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">

Hello tutor! I was just wondering if you could help me with this new concept we learned in my Calculus class yesterday I think I understand it, but I want to make sure.

</div></div>

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

Hello student! I would be more than happy to help you. What is it that you learned?

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

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

I believe it was about Linear Approximation. I understand how we find the equation of the tangent line, but I don't understand how we use it to etsimate the value of more difficult problems that would otherwise need a calculator.

</div></div>

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

Ok... first tell me what you do know and understand.

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

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

I know how to find the equation for a tangent line using the formula:
$y=f(a)+f'(a)(x-a)$
And when given a function I know how to find $L(x)$. But I don't understand how the Linear Approximation formula is used to find values for "ugly" numbers.

</div></div>

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

Ok. So, using the linear approximation of a function to estimate more difficult values of that function comes from looking at a graph. If we can identify a point on the tangent line that is really close to that equivalent x-value on the function, than that y-value on the tangent would be pretty close to that y-value on the function.

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

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

Oh... I see. Could you give me an example?

</div></div>

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

Sure. Let's say we have the function $f\left(x\right)=\sqrt[3]{x}$ and its derivative is $f'\left(x\right)=\frac{1}{3\sqrt[3]{x^2}}$ at $x=1000$. What is the Linear Approximation?

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

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

$L(x)=f(a)+f'(a)(x-a)$
$L(x)=\sqrt[3]{1000}+\left(\frac{1}{3\sqrt[3]{1000^2}}\right)(x-1000)$
$L(x)=10+\frac{1}{300}x-1000$

I think...

</div></div>

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

Yes! Now if I told you to give me the y-coordinate, $f(x)=\sqrt[3]{1003}$, it would be extremely difficult to do without a calculator. However, using the Linear Approximation of x, we can estimate a coordinate on the tangent line of the function that is really close to $\sqrt[3]{1003}$.

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

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

Mmm...

</div></div>

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

So, being that $f(x)=\sqrt[3]{1003}$ we know x, in the instance is equal to $1003$. We then plug that into our equation

$L(1003)=10+\frac{1}{300}(1003-1000)$

Then solve.
$L(1003)=10.01$ therefore, $f\left(1003\right)\sim10.01$

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

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

Ohhhhh...that makes sense. You're not calculating the exact answer, you're just estimating based on the tangent since graphically it is so close.

</div></div>

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

Exactly. Good job

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

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

Thanks for helping me understand.

</div></div>

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

No problem!

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

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