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: #dcd;
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/csDqAwM.png");
width:50px;
border-radius: 50%;
float:left;
}
.right{
content:url("https://i.imgur.com/nidMjPr.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">

Hey happy emoji! I had a question for you. Can you explain to me how to find local linearization? I do not quite understand it.

</div></div>

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

Of course panic emoji! First you have to understand that local linearization just means to find an equation by using the tangent line equation.

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

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

I see, but how do we go from the tangent line equation to the local linearzation equation?

</div></div>

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

I really do not know how we get from $y-y_{o}=m\left(x-x_{o}\right)$ to $L\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$

</div></div>

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

These are the steps to get to local linearization:
$y-y_{o}=m\left(x-x_{o}\right)$
$y=y_{o}+m\left(x-x_{o}\right)$
Then you change the variables to the function, and the derivitive of the function. So:
$L\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$
$a=$ the point of the function

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

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

Okay! Great! I am starting to understand. can you give me an example problem?

</div></div>

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

Sure thing. Find the linear approximation to $h\left(t\right)=t^{4}-6t^{3}+3t-7$ at $t=-3$
First we can find $h(t)$ by using the function. So $h(-3)=227$
Then we find $h'(-3)$ by using the derivitive definition, which you should already know so I will cut to the chase. $h'(-3)=-267$. Now we can plug in what we know to the formula: $L\left(x\right)=f\left(a\right)+f'\left(a\right)\left(x-a\right)$
So the formula will now look like this:
$h\left(t\right)=h\left(-3\right)+h'\left(-3\right)\left(x-(-3)\right)$
And then:
$h\left(t\right)=227-267\left(x+3\right)$
This will be our formula for linear approximation. Leave it as is, do not simplify.

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

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

Okay! I think I get it. Any more tips before I take this test?

</div></div>

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

Yes! Please remember that the equation you find is only good for points close to the number you found it for. For example the equation I showed you: $h\left(t\right)=227-267\left(x+3\right)$
Would only be good for numbers close to -3, like -3.1 and so on. If you input a number far from -3, like 14 then you are going to get a pretty hefty percent error.
I hope my tips helped you out! Good luck on your test!

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

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