Math 181 Miniproject 3: Texting Lesson.md
---
My lesson Topic
===

<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, are you the math tutor? I could really use some help.

</div></div>

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

Yes! What are you working on?

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

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

I am working on a homework assignment. The instructions say that I need for find the linear approximation of $f(x)=x^8$ at x=2 give the derivative $f'(x)=8x^7$

Then, I need to use the linear approximation to estimate the value of $2.5^8$

</div></div>

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

Okay, lets start with with finding the linear approximation of f(x). What do you have so far?

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

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

I know that I am supposed to use the formula for Linear aapproximation;

$L(x)= f(x)+f'(x)(x-a)$

</div></div>

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

This is a good sarting point! A good thing to remember is that if a function y=f(x) is defferentiable at the (a,f(a)), the function f(x) when looked at closely, looks like the tangent line at that point. This means that if we can create a formula for the linear approximation that of f(x), we can use that formula to approximate calues that are close to x. Lets srat by using what is given to us and plug in f(x) and f'(x) into the formula.

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

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

Okay, I think this is correct.

$L(x)= (x^8)+(8x^7)(x-a)$

</div></div>

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

Great! What do you think is the next step that we should take?

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

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

Ummm I think we need to inclue our value $x=(2)$

$L(x)=(2^8)+(8x^7)(x-2)$

I remembered something important from my reading. It is important to distinguish (a) from (x) right? the value (a) is a constant that comes from our given values while (x) is an expression for the rule that defines the function.

</div></div>

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

Yes, exactly! So if we were to simplify L(x) furhter, what would that look like?

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

</div></div>

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

I got:

L(x)= $256+1024(x-2)$

So, we can use formual for the tangent line apporximation to estimate $2.5^8$ right?

</div></div>

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

Yes, we can say that $2.4^8$ $=f(2.5)$, which is aproximaly ~ $L(2.5)$

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

</div></div>

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

Okay, I think I get it now, imputing 2.5 into L(x), I get:

$L(2.5)= 256+1024(2.5-2$)
$L(2.5)= 256+1024(.5)$
$L(2.5)= 256+512$
$L(2.5)= 768$

</div></div>

<div><div class="alert blue">
Great job, It look like you got this down! If you get stuck again, feel free to reach out again. Im always happy to help.

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

</div></div>

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

Thank you! I'll try a few problems on my own, but if I get stuck im definitely asking for more help. Have a good day!

</div></div>

---

To submit this assignment click on the Publish button . Then copy the url of the final document and submit it in Canvas.