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: 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(70% - 50px);
width: -moz-calc(70% - 50px);
width: calc(70% - 50px);
font-weight: 500;
color: #fff;
border-color: #336699;
background-color: #337799;
}

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

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

What is Local Linerization?

</div></div>

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

Local Linerization is the generalization of tangent plane functions; one that can apply to multivariable functions with any number of inputs. You are just finding an equation for a tangent line.

</div><img class="right"/></div>

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

Okay.

</div></div>

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

Then what would be: Find the local linerization for $f(x)=1/x$ given that $f'(x)=-1/2^2$ at the point x=2.

</div></div>

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

What is needed is: $f(x)=1/x, f'(x)=-1/x^2$ at x=2. It does not matter the word problem as long as you can figure out the $f(x)$ and then the $f'(x)$ and identify at what point.

</div><img class="right"/></div>

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

Makes sense.....but how do I do it?

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

You will need the formula $L(x)= f(x)+f'(x)(x-x_1)$ The L stands for Local Linerization.

</div><img class="right"/></div>

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

So then it should be L(x)=f(2)f'(2)(x-2)

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

Yes that is right. The next step would be to plug in the equations. ****There should only be one x once everything is filled in****

</div><img class="right"/></div>

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

$f(x)=1/2+(-1/2^2)(x-2)$
=$1/2+(-1/4)(x-2)$
Is that it??? Or did I miss something
</div></div>

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

Yep that is it since all you needed to do was find that equation for the point x=2.

</div><img class="right"/></div>
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