Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
body {
background-color: #eeeeee;
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h1 {
color: maroon;
margin-left: 40px;
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.gray {
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margin-right: 29%;
font-weight: 500;
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display: inline-block;
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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;
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float:left;
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content:url("https://i.imgur.com/5ALcyl3.png"); width:50px;
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<div id="container" style=" padding: 6px;
color: #fff;
border-color: #336699;
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display: flex;
justify-content: space-between;
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<div>
<i class="fa fa-envelope fa-2x"></i>
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<div>
<i class="fa fa-camera fa-2x"></i>
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<div>
<i class="fa fa-comments fa-2x"></i>
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<i class="fa fa-address-card fa-2x" aria-hidden="true"></i>
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<i class="fa fa-phone fa-2x" aria-hidden="true"></i>
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<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>
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<div><img class="left"/><div class="alert gray">
Hello, Im a calculus student and I've been having trouble understanding the concept and importance for linear approximation.

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Hello, the importance for why linear approximation is used is when the function of $f(x)$ is to complicated to solve to find the value of an exact point. By solving the linear approximation you are able to find the tangent line, a much easier function to solve for.

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

So, how is linear approximation used?

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Linear approximation is a method of estimating the value of funciton, $f(x)$, near a point; $x=a$, using this formula:
$$L(x)=f(a)+f'(x)(x-a)$$

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Can we work through an example?

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Yes, so for instance a question would be asked like: Use the derivative formula $f'(x)= 4x^3$ to find the linear approximation of $f(x)=x^4$ at $x=2$.

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Okay, so I know the formula is $L(x)=f(a)+f'(x)(x-a)$. So since $f'(x)=4x^3$ and $f(x)=x^4$ and $a=2$. It would be:
$L(x)=(2)^4+ 4(2)^3(x-a)$
$=16+32(x-2)$

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Correct, $15+32(x-2)$ is the tangent line that we will use to eastimate any value instead of using the function for $f(x)$.

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So now use the linear approximation to estimate the value of $2.05^4$.

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Okay, that is the part I'm having trouble with. Where do I plug in $2.05^4$ to estimate the value?

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

Okay, so we now have $15+32(x-2)$, which is easier to solve for than $f(x)$.
$2.05^4=f(2.05)$
$\sim L\left(2.05\right)$
$L(2.05)=12+ 32+(2.05-2)$

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

Oh okay, so $2.05$ is the $x$ of $$L(x)=f(a)+f'(x)(x-a)$$
So,now that $L(2.05)=12+ 32+(2.05-2)$
$=16 +32(0.05)$
$=16+1.6$
$=17.6$
The estimate of the value $2.05^2$ is 17.6

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

Correct! For exams I suggest making your own practice problems of all the concepts and explainging those problems to another person or going over them to yourself outloud.

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<div><img class="left"/><div class="alert gray">
Thank you! I will make my own cheat sheat and make up my own problems to study for the exam.

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