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: 40px;
}

.gray {
margin-left: 50px ;
margin-right: ;
font-weight: 500;
color: #000000;
background-color: #cccccc;
border-color: #aaaaaa;
}

.blue {
display: inline-block;
margin-left: 65px;
margin-right: 15%;
width: -webkit-calc(70% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: FFFfff;
border-color: #336699;
background-color: #CCFF;
}

.left {
content:url("https://i.imgur.com/rUsxo7j.png");
width:50px;
border-radius: 50%;
float:left;
}
.right{
content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
border-radius: 50%;
display: inline-block;
vertical-align:top;
}
</style>

<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
background-color: #337799;
display: flex;
justify-content: space-between;
margin-bottom:3px;">
<div>
<i class="fa fa-envelope fa-2x"></i>
</div>
<div>
<i class="fa fa-camera fa-2x"></i>
</div>

<div>
<i class="fa fa-comments fa-2x"></i>
</div>
<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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<div>
<i class="fa fa-list-ul fa-2x" aria-hidden="true"></i>
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<div>
<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
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</div>

<div><img class="left"/><div class="alert gray">

Hi peer tutor! I need some MAJOR help with calculus! Do you know anything about the limit definition of a derivative?

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<div><div class="alert blue">

Yes, indeed! The derivative of *f(x)* with respect to *x* is the function *f'(x)* and is defined as $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

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<div><img class="left"/><div class="alert gray">

Ok... in simple terms?

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<div><div class="alert blue">

The derivative of a function, when the limit exists, is the slope of the tangent line to the graph of the function. The tangent line is the best linear approximation of the function near that input value. Derivatives are important because we can use them to find the slope of (almost) any curve at (almost) any point. Now, I say almost because there are some exceptions, as mentioned before, if the limit does not exist. Make sense?

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<div><img class="left"/><div class="alert gray">

It makes so much sense when you explain it! Can you be my tutor this whole semester?

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<div><div class="alert blue">

Of course! Is there a homework problem you want to work together on?

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<div><img class="left"/><div class="alert gray">

That would be great! I've been staring at this problem for an entire day and I know where to start but, when the algebra hits, I get very confused. The problem states, *find the derivative of the following function using the defintion of the derivative:* $$f(x)=2x^2-16x+35$$ So, I need to plug the function into the definition of the derivative. $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$$$f'(x)=\lim_{h\to0}\frac{2(x+h)^2-16(x+h)+35-(2x^2-16x+35)}{h}$$ Truthfully, this is where I get lost.

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<div><div class="alert blue">

No worries! Be careful and make sure that you properly deal with parenthesis, according to PEMDAS. Now, we know that we can't just plug in *h=0* since this gives us a division by zero error. So, we are going to have to do some work. In this case, that means multiplying everything out and distributing the minus sign through the second term. Doing this gives us, $$f'(x)=\lim_{h\to0}\frac{2x^2+4xh+2h^2-16x-16h+35-2x^2+16x-35)}{h}$$ Keeping up?

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<div><img class="left"/><div class="alert gray">

Sort of. Why did *16x* turn into a postive value and *35* into a negative value?

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<div><div class="alert blue">

Remember, we have to distribute the minus sign throughout the second term. A negative integar multiplied by a negative integar equals a postive value. A postive integar multiplied by a negative integar equals a negative value.
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<div><img class="left"/><div class="alert gray">

Ok I got that. Please continue.

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<div><div class="alert blue">

Perfect, we can cancel out the like terms in the eqaution and simplfy it just as:
$$f'(x)=\lim_{h\to0}\frac{4xh+2h^2-16h}{h}$$
Notice that every term in the numerator that didn't have an *h* in it cancelled out and we can now factor an *h* out of the numerator which will cancel against the *h* in the deniminator. This is how you factor out the *h*. $$f'(x)=\lim_{h\to0}\frac{h(4x+2h-16)}{h}$$ Since the *h* cancels out in the numerator and denominator, the simplified equation is $$f'(x)=\lim_{h\to0}4x-16$$ Making sense?

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<div><img class="left"/><div class="alert gray">

Yes! So this is the derivative of the function?

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<div><div class="alert blue">

Almost! The last step is to replace *h* with zero. We can now do this because there is no longer a fraction in the equation. The replacement looks like:
$$f'(x)=\lim_{h\to0}4x+2(0)-16$$ Which then simplifies to: $$f'(x)=4x-16$$The derivative of $$f'(x)=2x^2-16x+35$$ is $$f'(x)=4x-16$$
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<div><div class="alert blue">

It is very important for you to practice and practice. The homework problems won't always look like this example so please be wary of the steps and **WHY** they are played out like that. Make sure to look at your opposite sides and conclude if the integar is positive or negative. A small mishap like that will possibly make you lose points or get the whole problem wrong. Take your time and don't rush through these. Did you understand the material today?

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<div><img class="left"/><div class="alert gray">

Yes peer tutor. Untill next time!

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<div><div class="alert blue">

See ya!

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