionaries. But we have the cognitive power of logical deduction! This will be your best friend in mathematics as it will allow you to only remember a few formulae and deduce the other ones from them. Always try to remember the least amount of formulae that enables you to recover all the other ones. For this, understanding the concepts will help you a lot.

Standard forms for lines, parabolas, circles, ellipses and hyperbolas

adrian_radillo — 2016-08-21 22:31:25 UTC

line

\(y=ax+b\)

parabola

standard form

\(y=ax^2+bx+c\)

vertex form

\(y=a(x-h)^2+k\)
conics form
\(4p(y-k)=(x-h)^2\)
(for the parabolas that we will encounter in this lab)

circle of radius \(r >0\) centered at \((h,k)\):

\((x-h)^2+(y-k)^2=r^2\)

Cosine CheatSheet

adrian_radillo — 2016-08-22 14:26:29 UTC

All you need to know about the cosine function in one page!

First 2 days lab problems

adrian_radillo — 2016-08-22 14:27:14 UTC

TRY THEM OUT, BE METICULOUS!

Model answer for question 1c (with modified numbers)

adrian_radillo — 2016-08-22 16:41:36 UTC

Question
Use a sign chart (\(x\)-axis) to solve \(\frac{1}{x-1}+\frac{1}{2x+4} \geq 0\)

Answer
First we rearrange the inequality to have a product on the left. It is easier to just handle the left-hand side for a while.
(when you write a long chain of equalities, split it on several lines as below)
\begin{align}
\frac{1}{x-1}+\frac{1}{2x+4} &=\frac{1}{x-1}+\frac{1}{2(x+2)} \\
&=\frac{2(x+2)+x-1}{2(x-1)(x+2)} \text{This step is important, did you get it?}\\
&=\frac{3x+3}{2(x-1)(x+2)}\\
&=\frac{3}{2}\frac{(x+1)}{(x-1)(x+2)}
\end{align}
Recall that we are interested in the sign of the last expression. Removing the factor \(\frac{3}{2}\) doesn't affect the sign of the expression (is it clear to you why?). Furthermore, for any \(a,b \in \mathbb{R}\), if \(b \neq 0\), then \(\frac{a}{b}\) and \(ab\) have same sign (again, is this clear to you and do you see how it relates to our problem?). We might as well, therefore, study how the sign of the product \((x+1)(x-1)(x+2)\) varies with the values of \(x\).
Be aware of an important difference between the fraction and the product though: The product may be defined \(\forall x \in \mathbb{R}\) whereas the fraction is defined on \(\mathbb{R} \setminus \{1,-2\}\); you must exclude from \(\mathbb{R}\) the points \(1\) and \(-2\) at which the denominator vanishes. Is it clear?
Later in this class you will learn how to quickly interpret the product \((x+1)(x-1)(x+2)\) as a cubic polynomial, but for now, you can summarize your results in a table as shown in the attached image.

How did I produce this table?
Well, each factor is linear in \(x\). Studying the sign of each factor amounts to finding out for which values of \(x\) a specific line is above or below the \(x\)-axis. Just as we did in class!!!!
Once you have produced the table, write out the solution in one of the equivalent forms:

\(\frac{1}{x-1}+\frac{1}{2x+4} \geq 0 \Longleftrightarrow x\in (-2,-1]\cup (1,+\infty)\)
The initial inequality is TRUE precisely when \(-2

Ask out anything that is not clear to you about this answer. I expect you to understand all my reasoning in this answer. In your written answer, I expect you to transform the left-hand side of the initial inequality as I did, and to state in a sentence that you produce the table in order to study the sign of the expression. I also expect you to say explicitly which values of \(x\) are excluded from your consideration (with me - and with me only - if you use the double bar in the table that will be enough to tell me that you exclude those points). And finally, conclude with one of the two bullet points I showed you above. I encourage you to practice the interval notation.

Quantifiers

adrian_radillo — 2016-08-23 12:54:06 UTC

\(\forall\) - "for all"
Very useful every time one wants to makes a claim about all the elements in a set.
\(\exists\) - "there exists"
We use this quantifier every time we want to make an existence claim.

Stuck?

adrian_radillo — 2016-08-23 13:58:02 UTC

It is okay to get stuck in math. It happens to everybody. However, in this course, never remain stuck for too long. The pace is fast, so unfortunately, you cannot afford to spend an entire week on a single problem.
Here is what I advise you to do:

Try honestly a question for a few minutes. If you realize that the question involves notions or concepts that your don't understand, fill those gaps in your knowledge first. That is, go and find the meaning of those concepts.
Now, the amount of information online about math can be overwhelming. This is why, if after 5 minutes of browsing you feel that you don't get the info that you need, log in to this padlet or to the CASA discussion board and post your question!

THERE IS NO DUMB QUESTION, ASK ANYTHING THAT IS MATH RELATED!

Practice, practice, practice!

adrian_radillo — 2016-08-23 14:15:33 UTC

You need to practice. You need to do many exercises. But NEVER sacrifice the quality for the quantity. Don't do 10 exercises on the same topic if the first one was not clear! Before starting the second exercise, post your question, and go on to another topic. Once you get your answer, carry on with the first topic.

Textbook exercises section 1.2

adrian_radillo — 2016-08-23 14:19:06 UTC

Textbook exercises section 1.3

adrian_radillo — 2016-08-23 14:19:19 UTC

Model answer for 2d

adrian_radillo — 2016-08-24 01:32:16 UTC

I will only mention the reasoning that justifies the graph.
The expression, that depends on \(x\) is of the form \(\frac{f(x)}{|f(x)|}\), where \(f\) is a function (you should know which one in this exercise). The first thing to notice is that the denominator depends on \(x\) so we should IMMEDIATELY check for the zeros of the denominator. Say that we find out that \(x^*\) is the only real number that makes \(|f(x)|\) vanish - in the exercise find the specific \(x^*\). Then we are assured that \(\frac{f(x)}{|f(x)|}\) is well-defined on \(\mathcal{D}:=\mathbb{R}\setminus \{x^*\}\).
The next step is to analyze the sign of \(f\) as a function of \(x\). Suppose you are able to characterize completely the following two sets:
\(P:=\{x \in \mathcal{D}\ :\ f(x)>0\}\) and \(N:=\{x \in \mathcal{D}\ :\ f(x)<0\}\)
Do you understand what the definition of these sets mean? Can you find them in this specific exercise?
Then we are assured that:
\(\frac{f(x)}{|f(x)|}=1\), \(\forall x\in P\) and \(\frac{f(x)}{|f(x)|}=-1\), \(\forall x\in N\).
Again, why?
Can you sketch the graph now?
And if I ask you to graph \(\frac{|9x-12|}{9x-12}\), can you do it?

Do the math in your head!

adrian_radillo — 2016-08-24 14:41:19 UTC

Every time you have to multiply and divide integers, do it in your head. This is a worthwhile practice! Trust me !

Question on limits

2016-08-25 01:44:32 UTC

How to find?
\begin{equation}\lim_{x \to -1}\frac{x+1}{\sqrt{x+2}-1}\end{equation}

Answer

adrian_radillo — 2016-08-25 13:14:52 UTC

Hi, first of all thanks for asking. I took the liberty to rewrite your question in mathematical typeset and modify the title slightly. If you have any objection, let me know.
Here is how to answer this question:

Since the denominator vanishes at the limit point, you cannot just plug-in the value and evaluate.
You have a fraction with a square root at the denominator. Therefore, it will prove useful to rationalize. I do that in the next bullet point.
Do you recall that \(a^2-b^2=(a+b)(a-b)\)? Well it is time to use it. Call \(a:=\sqrt{x+2}\) and \(b:=1\). The denominator is now \(a-b\). The idea is to multiply by \(a+b\) in order to make \(a^2-b^2\) appear. Since we are dealing with a fraction, we need to multiply by \(a+b\) on top and bottom at the same time. Hence, as long as \(x\geq -2\) the following equalities are TRUE:

\begin{align}
\frac{x+1}{\sqrt{x+2}-1} &=\frac{(x+1)\left(\sqrt{x+2}+1\right)}{\left(\sqrt{x+2}-1\right)\left(\sqrt{x+2}+1\right)}\\ &= \frac{(x+1)\left(\sqrt{x+2}+1\right)}{x+1}\\ &=\sqrt{x+2} +1
\end{align}

Finally, we conclude (and we can because the limit point satisfies \(-1>-2\)):

\begin{align}
\lim_{x \to -1}\frac{x+1}{\sqrt{x+2}-1}
&=\lim_{x \to -1}\sqrt{x+2}+1\\
&=2
\end{align}

Ask about anything that remains unclear.

Sign of roots of a quadratic

adrian_radillo — 2016-08-26 14:40:17 UTC

Suppose that you are given a quadratic expression \(ax^2+bx+c\) and that you know that it has two distinct real roots (that is \(b^2-4ac>0\)). Then the sign of \(\frac{c}{a}\) is the sign of the product of the two roots.
This can be seen using the quadratic formula. Can you do it?

Question on unit conversions

2016-08-26 21:25:10 UTC

I'm stuck on two questions. Can anyone help me out?
1. It takes 1,680 gallons of water to produce one pound of grain-fed beef. How many cubic feet of water are needed to produce 7 kilograms of beef?
2. I have to convert 65gal/min into cubic inches/second.

I've worked both problems multiple times, and even though they should be easy I keep getting the wrong answer.

Answer

adrian_radillo — 2016-08-27 00:06:02 UTC

1. Your system takes a volume unit (Gallons) as input and produces an output which has a mass unit (lb). It is better to stick to these units in your calculation, and then convert for the answer.

You know you want 7kg as output. Let's convert this to (lb) since this was the unit you were first given. For this you need to know how many (lb) is 1 kg. We have 1 kg=2.2046 lb, so 7 kg=15.4324 lb. Clear so far?

Let's compute the required input volume in Gallons first, to get the 15.4324 lb of beef. Since 1,680 Gal produce 1 lb, 1,680x15.4324=25,926.432 Gal will produce 15.4324 lb of beef.

Finally, convert this volume into ft^3: 1 Gal = 0.1337 ft^3 so 25,926.432 Gal = 3,466.364 ft^3.

This is your answer. You need 3,466.364 ft^3 to produce 7 kg.

2. Here your unit is in the form Volume/Time.
How many cubic inches is one galon? 231
So 65 gal = 231x65=15,015 inch^3.

So 65gal/min = 15,015 inch^3/min
But this is not the end of the story, you still have to deal with the time unit. How many seconds is 1 minute? 60

But your unit is \(min^{-1}\), not quite min. So you need to divide by 60, instead of multiplying. Namely, your 65gal/min become 15,015/60=250.25 inch^3/sec.

How to present the solutions of an equation in your exercises

adrian_radillo — 2016-08-27 00:35:23 UTC

Say I want to solve \(3x+4=0\) on \(\mathbb{R}\). An easy way to present your final result is,\begin{equation} 3x+4=0 \Leftrightarrow x=-\frac{4}{3}\end{equation}
Why this \(\Leftrightarrow\) symbol? Well, it means that if the equation is TRUE, then \(x\) must have the value \(-\frac{4}{3}\), and if \(x=-\frac{4}{3}\), then the equation is TRUE. The equivalence sign \(\Leftrightarrow\) gives you a succinct way of presenting the solutions of the equation.

Another example is the following. Say we want to express the solutions to the equation \(\sin(y)=0\) for \(0\leq y \leq \frac{11\pi}{5}\). We saw in class that the solutions are \(\{0,\pi,2\pi\}\). Either you just write what I just wrote, which is fine, or you may write the following:\begin{align}\left\{ \begin{array}{l} \sin(y)=0\\
0\leq y \leq \frac{11\pi}{5}\end{array}\right.
\Leftrightarrow y\in \{0,\pi,2\pi\}\end{align}
Same reasoning as above.

Question on limits

2016-08-27 16:28:30 UTC

Answer

adrian_radillo — 2016-08-27 17:03:27 UTC

Try first to factor the numerator of the fraction. Clearly, \(5\) is a common factor. What is \(135\) divided by \(5\)? You MUST BE ABLE to compute this in your head.
Here is how I do it.
\(135\) is \(125+10\). I know that \(125=25\times 5\) and that \(10 = 2 \times 5\). Therefore \(135=27\times 5\). Does that make sense to you???
Hence the numerator reads now \(5(x^3-27)\). Can we simplify further? YES because \(27=9\times 3=3\times 3\times 3\). In other words \(3\) is a root of the polynomial \(x^3-27\). This implies (read my post on factoring a quadratic expression in the "trick or treat" section) that \(x^3-27=(x-3)Q(x)\) where \(Q(x)\) is an unknown polynomial for now. But it should not be too hard to find it. What polynomial \(Q(x)\) makes the previous equation true? The only answer is \(Q(x)=x^2+3x+9\). I will try to post a video here later to explain how I found this \(Q\). Anyway, we now reduced the problem to finding the following limit:
\begin{equation}
\lim_{x\to 3}\left( \frac{5(x-3)(x^2+3x+9)}{x-3}\right)
\end{equation}
What should you do? The value of any function \(f\) at a specific \(x\)-value \(x^*\) doesn't matter when one evaluates the limit \(\lim_{x\to x^*}f(x)\).
In our example, \(f(x)=\frac{5(x-3)(x^2+3x+9)}{x-3}\). Is it clear to you that for any \(x\neq 3\), the \(x-3\) factor cancels out? In other words, for any \(x \neq 3\),
\begin{equation}
f(x)=5(x^2+3x+9)
\end{equation}
It should now be clear that,
\begin{equation}
\lim_{x\to 3}f(x)=5(3^2+3\times 3+9)=5\times 27 = 135
\end{equation}

Ask about anything that is unclear in this answer.

2016-08-27 18:46:32 UTC

Thank you for answering, I understand it now! I was just stuck on the factoring part. Thank you again :)

Important limits for \(\sin\) and \(\cos\)

adrian_radillo — 2016-08-29 13:35:28 UTC

Very good link to the proof of \begin{align}&\lim_{t\to 0}\frac{\sin(t)}{t}=1\\
&\lim_{t\to 0}\frac{1-\cos(t)}{t}=0\end{align}

LEARN YOUR TABLES

adrian_radillo — 2016-08-29 13:38:22 UTC

Guys, if you have any doubts about computing in your head multiplications like \(6 \times 8\) or \(7 \times 9\), get rid of them as soon as you can by re-learning your tables. No shame in this, and you don't have to tell anybody that you are doing this, BUT DO IT!!!!!!

adrian_radillo — 2016-08-29 13:46:11 UTC

Lab problems - week2

adrian_radillo — 2016-08-29 13:47:05 UTC

As usual guys, you should remain with no doubts at all in doing these exercises. Ask out anything that is not clear to you.

And more exercises

adrian_radillo — 2016-08-31 23:07:52 UTC

And more exercises!!!

adrian_radillo — 2016-08-31 23:08:07 UTC

More exercises

adrian_radillo — 2016-08-31 23:08:16 UTC

Question on continuity

2016-09-01 21:52:32 UTC

sketch the graph and find the x-values (if any) at which f is not continuous. Classify the discontinuities as being removable or essential. If the latter, state whether it is a jump discontinuity, an infinite discontinuity, or neither.

Answer

adrian_radillo — 2016-09-01 23:39:57 UTC

Without the formula for the function, I can't really sketch the graph. The following should help you answer though:

If your function is a rational function, that is if it is expressed as a ratio of polynomials in \(x\), say \(f(x)=\frac{P(x)}{Q(x)}\), then the function \(f\) will be discontinuous at all the roots of \(Q\).
Still for rational functions, try to factor the numerator and the denominator. For instance \(f(x)=\frac{x^2+x}{x^2-x-2}=\frac{x(x+1)}{(x-2)(x+1)}\). From the previous bullet points, we see that this particular \(f\) is discontinuous at \(x=2\) and \(x=-1\). However, the factor \(x+1\) is a bit peculiar because it appears in both the numerator and the denominator. This makes the discontinuity at \(x=-1\) a removable discontinuity. On the other hand, since the factor \(x-2\) only appears in the denominator, the discontinuity at \(x=2\) is infinite.

Let me know if you have more questions, and give me the formula of the function if you have questions about the sketching.

Question on how to solve this piecewise function

2016-09-02 16:40:29 UTC

question

2016-09-02 21:45:21 UTC

how would I solve this problem:\begin{equation}
\lim_{x\to 4^+}\frac{|x-4|}{x-4}
\end{equation}

Answer

adrian_radillo — 2016-09-02 23:57:20 UTC

By the way the function is defined, for any \(A\) and \(B\), the function will be continuous on the intervals \((-\infty,2)\), \((2,3)\) and \((3,\infty)\). So the only potentially problematic points are \(x=2\) and \(x=3\).
How to render the function continuous at these points?
Well, we need to compute left and right limits at each of these points.
To compute the left limit at \(x=2\), I use the definition of \(f\) on \((-\infty,2)\), namely,
\begin{equation}
\lim_{x\to 2^-}f(x)=2A-4
\end{equation}
Is it clear how I got this?
Now, for the right limit at \(x=2\) I use the formula for \(f\) on \((2,3)\), namely,
\begin{equation}
\lim_{x\to 2^+}f(x)=2B
\end{equation}
So far, we see that the function \(f\) will have same left and right limit at \(x=2\) if and only if \(2A-4=2B\). Clear so far?

Let's deal with the other point now. With similar strategy I find,
\begin{align}
\lim_{x\to 3^-}f(x)&=3B\\
\lim_{x\to 3^+}f(x)&=9B+2A
\end{align}
So the function \(f\) will have same left and right limit at \(x=3\) if and only if \(9B+2A=3B\).

At this point, we have established the following:\begin{align}f \text{is continuous at }2\text{ AND }3 &\Leftrightarrow\left\{ \begin{array}{l} \lim_{x\to 2^-}f(x)=\lim_{x\to 2^+}f(x)=f(2)\\
\lim_{x\to 3^-}f(x)=\lim_{x\to 3^+}f(x)=f(3)\end{array}\right.\\
&\Leftrightarrow \left\{ \begin{array}{l} 2A-4=2B\\
9B+2A=3B\end{array}\right.
\end{align}
So the function \(f\) will be continuous if and only if \(A\) and \(B\) solve the system of equations above.
Subtracting the first line from the second yields \(9B+4=B\). Solving for \(B\) yields \(B=-\frac{1}{2}\), which in turn implies \(A=\frac{3}{2}\).

As usual, read everything to make sure that it makes sense to you, and ask me if things remain unclear. I write all these details for you not to miss any of the important ideas. On your answers, I don't expect you to write that much. But I expect you to give the important justifications in this argument. You should absolutely mention the following:
1. \(f\) is continuous away from \(2\) and \(3\)
2. \(f\) is continuous at \(2\) and \(3\) if and only if its left and right limits agree at these points and are equal the value of the function itself.
3. The above condition is satisfied if and only if \(A\) and \(B\) solve the particular system of equations I derived.

Question on limits

2016-09-03 16:04:18 UTC

Hey Adrian I'm very confused on this question and was wondering if you could help.

Given that

f(x)= [(√x+7)−3] / (x−2)

, define the function

f(x)

at 2 so that it becomes continuous at 2.

Edit: Thank you very much! :) That helped a lot, I didn't think about multiplying by the conjugate of the numerator; instead I tried to get rid of the denominator somehow.

Answer

adrian_radillo — 2016-09-03 17:19:05 UTC

Try to find my post a few posts to the left that is called "Model answer for 2d". Everything is explained in details there. Ask me if there are things that you don't understand there.

For a quick answer read the following:
What is the sign of \(|x-4|\) when \(x>4\)? It is positive, therefore \(|x-4|=x-4\) for \(x>4\). Hence,
\begin{equation}
\lim_{x\to 4^+}\frac{|x-4|}{x-4}=\lim_{x\to 4^+}\frac{x-4}{x-4}=1
\end{equation}

Answer

adrian_radillo — 2016-09-03 17:55:07 UTC

Hi, I believe you made a typo when copying. The correct question should be:

Given that \begin{equation}
f(x)=\frac{\sqrt{x+7}-3}{x-2},\end{equation}define the function at \(x=2\) so that \(f\) becomes continuous at 2.

You have to rationalize, multiplying top and bottom by \(\sqrt{x+7}+3\). Let me know if this helps.

Model answers to problems 7, 10, 13, 14, 15 (for pb 5, see post on the left) from written HW 1

adrian_radillo — 2016-09-03 21:39:18 UTC

Some feedback on your work:

Always justify your answers
DO NOT WRITE any of the following as it does not make mathematical sense:\begin{align}&\lim_{x\to 4}=\frac{1-x}{1+x}\\ &\lim_{x\to 4}\frac{x(x+1)}{(x+1)(x+2)}=\frac{x}{x+2}\\ &\frac{x+2}{x}=\frac{0+2}{0}\end{align}
Learn how to write by hand the greek letter delta: \(\delta\)

Limits Question

2016-09-04 19:06:15 UTC

I can't figure out what to do with the Tan in the denominator.

Answer

adrian_radillo — 2016-09-05 15:16:35 UTC

First, recall that \(\tan(4x)=\sin(4x)/\cos(4x)\):\begin{align}\frac{\sin(x)}{3x}+\frac{2x}{\tan(4x)}=\frac{\sin(x)}{3x}+\frac{2x\cos(4x)}{\sin(4x)}\end{align}
Then, you need to transform your expression to use both the facts that \(\lim_{x\to 0}\frac{\sin(ax)}{ax}=\lim_{x\to 0}\frac{ax}{\sin(ax)}=1\) and \(\lim_{x\to 0}\cos(ax)=1\). We have, \begin{align}\frac{\sin(x)}{3x}+\frac{2x\cos(4x)}{\sin(4x)}&=\frac{1}{3}\frac{\sin(x)}{x}+\cos(4x)\frac{1}{2}\frac{4x}{\sin(4x)}\end{align}
Using the known limits I mentioned above and the algebraic rules for limits we get the final answer:\begin{equation}\lim_{x\to 0}\left(\frac{\sin(x)}{3x}+\frac{2x}{\tan(4x)}\right)=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\end{equation}

Determine if the function \(f(x) = |4 - x^2|\) is continuous at the point where \(x = 2\).

2016-09-05 20:40:05 UTC

I keep getting this answer wrong. can you help me determine what type of discontinuity this function is and how to determined it. Thanks

Answer

adrian_radillo — 2016-09-05 22:07:33 UTC

This function is continuous at every point on \(\mathbb{R}\). Try to plot it.
Here are hints for being able to quickly plot this function.
Let's first define \(g(x)=4-x^2\) on \(\mathbb{R}\). What does the graph of \(g\) look like?
It is a downwards-facing parabola with axis of symmetry the \(y\)-axis and \(y\)-intercept \(4\).
What are the roots of \(g\)? They are \(-2\) and \(2\). In particular, we deduce that \(g(x)<0 \Leftrightarrow x \notin [-2,2]\).

How does \(f\) relate to \(g\)? The answer is \(f=|g|\). This means nothing else than the following: \(f(x)=g(x) \Leftrightarrow g(x)\geq 0\) and \(f(x)=-g(x) \Leftrightarrow g(x)\leq 0\). Therefore, the graph of \(f\) is identical to the graph of \(g\) whenever \(g(x)\geq 0\), that is to say when \(x\in [-2,2]\), and it is a reflection of the graph of \(g\) about the \(x\)-axis whenever \(g(x)\leq 0\), that is when \(x\notin [-2,2]\).

Does this help? Let me know about anything that is unclear.

adrian_radillo — 2016-09-05 22:59:01 UTC

adrian_radillo — 2016-09-05 22:59:23 UTC

Question

2016-09-05 23:37:57 UTC

Answer

adrian_radillo — 2016-09-06 12:25:09 UTC

As usual, transform the expression until you get the ratio \(\frac{\sin(ax)}{ax}\) to appear. Recall that \(\tan^2(9x)=\sin^2(9x)/\cos^2(9x)\). Hence,
\begin{align}
\frac{2x^2}{\tan^2(9x)}=\frac{2x^2\cos^2(9x)}{\sin^2(9x)}=\frac{2}{9}\cos^2(9x)\frac{9x^2}{\sin^2(9x)}
\end{align}
At this point, use the fact that,\begin{equation}\lim_{x\to 0}\frac{(ax)^2}{\sin^2(ax)}=1, \end{equation} to get: \begin{equation} \lim_{x\to 0} \frac{2x^2}{\tan^2(9x)}=\frac{2}{9}\end{equation}

Isn't this impossible? The directions have no indication that you wouldnt be able to do it.

2016-09-06 22:50:56 UTC

Answer

adrian_radillo — 2016-09-07 02:08:39 UTC

You are right in saying that very little information is given. Here are hints for answering this question:

What is the domain of definition of the function \(f\)?
You should find out that it makes sense to study at least the limit from the left or the limit from the right at \(x=5\).
The function will be continuous at \(x=5\) if and only if \(f(5)\) equals the one-sided limit mentioned in the previous bullet point.
So try to rationalize in some way in order to become able to compute the one-sided limit at \(x=5\), and then set \(f(5)\) to this value.

Let me know if it is unclear.

Trig Limit Question

2016-09-07 05:23:32 UTC

Not sure what to do after converting the tan and sec.

Answer

adrian_radillo — 2016-09-07 13:15:22 UTC

Remember, you have to use the known limits,
\begin{equation}
\lim_{x\to 0}\frac{\sin(ax)}{ax}=\lim_{x\to 0}\frac{ax}{\sin(ax)}=1
\end{equation}
Can you make those terms appear? Here we have,
\begin{align}
\frac{\tan(4t) \sec(5t)}{t}&=\frac{\sin(4t)}{t\cos(4t)\cos(5t)}\\
&=\frac{4}{\cos(4t)\cos(5t)}\frac{\sin(4t)}{4t}
\end{align}
Can you find the limit now? You should find \(4\).

Model answer to quiz 1

adrian_radillo — 2016-09-07 14:08:57 UTC

Exercises for week 3

adrian_radillo — 2016-09-07 14:09:43 UTC

Make sure you can do all of them

Question on solving an inequality

2016-09-07 22:20:06 UTC

This type of question is on the online quiz that's due Friday and it seems really simple but I keep getting it wrong

Answer

adrian_radillo — 2016-09-08 00:41:37 UTC

Have a look at my answer named "model answer for question 1c" in the section "Lab exercises week 1" in this padlet, at the top right from here...
Let me know if you still have doubts after this.
Note that there may be mistakes on the online quizzes.

Find f'(x) using the difference quotient

2016-09-09 20:26:54 UTC

Answer

adrian_radillo — 2016-09-10 02:21:52 UTC

Just as a precaution, I will solve the question with different numbers.
Say \(f(x)=2\sqrt{x+6}\), then we start by recalling the difference quotient definition of the derivative:
\begin{equation}
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
\end{equation}
With the given formula for \(f\) we rewrite and transform this difference quotient as follows (assuming that \(h\neq 0\) for now):
\begin{align}
\frac{f(x+h)-f(x)}{h}&=\frac{2\sqrt{x+h+6}-2\sqrt{x+6}}{h}\\
&=\frac{2}{h}\frac{x+h+6-x-6}{\sqrt{x+h+6}+\sqrt{x+6}}\quad \text{(I rationalized)}\\
&=\frac{h}{h}\frac{2}{\sqrt{x+h+6}+\sqrt{x+6}}\\
&=\frac{2}{\sqrt{x+h+6}+\sqrt{x+6}}
\end{align}
We find the derivative by taking the limit of the last expression as \(h \to 0\),
\begin{equation}
f'(x)=\lim_{h\to 0}\frac{2}{\sqrt{x+h+6}+\sqrt{x+6}}=\frac{1}{\sqrt{x+6}}
\end{equation}

Let me know if things remain unclear. MAKE SURE YOU UNDERSTAND EVERY STEP.

Question on Derivates

2016-09-10 14:05:40 UTC

I'm not sure if we went over this, but this question asks to: Solve for \(f '(x)>0\).

Answer

adrian_radillo — 2016-09-10 19:47:57 UTC

I will solve it with modified values: \(f(x)=3x^2-x+4\).
The function \(f\) is a polynomial and you have to find its derivative. It is,
\(f'(x)=8x-2\), namely a line. When is this line above the \(y\)-axis? The answer is, if and only if, \(x>\frac{1}{4}\).

Ask me if anything remains unclear.

Model answers to graded problems in written homework 2

adrian_radillo — 2016-09-10 19:52:24 UTC

Some feedback:

Use a ruler to draw the axes of your sketches
Computer-produced graphics are not allowed in questions about sketching graphs.
Do not write \(3\cdot -5\), either write \(3\cdot (-5)\) or \(3\times (-5)\), always use parentheses in these cases.
The following two expressions are not equal, so DO NOT FORGET YOUR PARENTHESES:\begin{equation}1-\cos(x)\cos^2(x)\neq (1-\cos(x))\cos^2(x).\end{equation}

Model answer to quiz 2

adrian_radillo — 2016-09-10 20:01:15 UTC

Some feedback on this quiz:

Make sure you do not make mistakes in the following simplification:\begin{equation}\frac{2xh+h^2-2h}{h}=\frac{h(2x+h-2)}{h}=2x+h-2,\qquad \text{provided }h\neq 0.\end{equation}
Make sure you know the following identity by heart, especially don't miss the 2 in your expansion:\((a+b)^2=a^2+2ab+b^2\).
Make sure you know how to distribute a minus sign and don't make mistakes like this anymore. The following is correct:\(-\left(x^2-2x\right)=-x^2+2x\).

Question on Derivates

2016-09-10 20:32:33 UTC

The process seems fairly simple for quadratic functions, but how about functions with degrees greater than two? Most likely there will be more than one answer so how do I determine which answer is the most correct? The question asks to: Solve \(f '(x)<0\)

Answer

adrian_radillo — 2016-09-10 22:07:29 UTC

Ok, again, I will solve it for modified values:
\(f(x)=2x^5-x^3+2x-3\).

Taking the derivative we get: \(f'(x)=10x^4-3x^2+2\). Note that the powers of \(x\) in the question were no coincidence. Whenever all the powers are even, like for \(f'\) here, make a substitution \(\alpha=x^2\). That is, rewrite the degree \(4\) polynomial as the following quadratic: \(10\alpha^2-3\alpha+2\).
Now, the question is reduced to studying the sign of this quadratic on the domain \(\alpha \geq 0\) (this is because \(\alpha=x^2\geq 0, \quad \forall x\in \mathbb{R}\)). The discriminant \(b^2-4ac=-71<0\) is negative (check it). The quadratic is therefore always positive for \(\alpha \geq 0\). This implies that \(f'(x)>0\) for every value of \(x\). Hence, the inequality \(f'(x)<0\) has no solution.

Make sure you understand everything, and ask if it is not clear.

Trig Identity

2016-09-10 23:51:48 UTC

Hey Adrian, I was stuck on how this identity was derived. I know it's a rule but I was wondering how it came to be. It's the one where cos(x+h) = cosxcosh - sinxsinh. Can you please help me out? Thank you

Answer

adrian_radillo — 2016-09-11 00:34:45 UTC

You might find the following Wikipedia article useful.
The easiest derivation that requires complex numbers is the following: For all \(x\in \mathbb{R}\),
\(e^{ix}=\cos(x)+i\sin(x)\). Hence, \begin{align}\cos(x+y)+i\sin(x+y)&=e^{i(x+y)}=e^{ix}e^{iy}\\
&=(\cos(x)+i\sin(x))(\cos(y)+i\sin(y))\\
&=\cos(x)\cos(y)-\sin(x)\sin(y)+i(\cos(x)\sin(y)+\cos(y)\sin(x))\end{align}

But this is definitely more advanced than Cal I.

Help with Derivatives

2016-09-11 16:48:27 UTC

Im having trouble with this problem

Answer

adrian_radillo — 2016-09-11 17:49:17 UTC

I will solve the question, with the modified values:\begin{equation}f(x)=\left\{\begin{array}{cr}-3-x^3,& x\leq -1\\
Bx-C,& x>-1\end{array}\right.
\end{equation}
The first thing to notice is that \(f\) is piecewise defined as polynomials on \((-\infty,-1]\) and \((-1,\infty)\) respectively. It implies that \(f\) is, with whatever fixed values of \(B\) and \(C\), continuous and differentiable on \((-\infty,-1)\) and \((-1,\infty)\). We therefore only have to ensure continuity and differentiability at \(-1\), and this will be by finding the right values of \(B\) and \(C\). To find these values, we need to derive equations involving \(B\) and \(C\) and then solve these equations for \(B\) and \(C\). I derive the equations below. One comes from the continuity condition, the other one from the differentiability condition.

Continuity
We first need to ensure continuity of \(f\) at \(x=-1\), that is to say, the following equations must be true:
\begin{equation}
\lim_{x\to -1^-}f(x)=f(-1)=\lim_{x\to -1^+}f(x).
\end{equation}
Let's compute the three terms above:
\begin{align}
\lim_{x\to -1^-}f(x)&=-3-(-1)^3=-2 \qquad \text{(make sure you can do this)}\\
\lim_{x\to -1^+}f(x)&=B(-1)-C=-B-C\\
f(-1)&=-3-(-1)^3=-2.
\end{align}
We can now rewrite the continuity equation with specific values:
\begin{equation}
-B-C=-2. \qquad \text{(continuity equation)}
\end{equation}
Differentiability
To ensure differentiability of \(f\) at \(x=-1\), the following equations must hold:
\begin{equation}
\lim_{x\to -1^-}f'(x)=\lim_{x\to -1^+}f'(x).
\end{equation}
On \((-\infty,-1)\), \(f'(x)=-3x^2\) and on \((-1,\infty)\), \(f'(x)=B\) (again, make sure you know how to find this). We may hence compute the one-sided limits for the derivative of \(f\):
\begin{align}
\lim_{x\to -1^-}f'(x)&=-3(-1)^2=-3 \\
\lim_{x\to -1^+}f(x)&=B.
\end{align}
We can now rewrite the differentiability equation with specific values:
\begin{equation}
B=-3. \qquad \text{(differentiability equation)}
\end{equation}
Concluding
The differentiability equation gave us a value for \(B\). With the continuity equation we may hence find the value for \(C\):
\begin{equation}
-(-3)-C=-2 \Rightarrow C=5.
\end{equation}
The required values for \(B\) and \(C\) are \(B=-3, C=5\).

Ask about anything that is unclear.

Consider the function \(f\), find the \(x\)-value(s) where the tangent line is horizontal. Find the equation of each horizontal tangent line.

2016-09-11 19:46:08 UTC

\(f(x)=x^2-(16/x)\)

solve for \(x\) when \(f'(x) <0\)

2016-09-11 20:14:36 UTC

Question on difference quotient

2016-09-11 21:53:16 UTC

On the problem \(3\sqrt{x+5}\), I end up getting \(9\) for \(f'(x)\), but I don't think the answer is that simple. Did I do something wrong with the multiplication?

Answer

adrian_radillo — 2016-09-11 21:54:06 UTC

I will solve it with the modified values: \(f(x)=6x^6-10x^4+3\).
The derivative is \(f'(x)=36x^5-40x^3\).
We factor it: \(f'(x)=4x^3(9x^2-10)\). Now, as always, in order to study the sign of a product, study first the sign of its factors.
Here, \(4x^3\) is negative if and only if \(x<0\) and positive if and only if \(x>0\).
What about \(9x^2-10\)? This a quadratic; in this present example, it is parabola with a 'cup' form and two real roots, so it is negative strictly between its roots and positive strictly outside of them. The roots are \(\pm\sqrt{10/9}\), so the quadratic \(9x^2-10\) is negative if and only if \(-\sqrt{10/9}We now combine our information, the product of two factors is negative if and only if the factors are of opposite sign, hence, \(f'(x)<0\) if and only if \begin{equation}
x\in (-\infty,-\sqrt{10/9}) \cup (0,\sqrt{10/9})
\end{equation}

Answer

adrian_radillo — 2016-09-11 22:11:34 UTC

I will solve it with modified values: \(f(x)=-x^9+3x^7-x^5+2x+5\); \(c=1\). The derivative is: \(f'(x)=-9x^8+21x^6-5x^4+2\).
Slope: you first need to provide the slope of the tangent. It is nothing else than the value of the derivative evaluated at \(c\). With my example the slope is \(f'(c)=f'(1)=-9+21-5+2=9\).
Equation of tangent line: now you only have to find the equation of the tangent. There are several ways of doing it. Here is one way. The equation will always be of the form \(y=ax+b\). Since we know the slope, \(a=9\), we only have to find \(b\). To do this, we use the fact that the tangent line passes through the point \((c,f(c))\) (see the question). We hence rewrite the equation of the tangent line, but instead of \(y\) we write \(f(c)=f(1)\) (and compute it), and instead of \(x\), we write \(c=1\). Let's first quickly compute \(f(c)=f(1)=-1+3-1+2+5=8\). The equation of the tangent line evaluated at \((1,f(1))\) now reads: \(8=9\cdot 1+b\). We solve for \(b\) and get \(b=-1\). We now only have to rewrite the equation of the tangent in its general form with the values of \(a\) and \(b\) that we found:
\begin{equation}
y=9x-1.
\end{equation}

Question

adrian_radillo — 2016-09-11 23:00:54 UTC

Given \(f\) and \(c\in \mathbb{R}\), find the slope of the tangent to the graph of \(f\) at the point \((c,f(c))\) and then find the equation of the mentioned tangent.

Answer

adrian_radillo — 2016-09-11 23:05:35 UTC

I will solve the problem for the modified function: \(f(x)=-2x^2+9/x\).

1. Find the derivative
Here it is: \(f'(x)=-3x-\frac{9}{x^2}\).

2. Solve the equation \(f'(x)=0\) (because this is what it means to say that the tangent is horizontal):
The derivative is not defined at \(x=0\). When \(x\neq 0\) we have:
\begin{align}
&-3x-\frac{9}{x^2}=0& \\
\Rightarrow & -3x^3-9=0& \\
\Rightarrow & x^3=-3&\\
\Rightarrow & x=-3^\frac{1}{3}&
\end{align}

3. The previous equation only had 1 solution, so there will only be one horizontal tangent line to the graph of \(f\). This tangent must have an equation of the form \(y=c\), where \(c\) is a constant number. Since the tangent passes through the point \(\left(-3^\frac{1}{3},f\left(-3^\frac{1}{3}\right)\right)\), we must have \(c=f\left(-3^\frac{1}{3}\right)\). In your exercise, compute explicitly the right-hand side of this last equation.

Answer

adrian_radillo — 2016-09-11 23:20:05 UTC

Your result should still depend on \(x\), try to redo my answer below on your own to make sure that you can do every step correctly.

Need Help

2016-09-13 04:21:03 UTC

Given \(f(x)=5cos(x)\) and \(c=7π/6\), determine whether \(\lim_{h\to 0}\frac{f(c+h)−f(c)}{h}\)exists, and give the value if the limit does exist.

Answer

adrian_radillo — 2016-09-13 15:00:23 UTC

I will answer with modified values. Say \(f=-2\cos(x)\) and \(c=\pi/3\), then the first thing to do is to compute the difference quotient:
\begin{align}
\frac{f(c+h)-f(c)}{h}&=\frac{-2\left(\cos\left(\frac{\pi}{3}+h\right)-\cos\left(\frac{\pi}{3}\right)\right)}{h}\\
&=-\frac{2}{h}\left(\cos\left(\frac{\pi}{3}\right)\cos(h)-\sin\left(\frac{\pi}{3}\right)\sin(h)-\cos\left(\frac{\pi}{3}\right)\right)\\
&=-\frac{2}{h}\left(\frac{1}{2}(\cos(h)-1)-\frac{\sqrt{3}}{2}\sin(h)\right)\\
&=\frac{1-\cos(h)}{h}+\sqrt{3}\frac{\sin(h)}{h}
\end{align}
We deduce that the limit is \(\sqrt{3}\). Does this make sense to you?

Question

2016-09-13 20:14:02 UTC

If \(f(1)=2\) and \(f'(1)=5\), what is the equation of the line that is tangent to f(x) at x=1?

Question on Trig. Limits

2016-09-13 21:12:59 UTC

How do I go about solving this problem?

Answer

adrian_radillo — 2016-09-14 02:05:52 UTC

The following is very important:
If you know that a line goes through the point \((a,b)\) and has slope \(m\), are you able to write down the equation of the line?
You must learn the answer. The answer is yes and the equation is:\begin{equation}\frac{y-b}{x-a}=m.\end{equation} You may rearrange the previous equation in standard form (\(y=\alpha x+\beta\)) if you like.

Do you see how this applies to your question? The point is \((1,2)\) and the slope is \(5\).

Answer

adrian_radillo — 2016-09-14 02:06:07 UTC

I will solve the question for modified values: \begin{equation}\lim_{x\to 0}\frac{\cos(4x)-1}{7x}.\end{equation} The term \(\cos(4x)-1\) on top can be re-written \(-(1-\cos(4x))\). Hence, \begin{align}\frac{\cos(4x)-1}{7x}&=-\frac{1-\cos(4x)}{7x}.\end{align}
Are you with me so far? Then a last transformation into, \begin{equation}-\frac{4}{7}\cdot\frac{1-\cos(4x)}{4x},\end{equation} allows us to use the known limit \(\lim_{x\to 0 }\frac{1-\cos(ax)}{ax}=0\) to deduce that, the limit of your initial expression is \(0\).

question

2016-09-14 23:15:30 UTC

If \(f(x)=\cos^2(2x)\), find \(f'(\pi/3)\)

Answer

adrian_radillo — 2016-09-15 00:06:22 UTC

I will modify the question to the following: if \(f(x)=\sin^2(-x)\), find \(f'(\pi/6)\).
To answer, you have to take two steps:
1. First compute the derivative of \(f\). With my example, let's call \(u:=\sin(-x)\). Then \(f'=\frac{d}{dx}\left(u^2\right)\). Recall that, since \(u\) is a function of \(x\), you want to use the rule \(\frac{d}{dx}\left(u^n\right)=nu'u^{n-1}\). So here, since by the chain rule \(u'=-\cos(-x)\), we conclude that \begin{equation}f'(x)=-2\cos(-x)\sin(-x).\end{equation}

If you want, taking into account the fact that \(\cos\) is an even function and \(\sin\) is odd, you may further simplify: \(f'(x)=2\sin(x)\cos(x)=\sin(2x)\).

2. The second step is to substitute \(x\) for \(\pi/6\) to get:\(f'(\pi/6)=\sin(\pi/3)=\sqrt{3}/2\).

Ask about anything that does not make sense to you.

Classwork from 14 Sept 2016

adrian_radillo — 2016-09-15 00:22:19 UTC

I'll try to upload solutions to these asap.

Homework Question

2016-09-15 17:03:32 UTC

I think that first you take the derivative of f(x), which is f'(x) = 2x, but I'm not sure what the next step is

Answer

adrian_radillo — 2016-09-15 18:46:09 UTC

I will solve the question for the modified function \(f (x)=-2x^2+7\) and line \(y=6x+1\).
First compute \(f'(x)=-4x\). Then, since we want a tangent parallel to the given line, we must have \(f'(x)=6\) (because the slopes of 2 parallel lines are always equal).
We then solve for \(x\) the last equation that yields: \(x=-3/2\). The last step is to compute \(f (-3/2)=5/2\).
The final answer is that there is only one point meeting the condition:\((-3/2,5/2)\).

Rules to know for differentiation

adrian_radillo — 2016-09-15 18:56:12 UTC

You must also know \(\cos' (x)=-\sin (x)\), \(\sin'(x)=\cos (x)\) and \(\tan'(x)=\sec^2 (x)\).

find f'(x)

2016-09-15 19:33:03 UTC

Quiz

2016-09-15 23:20:40 UTC

What will the quiz tomorrow be covering?

Answer

adrian_radillo — 2016-09-16 11:04:56 UTC

Today's quiz will be on sections 2.3 and 2.4 of the textbook.

Answer

adrian_radillo — 2016-09-16 11:05:26 UTC

I will modify the question, as usual,
\(f(x)=\cos\left(\frac{-x}{3x-2}\right)\).
We need chain rule and quotient rule.
Let's work on this piece by piece.
The quotient rule gives,
\begin{align}
\frac{d}{dx}\left(\frac{-x}{3x-2}\right)&=\frac{-1\cdot(3x-2)+x\cdot 3}{(3x-2)^2}=\frac{2}{(3x-2)^2}
\end{align}
Now the chain rule gives,
\begin{align}
\frac{d}{dx}\left(\cos\left(\frac{-x}{3x-2}\right)\right)&=-\frac{2}{(3x-2)^2}\sin\left(\frac{-x}{3x-2}\right)
\end{align}
At this point we are done.

Solution to classwork 3

adrian_radillo — 2016-09-16 14:19:53 UTC

Review session for test 2 on Monday 19 Sept. in room ALC, Cougar Place 1020

adrian_radillo — 2016-09-16 16:42:14 UTC

find the equations of the tangent and normal lines to the curve f at the given point

2016-09-16 23:06:37 UTC

Test coming up is:

adrian_radillo — 2016-09-17 18:13:58 UTC

50min
part mc, part fr
mc---49 points total
fr--51 points total

Answer

adrian_radillo — 2016-09-17 18:14:49 UTC

I will solve the question for \(f(x)=\csc(4x)\) and \(x=\frac{-\pi}{12}\). First, the derivative is \(f'(x)=-4\csc(4x)\cot(4x)=-4\frac{\cos(4x)}{\sin^2(4x)}\). Next, the slope of the tangent line at the given point will be \[
f'\left(\frac{-\pi}{12}\right)=4\cdot \frac{\frac{1}{2}}{\left(\frac{\sqrt{3}}{2}\right)^2}=\frac{8}{3},
\]
we deduce that the slope of the normal line, which is always perpendicular to the tangent, is \(-\frac{3}{8}\).
It now only remains to find the point \((x,f(x))\) through which both lines pass. We have,
\[
f\left(\frac{-\pi}{12}\right)=\frac{1}{\sin\left(\frac{-\pi}{3}\right)}=-\frac{2}{\sqrt{3}}
\]
The equation of the tangent line is therefore:
\[
y=\frac{8}{3}\left(x+\frac{\pi}{12}\right)-\frac{2}{\sqrt{3}}.
\]
And the equation of the normal line is:
\[
y=-\frac{3}{8}\left(x+\frac{\pi}{12}\right)-\frac{2}{\sqrt{3}}.
\]
It would be good to rearrange these equations in standard form in an exam.

Model Answer Quiz from Friday 16 Sep 201

adrian_radillo — 2016-09-17 19:42:21 UTC

Classwork 4

adrian_radillo — 2016-09-17 19:44:42 UTC

Game for derivatives of trig functions

adrian_radillo — 2016-09-17 19:46:41 UTC

adrian_radillo — 2016-09-18 00:46:16 UTC

HW question

2016-09-18 01:21:03 UTC

I am not sure what to do after finding the derivative of f(x). What would be the next step?

find the tangent line to the graph of the given equation at the indicated point

2016-09-18 18:55:13 UTC

Find \(f'(x)\).

2016-09-18 22:43:20 UTC

\(f(x)=tan^2(x^4+1)\)

Find \(d^2y/dx^2\).

2016-09-19 00:44:10 UTC

\(x^2y^2-2x=3\)

adrian_radillo — 2016-09-19 14:20:25 UTC

Answer

adrian_radillo — 2016-09-19 14:24:09 UTC

I will answer the three questions below in class

Practice Test 2 Question

2016-09-19 22:46:34 UTC

How do I exactly solve this problem?

Answer

adrian_radillo — 2016-09-21 16:31:26 UTC

The numerator may be rewritten:
\(\frac{\cos^2(5x)-1}{\cos^2(5x)}=\frac{-\sin^2(5x)}{\cos^2(5x)}\),
so the whole expression reads:
\(-\left(\frac{\sin(5x)}{3x}\right)^2\cdot \frac{1}{\cos^2(5x)}\),
and using the known trig limits we get,
\(\lim_{x\to 0}\frac{1-\sec^2(5x)}{(3x)^2}=-\frac{25}{9}\).

Answer to one of the previous questions below

adrian_radillo — 2016-09-21 16:46:52 UTC

The calculations are very lengthy, you will need to derive twice with respect to \(x\). The first derivation yields,
\begin{align}
2xy^2+2yy'x^2-2&=0\\
\Rightarrow xy^2+y'yx^2&=1\\
\Rightarrow y'&=\frac{1-xy^2}{x^2y}\\
\Rightarrow y'&=\frac{1}{x^2y}-\frac{y}{x}
\end{align}
Now we derive a second time with respect to \(x\) (my calculations might contain some typos):
\begin{align}
\frac{d^2y}{dx}&=\frac{d}{dx}(y')\\
&=-\frac{2xy+x^2y'}{(x^2y)^2}-\frac{y'x-y}{x^2}\\
&=-\left[\frac{2}{x^3y}+\frac{y'}{x^2y^2}+\frac{y'}{x}-\frac{y}{x^2} \right]\\
&=-\left[\frac{2}{x^3y}+\frac{1}{x^4y^3}-\frac{y}{x^3y^2}+\frac{1}{x^3y}-\frac{y}{x^2} -\frac{y}{x^2}\right]\\
&=-\frac{2}{x^3y}-\frac{1}{x^4y^3}+\frac{2y}{x^2}
\end{align}

Answer to another one of the questions below

adrian_radillo — 2016-09-21 17:52:20 UTC

Given \(f(x)=\tan^2(x^4+1)\), we use the chain rule to find the derivative, with:
\begin{align}
&u(x)=\tan^2(x)\qquad u'(x)=2\sec^2(x)\tan(x)\\
&v(x)=x^4+1\qquad v'(x)=4x^3\\
&f(x)=(u\circ v)(x) \qquad f'(x)=v'(x)\cdot (u'\circ v)(x)\\
&f'(x)=8x^3\sec^2\left(x^4+1\right) \tan\left(x^4+1\right)
\end{align}

Answer to still another one of the questions below

adrian_radillo — 2016-09-21 18:01:47 UTC

If \(f(x)=(1+x^2)^{-2}\), then you should find that,
\begin{equation}
f'(x)=\frac{-4x}{(1+x^2)^{3}}.
\end{equation}
Note that it is a quotient of the form \(a/b\), with \(b\) always positive. So the sign of this derivative is the same as the sign of \(-x\), that it is, it is \(0\) when \(x=0\), it is positive when \(x<0\) and it is negative when \(x>0\)

Question on Derivates

2016-09-23 18:34:43 UTC

How would I find \(\frac{dy}{dx}\) in this equation?

\(y=x-\frac{3}{x}\)

Anser

adrian_radillo — 2016-09-24 00:43:46 UTC

I will solve the modified problem: \(y=2x+\frac{7}{x}\).
\(y\) is a sum of two terms, so \(dy/dx\) is the sum of the derivatives of each term. The first term has derivative \(2\) and the second term has derivative \(-\frac{7}{x^2}\), so \(\frac{dy}{dx}=2-\frac{7}{x^2}\).

Answer to the first teaser question

adrian_radillo — 2016-09-24 00:46:14 UTC

The question was to find \(\frac{dy}{dx}\) given that \begin{equation}y=\sqrt{x+\sqrt{x+\sqrt{x}}}.\end{equation}
To answer, I chose to break up the problem in parts. Let \(u(x)=\sqrt{x}\), \(v(x)=x+\sqrt{x}\) and \(w(x)=(u\circ v)(x) =\sqrt{x+\sqrt{x}}\). Then, we have \(y=u(x+w(x))\) and by the chain rule \begin{equation}\frac{dy}{dx}=(1+w'(x))\cdot u'(x+w(x)).\end{equation}
Let's find the terms involved.
First, \(u'(x)=\frac{1}{2\sqrt{x}}\) and \(v'(x)=1+u'(x)\), so that by the chain rule,
\begin{equation}
w'(x)=v'(x)\cdot (u'\circ v)(x)=\left(1+\frac{1}{2\sqrt{x}} \right)\cdot \frac{1}{2 w(x)}.
\end{equation}
If you want, you may rearrange this into,
\begin{equation}w'(x)=\frac{2\sqrt{x}+1}{4\sqrt{x^2+x\sqrt{x}}},\end{equation}
or into,
\begin{equation}
w'(x)=\frac{2x^\frac{1}{2}+1}{4\left(x^2+x^\frac{3}{2}\right)^\frac{1}{2}}.
\end{equation}
We can now come back to \(dy/dx\):
\begin{align}
\frac{dy}{dx}=\left(1+\left(1+\frac{1}{2\sqrt{x}} \right)\cdot \frac{1}{2 w(x)}\right)\cdot \frac{1}{2\ y}
\end{align}

Unfortunately, the expression doesn't get much nicer than this. In math, sometimes, one has to accept that formulae are ugly and remain like this. The only way to keep the formulae readable is to rename variables to replace big expressions by single letters, as I did above with \(w(x)\) and \(y\).

Answer to the second teaser question

adrian_radillo — 2016-09-24 01:08:17 UTC

The question was, "How many tangent lines to the curve \(y=x/(x+1)\) pass through the point \((1,2)\)? At which points do these tangent lines touch the curve?"

This question is designed to test two things: 1. your ability to think about the geometry of the graph of a function and 2. your ability to translate a geometrical problem into an algebraic one, i.e., an equation to solve.

Here is one approach to the solution. The first thing that I strongly encourage you to do is to try to sketch the curve, without any computer or calculator. Here are a few hints about the graph:
1. The limit of \(y\) as \(x\) tends to both \(+\infty\) and \(-\infty\) is \(1\), therefore the graph has a horizontal asymptote at both ends of the \(x\)-axis.
2. The function \(f(x)=x/(x+1)\) is a rational function with denominator that vanishes at \(-1\), the graph will therefore have a vertical asymptote at \(x=-1\).
3. Then, analyze the sign of \(y\) for values of \(x\) close to \(-1\). When \(x<-1\), both numerator and denominator of \(f(x)\) are negative, so \(f(x)\) is positive and this means that \(\lim_{x\to -1^-}f(x)=+\infty\). On the other hand, for values of \(x\) close to \(-1\) but satisfying \(x>-1\), the denominator will become positive while the numerator will remain negative, hence \(\lim_{x\to -1^+}f(x)=-\infty\).

Can you sketch the curve now?

Do you understand geometrically what the problem is asking?

From your sketch, can you get an idea of the answer?

Now, here is how to solve the question formally. The derivative of \(f\)
is \(f'(x)=\frac{1}{(1+x)^2}\) (quotient rule). We will now consider the family of tangent lines to the graph of \(f\). To each point \((\alpha,\beta)\) of the graph of \(f\) corresponds a tangent to this graph (at that point). Call this tangent line \(L_\alpha\). Can you find the equation of \(L_\alpha\)?

The line \(L_\alpha\) has slope \(f'(\alpha)\) and passes through the point \(\alpha,\beta\), by construction. Therefore the line has equation:
\begin{equation}
y-\beta=f'(\alpha)(x-\alpha).
\end{equation}
Note that, by assumption, \(\alpha,\beta\) is on the graph of \(f\), which precisely means that \(\beta=f(\alpha)=\alpha/(1+\alpha)\). Therefore, using all the information listed so far, the equation of the line \(L_\alpha\) now becomes,
\begin{equation}
y-\frac{\alpha}{1+\alpha}=\frac{1}{(1+\alpha)^2}(x-\alpha).
\end{equation}
Are you with me so far? To summarize, for each point \((\alpha,\beta)\) lying on the graph of \(f\) we can express the equation of the tangent line to the graph through that point as:
\begin{equation}
y=\frac{x-\alpha}{(1+\alpha)^2}+\frac{\alpha}{1+\alpha}.
\end{equation}

Now the problem may be rephrased as follows: For how many values of \(\alpha\) does the line \(L_\alpha\) pass through the point \((1,2)\)?

Well, the point \((1,2)\) belongs to \(L_\alpha\) if and only if the coordinates of the point satisfy the equation of the line, that is, if and only if the following equation is true:
\begin{equation}
2=\frac{1-\alpha}{(1+\alpha)^2}+\frac{\alpha}{1+\alpha}.
\end{equation}

So all that is left to do is to solve the previous equation for \(\alpha\).
Recall that we must have \(\alpha \neq -1\), so we may multiply left and right-hand sides of the previous equation by \((1+\alpha)^2\) to get,
\begin{align}
2(1+\alpha)^2&=1-\alpha+\alpha(1+\alpha)\\
\Leftrightarrow 2+2\alpha^2+4\alpha&=1+\alpha^2
\Leftrightarrow \alpha^2+4\alpha+1&=0.
\end{align}
The previous quadratic has two real roots:
\begin{equation}\alpha=\frac{-4\pm\sqrt{16-4}}{2}=-2\pm \sqrt{3}.\end{equation}

So the only two points on the graph of \(f\) at which the tangent line passes through \((1,2)\) are \(\left((-2+\sqrt{3},f\left(-2+\sqrt{3}\right)\right)\) and \(\left(-2-\sqrt{3},f\left(-2-\sqrt{3}\right)\right)\).

2016-09-24 02:14:46 UTC

Hey Adrian,
In our written homework, the problem specifies "Water is leaking from the bottom of a canonical cup that is 6 inches across and 8 inches deep." What in the world is a canonical cup? Is it a cone? A cylinder? Something more complex?

Model answers to whw4

adrian_radillo — 2016-09-24 18:05:29 UTC

About the canonical cup

adrian_radillo — 2016-09-28 12:17:59 UTC

I spoke with the instructors and the unfortunate expression "canonical cup" will be changed to "conical cup" in the textbook, as we discussed in class. Also, the rate at which water is leaking should be specified in volume units per seconds, rather than in height units per seconds.

Question on Related Rates

2016-10-01 22:07:26 UTC

How would I solve this problem?

Answer

adrian_radillo — 2016-10-02 21:18:28 UTC

The free fall equation, with \(v_0\) given, is:
\begin{equation}
h(t)=-16t^2+4t+h_0.
\end{equation}
We are told that at a specific time \(t^*\) the height is \(h(t^*)=90\) and 2 seconds later, the height is \(h(t^*+2)=10\). You are asked to find \(h_0\).
Here is how I do it. First find \(t^*\). In order to do this, we use the following equation (do you understand why it is true?)
\begin{equation}
\frac{h(t^*+2)-h(t^*)}{2}=\frac{10-90}{2}=-40.
\end{equation}
The numerator of the left-hand side simplifies to
\begin{equation}
h(t^*+2)-h(t^*)=-16(4+4t^*-1/2).
\end{equation}
So we want to solve the following equation for \(t^*\):
\begin{equation}
\frac{-16(4+4t^*-1/2)}{2}=-40.
\end{equation}
The solution is \(t^*=3/8\) seconds.

The next step is to find \(h_0\) using the height equation:
\begin{equation}
h(t^*)=90=-16(t^*)^2+4t^*+h_0.
\end{equation}
Solving for \(h_0\) yields \(h_0=363/4\) feet.

Does it make sense?

Model answer to whw 5

adrian_radillo — 2016-10-02 21:26:35 UTC

I'm reviewing for the test by doing some odd problems from the textbook, could you explain how to solve this related rates problem?

2016-10-06 17:53:37 UTC

Answer

adrian_radillo — 2016-10-06 23:42:37 UTC

Differentiate the equation for the area, with respect to \(r\), on both sides to get:
\begin{equation}
\frac{dA}{dr}=\frac{d}{dr}\left( \frac{1}{2}r^2\theta\right).
\end{equation}
Since the area is constant, its derivative is zero. Using the product rule on the right-hand side we get:
\begin{equation}
0=\frac{1}{2}\left(2r\theta + r^2\frac{d\theta}{dr}\right).
\end{equation}
Rearranging to get the change rate for \(\theta\) on one side we get,
\begin{equation}
\frac{d\theta}{dr}=\frac{-2\theta}{r}.
\end{equation}
Now, we notice that \(\theta\) itself may be expressed in terms of \(A\) when \(r=5\), using the equation given in the question. That is, we find,
\(\theta=2A/25\). Using this fact, together with \(r=5\), we find,
\begin{equation}
\frac{d\theta}{dr}=-\frac{2A}{125}.
\end{equation}
We can't go much further, unless you know how to solve first order linear differential equations...
Let me know if this does not make sense.

Solving a question about extreme values

adrian_radillo — 2016-10-06 23:55:31 UTC

QUESTION: Find any extreme values for \(f(x)=x/(1+x^2)\) and label them as local maxima or minima.

ANSWER: We first compute the derivative of \(f\) (using the quotient rule), which is defined everywhere since the denominator of \(f\) never vanishes:
\begin{equation}
f'(x)=\frac{1+x^2-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}.
\end{equation}
Remember that for extreme values, the signs of the first derivative is what matters. Given the formula for \(f'\) above, we notice that \(f'(x)\) has always the same sign as its numerator (because the denominator is always positive). The numerator of \(f'\) is a quadratic which is zero at \(-1\) and \(1\), positive on \(-1,1\) and negative otherwise. Hence, the derivative of \(f\) changes sign at \(-1\) from negative to positive, and at \(1\) from positive to negative.
The first derivative test tells us that \(f(-1)=-1/2\) is a local minimum and \(f(1)=1/2\) is a local maximum.

Model answers to whw 6

adrian_radillo — 2016-10-09 23:59:36 UTC

Make sure you know your definitions!

For each function, find the interval(s) on which the graph is concave up (if any), find the interval(s) on which the graph is concave down (if any), find the points of inflection (if any).

2016-10-10 00:30:06 UTC

\(f(x)=2cos^2(x)-x^2\)
I've found the derivative to be \(f'(x)=-2(x+sin2x)\), but am not sure how to proceed when I set it equal to zero.

Question from Practice Test 3

2016-10-12 20:06:14 UTC

I don't think we did a problem like this before so how should I solve this problem?

Related Rates

2016-10-12 21:52:47 UTC

I assume that the pythagorean theorem should be used to relate the variables, but would the rates of change for the cars be negative or positive? if you could solve this it would be much appreciated!

Answers

adrian_radillo — 2016-10-13 02:51:15 UTC

I will try to answer your questions by Friday

Classwork on curve sketching

adrian_radillo — 2016-10-13 02:51:44 UTC

Answer

adrian_radillo — 2016-10-16 16:30:40 UTC

You are right about the formula for the first derivative, but since the question is about concavity, you should compute the second derivative: \(f''(x)=-2(1+2\cos(2x))\), which may be rearranged into, \[f''(x)=-8(\cos x +1/2)(\cos x -1/2).\] Can you analyze the sign of \(f''\) from there and answer the question?

Model Answers to whw 7

adrian_radillo — 2016-10-17 03:55:45 UTC

https://dl.dropboxusercontent.com/u/90951391/key_whw7_AdrianRadillo.pdf

Question

2016-10-18 17:10:02 UTC

Hi Adrian,

this was a question from an online quiz and I just don't get what to do or where to start.

Answer

adrian_radillo — 2016-10-19 02:20:31 UTC

Let's start by deriving g, we get, \(g'=-\frac{\left(f^{-1}\right)'}{\left(f^{-1}\right)^2}\). Plugging in 6 yields:\(g'(6)=-\frac{\left(f^{-1}\right)'(6)}{\left(f^{-1}(6)\right)^2}\). The denominator is \((-7)^2=49\) and the numerator is \(3/4\). So the final answer is \(g'(6)=\frac{3}{4\cdot 49}\)

How to solve for \(y\) in \(y^2=39^2-36^2\) without a calculator

adrian_radillo — 2016-10-19 02:27:38 UTC

Recall my favorite formula: \(a^2-b^2=(a+b)(a-b)\). Hence, \(y^2=(39+36)(39-36)=3\cdot(13+12)\cdot 3=9\cdot 25\). So \(y=\pm 3\cdot 5=\pm 15\).

Question on an EMCF that I need help solving

2016-10-27 16:07:09 UTC

Answer

adrian_radillo — 2016-10-30 00:08:45 UTC

Rewrite: \(f(x)=e^{2x\ln x}\), so that \(f(x)=e^u\). Now recall that \(f'(x)=u'e^u\). We first compute \(u'\). By the product rule we get:
\(u'=2\ln x +\frac{2x}{x}=2(\ln x +1)\). Hence, since \(e^u=f(x)=x^{2x}\) we find,
\begin{equation}
f'(x)=2(\ln x+1)x^{2x}.
\end{equation}
Evaluating the latter at \(x=1\) yields \(f'(1)=2\).

Model answer to written homework 9

adrian_radillo — 2016-10-30 00:15:35 UTC

Section 4.1, Problem 14
The derivative is \(f'(x)=3x^2-12x+12\) and is defined on \(\mathbb{R}\). Its sign is the same as the sign of \(x^2-4x+4=(x-2)^2\), which is positive on \(\mathbb{R}\setminus \{2\}\) and vanishes at \(x=0\). This tells us that \(f\) is strictly increasing on \(\mathbb{R}\) and therefore has an inverse function.

Section 4.1, Problem 20
The function \(f\) is undefined at \(x=0\). For any \(a,b\neq 0\), we have,\begin{align}
f(a)=f(b)&\Rightarrow\frac{a+2}{a}=\frac{b+2}{b}\\
&\Rightarrow a(b+2)=b(a+2)\\
&\Rightarrow ab+2a=ba+2b\\
&\Rightarrow 2a=2b\\
&\Rightarrow a=b.
\end{align}
This tells us that \(f\) is one-to-one.
We can already tell that the domain of \(f^{-1}\) is the range of \(f\), namely \(\mathbb{R}\setminus \{1\}\) (try to graph \(f\); analyzing its asymptotes and derivative will help).

As for the inverse of \(f\), we derive its formula as follows:\begin{align}
x=\frac{y+2}{y} &\Rightarrow xy=y+2\\
&\Rightarrow xy-y=2\\
&\Rightarrow y(x-1)=2\\
&\Rightarrow y=\frac{2}{x-1}.
\end{align}

Section 4.2, Problem 16
We want to find the derivative of \(f(x)=e^{x^2\sin x}\). We use logarithmic differentiation. For this we take the natural log of both sides of the equation of \(f\) which has the effect of annihilating the exponential: \(\ln f(x)=x^2\sin x\). Next, we differentiate both sides with respect to \(x\):
\begin{equation}
\frac{f'(x)}{f(x)}=\frac{d}{dx}\left(x^2\sin x\right)=2x\sin x + x^2\cos x.
\end{equation}
Rearranging yields the result:
\begin{equation}
f'(x)=f(x)\left(2x\sin x + x^2\cos x\right)=\left(2x\sin x + x^2\cos x\right)e^{x^2\sin x}.
\end{equation}

Section 4.2, Problem 20
We want to find the derivative of \(f(x)=\sin\left(e^{5x-1}\right)\). We need the chain rule, and the fact that \(\left(e^u\right)'=u'e^u\).
We get, \begin{equation} f'(x)=5e^{5x-1}\cos\left(e^{5x-1}\right).\end{equation}

Section 4.2, Problem 30
We want to find the slope of the tangent line to the graph of \(f\) at \(c=0\), where \begin{equation}f(x)=\frac{e^{2x}}{1+e^x}\end{equation}.
We use the quotient rule to find the derivative of \(f\):
\begin{align}
f'(x)=\frac{2e^{2x}(1+e^x)-e^{2x}e^x}{\left(1+e^x\right)^2}.
\end{align}
Substituting in \(0\) for \(x\) yields,
\(f'(0)=\frac{2(1+1)-1\cdot 1}{(1+1)^2}=3/4\).

Model Answer to Quiz 9

adrian_radillo — 2016-10-30 01:43:37 UTC

question on homework over optimization

2016-11-03 16:31:01 UTC

how do I solve this?

Answer

adrian_radillo — 2016-11-04 17:22:59 UTC

1. Make a sketch of what this mean. Could you?
2. Label the important variables in your sketch and identify the ones that are allowed to vary in the problem, and the ones that are fixed.
3. One of these variables has to be the area of the rectangle. Let's call it A. A will be a function of some other variables. You want to find the absolute maximum of A. But first find its domain.

Does all this help at all? Really try to do the steps I explained above in that order.

Question on Optimization

2016-11-04 18:47:51 UTC

I'm not sure how to include the costs in finding the dimensions of the cylinder

Simplifying

2016-11-05 04:51:13 UTC

Hello there Adrian, could you detail the process of how the equation on the left simplifies to the equation on the right in the image? Thanks.

Answer

adrian_radillo — 2016-11-05 13:52:53 UTC

Remember that \(a^2-b^2=(a+b)(a-b)\) and that for any expression the following holds:\(u (x)=\sqrt{u (x)}\sqrt{u (x)}\). Also try to compute \(11^2\).

About the cylinder cost function

adrian_radillo — 2016-11-05 14:00:33 UTC

What are the main dimension variables of a cylinder? The dimensions that characterize it uniquely? Well, there is the radius of its base, and its height right? Lets call them \(r\) and \(h\). For each value of these two variables, you will get a different cylinder, which will presumably have a different cost. So one way to start is to set up a cost function of 2 variables for the total cost of your cylinder: \(C(r,h)\). But remember that you know the total capacity of the cylinder. So use the formula for the volume of the cylinder to express \(h\) in term of \(r\). This way, \(C(r,h)\) may be reduced to a function of \(r\) only: \(C(r)\).

Next, you are given information about how the cost should be subdivided. The cost of the whole cylinder will be a sum of the costs of three subparts: bottom base, side, top base (do you picture those parts?). Presumably, the area of the bottom and top parts have nothing to do with the height of the cylinder, so their respective cost functions, call them \(C_{bot}(r)\) and \(C_{top}(r)\) are only functions of \(r\).

I will write \(C_{bot}(r)\) and you deduce the other one on your own.
\(C_{bot}(r)=20\cdot \pi r^2\).

Now, concerning the cost of the side, you will need both \(r\) and \(h\).
As for the bottom, find out the formula for the area of the side in terms of these two variables and multiply it by the corresponding cost for this part of the cylinder. And from what I said above, you should be able to replace \(h\) in the formula by an expression involving \(r\).

So, at this point, you should have \(C(r)=C_{bot}(r)+C_{top}(r)+C_{side}(r)\), and you should have explicit expressions i \(r\) for all terms.

Your final task is to find the local minimum of \(C\) and give the \(r\) and \(h\) values that correspond to this cost.

Does this make sense?

Question about Optimazation

2016-11-06 18:38:18 UTC

I know that this question was asked before, but I just don't know how to use the radius in the semicircle to find the area of the rectangle.

A link for advanced discussion on the power of a number

adrian_radillo — 2016-11-06 20:18:15 UTC

http://mathforum.org/library/drmath/view/76619.html

Answer on rectangle inscribed inside semi-circle

adrian_radillo — 2016-11-06 20:23:30 UTC

Ok, I see that this question is hard. I will give you hints, but not the full answer.
1. Make sure you draw on a sheet of paper a semi-circle (It is easier to draw the top-half of a circle for what comes next), and a rectangle inscribed in it.

Do you understand what inscribed means? Did you notice that your rectangle has to be symmetric about the vertical line (picture it in your head, or even better, draw it as well) that cuts the semicircle in half?

2. Give names to the height and length of your rectangle. I pick a and b.
3. Define an area variable, A. Can you express A as a function of a and b?
4. Now, do you notice that your rectangle can be "controlled" by another variable yet? It is an angle variable. It is the angle formed by the upper-right corner of your rectangle, with the horizontal baseline of the semicircle. Call that angle x for instance.

Can you figure out the allowable domain for x?

5. Your task now is to express a as a function of x, and b as a function of x. Hint: these functions are trig functions...but be careful, your circle is not a unit circle, so there is an additional mutliplicative constant somewhere.
6. If you managed to do all of the above, then you should be able to write A as a function of x only. Then it is just a matter of differentiating A with respect to x...
... Good luck. Think hard.

Model answers to written hw 10

adrian_radillo — 2016-11-07 00:24:24 UTC

Question on Hyperbolic Functions

2016-11-07 02:46:25 UTC

I don't know where to start in solving this problem

Answer

adrian_radillo — 2016-11-08 15:08:00 UTC

Compute \(y'\) and \(y''\) in terms of \(A,B,C,x\).
Then write down the three equations that the question imposes as constraint.
Derive from this what A, B and C have to be, remembering that C>0.

Question about Hyperbolic Functions

2016-11-09 02:15:16 UTC

How do I differentiate this equation?

Answer

adrian_radillo — 2016-11-09 03:26:47 UTC

Use the definition of cosh in terms of the exponential function and remember that the exponential function is the inverse of the natural logarithm. Take the derivative after performing the transformations and simplifications above.

Question on Optimization

2016-11-11 23:08:53 UTC

How do I solve this problem?

Use Differentials to Estimate the Value of the Expression

2016-11-12 03:54:18 UTC

Hello there Adrian, I was just a bit confused on what x-value I'm supposed to use in this differential problem. I'm tempted to use 64 since it's a perfect cube root (no other number is a perfect cube root close to 84) but isn't that too far of a number?

Expression: \(84^(1/3)\)

L' Hospital's Rule/Limits

2016-11-12 05:29:10 UTC

Do the concepts (1/0)=infinity and (1/infinity)=0 only apply to limits as x approaches to 0 or infinity, or does it apply to all limits? For example, in the limit below, would the answer be DNE or infinity since you get (1/0)?

Lim as x--> 1 of (1-lnx)/(x-1)

help simplifying

2016-11-12 20:02:27 UTC

I know how to find the derivative of this, it should be \(3e^{arcsin(3x)}/\sqrt{1-9x^2}\) but once I plug in the 1/6 I can't figure out how to simplify it correctly.

arcsin(1/2)

adrian_radillo — 2016-11-14 01:38:10 UTC

remember that arcsin is the inverse of the sin function on the domain (-pi/2,pi/2)

Model answer for whw 11

adrian_radillo — 2016-11-15 22:19:25 UTC

Answer

adrian_radillo — 2016-11-15 22:20:43 UTC

Remember that \(\ln 1=0\)

Question From Practice Test 4

2016-11-16 02:32:50 UTC

I don't know how to solve this problem completely.

Also, how should I use y'' - 4y = 0 in solving A, B and C?

Question from Practice Test 4

2016-11-17 02:11:08 UTC

I'm not sure how to find A, B, C so can you show the steps to solving problems like these?

Question from Inverses

2016-11-17 02:15:33 UTC

How do I solve this problem?

Question from Practice Test 4

2016-11-18 04:23:04 UTC

I don't know how to solve this problem so can show the steps in solving this problem?

Answer

adrian_radillo — 2016-11-18 18:44:33 UTC

I hope what we did in class today clarified your doubts. If not, ask again. I will try to check this website every day.

Question from Integrals

2016-11-23 02:55:40 UTC

I'm not sure how to solve this problem. Also, what does \(n = 6\) mean?

Question about Integrals

2016-11-23 03:27:37 UTC

How do I solve this type of problem?

Question over 6.1

2016-11-23 03:28:41 UTC

I don't know how to compute the lower Riemann sums for a trig. function.

Question from HW

2016-11-28 13:54:43 UTC

I don't know how to solve the problem.

Last question of the semester?

2016-12-06 03:21:14 UTC

I don't know how to use the picture to solve the problem. Also, will you take down this site once the next semester comes around?

P.S. I will definitely miss this class, it was a lot of fun doing math. The semester went by really fast since I've grown fond of the class. It's unfortunate that all good things must come to an end but why so quickly?