Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
body {
background-color: #eeeeee;
}
h1 {
color: maroon;
margin-left: 0px;
}
.gray {
margin-left: 50px ;
margin-right: 29%;
font-weight: 500;
color: #000000;
background-color: #cccccc;
border-color: #aaaaaa;
}
.blue {
display: inline-block;
margin-left: 29% ;
margin-right: 0%;
width: -webkit-calc(100% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: #fff;
border-color: #336699;
background-color: #337799;
}
.left {
content:url("http://pngimg.com/uploads/mr_bean/mr_bean_PNG36.png");
width:200px;
border-radius: 500%;
float:left;
}
.right{
content:url("https://i.pinimg.com/originals/a1/a5/f4/a1a5f43315f458ab3c5e439aee5ac9de.png"); width:200px;
border-radius: 500%;
float:right;
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</style>
<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
background-color: #337798;
display: flex;
justify-content: space-between;
margin-bottom:3px;">
<div>
<i class="fa fa-envelope fa-2x"></i>
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<div>
<i class="fa fa-camera fa-2x"></i>
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<div>
<i class="fa fa-comments fa-2x"></i>
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<div>
<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
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<div>
<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
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<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
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<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
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<div><img class="left"/><div class="alert gray">
hey buddy... so ummm I don't want to look stupid in class anymore...so could #tutor me on the concept in problem 2 of our exam 1 (read it out loud it will make sense).
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<div><div class="alert blue">
Don't call me buddy, I'm your bro remember bros before hos. I got you broseph.
Question 2 begins with what is the limit definition of derivative for f'(x) of the function f(x)? So just remember
$$f'(x)=lim_{h→0}\ [f(x+h)-f(x)]/h $$
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<div><img class="left"/><div class="alert gray">
Ok, but how do I use the formula?
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<div><div class="alert blue">
Do not worry your pretty little head all questions will "bean" answered soon young skywalker. To begin with we need to utilyze the second part of the problem 2. It is the "two towers" to your "fellowship". This question part b of question 2 says, "for the function $f(x)=3x^2-4x$ find the exact formula for f'(x). Use only the definition of derivative."
We will now utilize the formula $$f'(x)=lim_{h→0}\ [f(x+h)-f(x)]/h$$. The two f's in the formula are basically saying in simple terms to replace these two f's with the given function $f(x)=3x^2-4x$ but it is also asking us to replace each respective "x" found after writing it out with the respective $(x+h)$ or $(x)$. allow me to demonstrate this concept.
$$f'(x)=lim_{h→0}\ [f(x+h)-f(x)]/h$$
Here we have our formula but do not focus on the portion labeled h or 0 for now. Instead, replace the two f's in the formula with the associated function found in the problem description.
$$f'(x)=lim_{h→0}\ [(3x^2-4x)-(3x^2-4x)]/h$$
Now we will segment out the "x's" like so...
$$lim_{h→0}\ [3(x)^2-4(x)-3(x)^2-4(x)]/h$$
but remember that these are still two seperate smaller equations found in the larger equation and the the central most negative sign (-) is something that is to be distributed amongst the second smaller equation later on, but for now let us get back to replacing the "x's" with what the formula is asking. We will start with the left "x's" and then do the right ones.
$$lim_{h→0}\ [3(x+h)^2-4(x+h)-3(x)^2-4(x)]/h$$
bote that the x's did not change on the right side because replacing the x's on the right formula keeps that part of the equation the same.
We must now distribute the negative sign (-)
$$lim_{h→0}\ [3(x+h)^2-4(x+h)-3(x)^2+4(x)]/h$$
Now we can focus on the
$$3(x+h)^2$$ for a moment. This can be re-written as
$$3(x+h)*(x+h)$$
For this step we must foil
$$3(x^2 +2xh +h^2)$$
then we distribute the "3" giving us
$$3x^2 +6xh +3h^2$$
Now we must jump back to the main problem and distribute the rest of the way.
$$lim_{h→0}\ [3x^2 +6xh +3h^2-4x-4h-3x^2+4x]/h$$
now we collect like terms to try to reduce and cancel the load in the problem.
$$lim_{h→0}\ [6xh +3h^2-4x-4h+4x]/h$$
$$lim_{h→0}\ [6xh +3h^2-4h]/h$$
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<div><div class="alert blue">
At this point we can now factor out an h from the bracketed equation like this...
$$lim_{h→0}\ h[6x +3h-4]/h$$
and then cancel out the h from the top and the bottom of the equation leaving us with...
$$lim_{h→0}\ [6x +3h-4]$$
At this point we can finally do what our formula tells us to do and substitute a 0 in for h, this now becomes...
$$f'(x)=[6x +3(0)-4]$$
which gives us the final answer of...
$$f'(x)=6x-4$$
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<div><img class="left"/><div class="alert gray">
Wow, that is a lot of work...do we really need to write out $$lim_{h→0}\ $$ each time and the f'(x)=the answer at the end?
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<div><div class="alert blue">
Yes, it is a good habbit and will come in handy as the problems get more complex not to mention placing that f'(x) at the end helps both you and the reader know what you were doing the entire time. Remember what the instructor said, "Think of these problems like sentences in a paper, you should do your best to use proper grammer, spelling, and punctuation". Plus, you said, "I don't want to look stupid" so I suggest to practice you write the formula out every 10 minutes for the next 2 days and this will help commit it to memory. As for how to use it in problems take homework examples and class examples and 3 times a day for the next 3 days solve them making sure to check your answers for correctness. When you solve re-write them on a new sheet and solve again after some time has passed. Before you know it you will be a mathlete or a keyboard samurai slicing problems and taking names.
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<div><img class="left"/><div class="alert gray">
Thanks, again for using code to communicate the bro-code of math to me. It has been most real and tubular.
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<div><div class="alert blue">
Yes, no thanks is needed if you need any other help just drop me a line or text me on my cellphone telephone.
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