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">

Consider the function $h(x)=x^3+3x^2-3$
Find the formula of the tangent line when $x=2$.

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

In order to solve this problem it is essential to take the derivative of the function.

Looking at $h(x)$, we notice that the function is written in a way which we are able to apply the power rule ($h'\left(x\right)=nx^{n-1}$).
</div><img class="right"/></div>

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Functions are not always conveniently written in this form, so it essential to have some sort of grasp for solving the limit definition of the derivative for the function.
</div><img class="right"/></div>

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

$h'(x)=3x^2+6x$

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Correct

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

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Now we must use the point slope form of a line to slove the problem
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<div><div class="alert blue">

$y=h(x)+h'(x)(x-a)$
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Since the problem is asking for the formula of the tangent line at a specific input for $x$ ($x=2$) we must insert the number $2$ in the place of $x$ for both $h(x)$ and $h'(x)$

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

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

$L(x)=h(2)+h'(2)(x-2)$
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<div><img class="left"/><div class="alert gray">

$L(x)=[2^3+3(2)^2-3]+[3(2)^2+6(2)](x-2)$

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

$L(x)=[8+12-3]+[12+12](x-2)$

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

$L(x)=17+24(x-2)$

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Impressive

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With this particular formula, we are now able to approximate the outputs for $h(x)$ close to $x=2$, such as $x=2.01$.

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Keep note that inputs for $x$ closest to $x=2$ will provide the most correct approximation of the tangent line.

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

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Now that we know the formula for the tangent line for $h(x)$ at $x=2$, lets find the approximation for $x=2.02$.

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

$h(2.02)$ ~ $L(2.02)$

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

$L(2.02)=17+24(2.02-2)$

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

$L(2.02)=17+24(2.02-2)$

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

$L(2.02)=17.48$ ~ $h(2.02)$

</div></div>

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