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;
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.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;
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.left {
content:url("https://i.imgur.com/rUsxo7j.png");
width:50px;
border-radius: 50%;
float:left;
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.right{
content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
border-radius: 50%;
display: inline-block;
vertical-align:top;
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<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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<i class="fa fa-user-plus fa-2x" aria-hidden="true"></i>
</div>
</div>

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So... what exactly is this assignment? it looks tricky.
$a)$ Use the derivative formula $f'(x)=4x^3$ to find the linear approximation of $f(x)=x^4$ at $x=2$. $b)$ Use the linear approximation to estimate the value of $3.06^4$

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

Oh! Do you remember the lecture for 1.8 that we did in class friday?

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

I kind of remember... but can you refresh my memory?

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What is $L(x)$?

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So, in class for section 1.8 we talked about using the derivative of an equation to find the linear approximation which is $f(x)=L(x)=f(a)+f'(a)(x-a)$

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Oh right! Do we just plug in our values from the question since they give us the values in the question?

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Yes! you got it. :smile: For part $b)$ of the problem you would just plug in the given value for x into the linear approximation you just did.

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

Alright so the equation would look like this?
$L(x)=f(2)-f'(2)(x-2)$
So..
$L(x)=16+32(x-2)$

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Yes!
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Also, don't forget to plug in your value into your linear approximation to estimate the value of $3.06^4$

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

Like this? $L(x)=16+32(3.06-2)$
Which is $49.92$.

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

Awesome! you totally got this :100:

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