---
# System prepended metadata

title: My lesson Topic

---

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, Professor Lee. I get confused when approaching a problem that asks me to find the linear approximation, and if that part doesn't go right, I can never get to solving the next portion where I have to use that linear approximation to actually estimate some number. 

</div></div> 


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

Hello, Sam. Of course, I can.
Yes, it is important to ensure your linear approximation is correct in order to derive a correct answer for the second portion. 
  

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

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

Apart of just your assignment, linear approximation is actually important in real-life. You will not always have access to a calculator or some fancy contraption that will answer all your numerical problems for you. This method is important for making approximations of a value of a specific function at one particular point. When given a function at a point, it can be utilized to determine the linear approximation equation to solve for other points.

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

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

Oh, you're right.   
Could you help explain it with like an example?

</div></div> 


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

Sure. Let's start with this problem. 
Use the formula $f(x)=x^4$ and $f'(x)=x^2+20$ to find the linear approximation at x=3. 

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

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

So, we know that $L(x)=y_{1}+m\left(x-x_{1}\right)$. 
Now, switch out the $y_{1}$ with $f(x)$ and $m$ with $f'(x)$ which gives you
$=f(x)+f'(x)(x-x_{1})$.

From here, plug in their actual functions and x=3 to solve.
$=\left(x^{4}\right)+\left(x^{2}+20\right)\left(x-x_{1}\right)$
$=\left(3^{4}\right)+\left(3^{2}+20\right)\left(x-3\right)$
$=81+(9+20)(x-3)$
$=81+29(x-3)$

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

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

I like how you showed all the steps on how you solved that! I can see where you are plugging in fucntions and numbers. 

</div></div> 

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

It is super important to show all your work, so I can see where you may have made a mistake. You can utilized that to study as well, so you don't make the same mistake again on the exam! 
However, we are not finished. Now that we have found the linear approximation at x=3 (part a), we have to use that to estimate the value of $2.22^4$, for example (part b). 

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

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

If you look at $2.22^4$, you can see that it is just 2.22 inserted into the function from part a. 
That means $2.22^4=f(2.22)$, which then can be written ≈$L(2.22)$. Now, you can plug 2.22 in place of x of the formula we derived in part a. 

≈$L(2.22)
$=81+29(2.22-3)$
$=81+29(-0.78)$
$=81+22.62$
$=58.38$


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

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

Wow, it became more clear to me now! Thank you so much Professor Lee. I will continue practicing with the homework questions to better prepare for the exam. I think I know where I was getting confused. Seeing it broken down step-by-step definitely helped. 

</div></div> 

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

I am glad you are feeling more confident about attempting these problems now. 
Some tips that might be useful for you: 
- Brush up on your algebra because not all functions will only contain an exponent. You could see square roots and fractions, so understand how to tackle those expressions. 
- Be familiar with finding f'(x) if it is not given to you in the question. You will need the slope to accurately derive the linear approximation of a point. 
- Do not get confused with $f(x) and f'(x)$ and where they fit into $L(x)=y_{1}+m\left(x-x_{1}\right)$.

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

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

Let me know if you have any more questions or concerns. I would be happy to help. 
Happy studying! (: 

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


---

To submit this assignment click on the Publish button ![Publish button icon](https://i.imgur.com/Qk7vi9V.png). Then copy the url of the final document and submit it in Canvas.



