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>
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<div>
<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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What is this problem about?!

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You are trying to find the Linear approximation (or tangent line) of f'(x) using f(x), at x= 1000. Using the function $f(x)= ∛x$ and it's derivative $f'(x)=\frac{1}{4(∛x^{3})}$ use the Linear Approximation equation to find this equation of a tangent line at x= 1000!
Part b asks you to use Linear Approximation to estimate a specific value that's given to you. Estimate (∛1004).

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So what is the Linear approximation formula?

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Is there an explanation for this formula?

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A tangent line at the point (a,f(a)) to f(x) is $y-y_{1}=m\left(x-x_{1}\right)$. This reduces to $y- f(a)=f'(a)(x-a)$, then to y=f(a)+f'(a)(x-a). That equation is the linear approximation. $L(x)=f(a)+f'(a)(x-a)$. Linear approximation = local linearization. L(x) gives us this tangent line given the function, it's derivative and an x value (1000).

$f'(x)=\frac{1}{4(∛x^{3})}$ and $\ f\left(x\right)=∛x$ at x=1000

$L\left(x\right)=(\left(∛1000\right)+\left(\frac{1}{4(∛1000^{3})}\right))\left(x-1000\right)$

$L\left(x\right)=10+\frac{1}{4000}\left(x-1000\right)$

HINT: there can only be one x in the L(x) equation, or else it wouldn't be a line.

b). Estimate ∛1004.
∛1004= f(1004)
∛1004≃ L(1004)
∛1004= $10 + \frac{1}{4000}(1004-1000)$
∛1004= $10+\frac{4}{4000}$
∛1004= 10.001

The rounded estimate from the calculator for ∛1004 is 10.01 and we got 10.001. That's close enough!

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What is the reasoning behind using this method/importance?

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You use the function and it's derivative (which are given) to to find the tangent line of the derivative using a given value.
*Side fact: You can adjust a for the slope of the tangent line at a point.*
It's important because we use Linearization to approximate the value of a function at a particular point. The reason it's useful is because it's difficult to find the value of a function at a particular point.

In part b where it asks you to estimate using the Linear Approximation, this is important because you are evaluating similar to a calculator. It's an easy trick to use if you don't have a calculator near you.

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Are there any tips you have?

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My biggest tip is to memorize the L(x) formula! The whole problem isn't too mentally exausting if you know the formula and where to plug in it's pretty easy! The derivative, original function and x value to find the tangent line are given along with the value to estimate. Just plug in and simplify!

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