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">

So... I'm stuck on this problem on the math homework. The problem is: For the function $f(x)=18x-x^2$. Find the exact formula for $f'(x)$ using the limit definition. I'm confused on where to start with this one, can you help me?

</div></div>

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

Of course I can! Now, when trying to determine the formula for a function using the limit definition we know there's a specific template to use when trying to do this. The template we use is the formula:
$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$.
Now I know this may look daunting and sort of confusing but with this you're going to plug in your given equation for $f(x)$ and solve.
So with your given function it will look like:
$\lim_{h \to 0} \frac{18\left(x+h\right)-\left(x+h\right)^{2}-\left(18x-x^{2}\right)}{h}$

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

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

Okay wow that does seem sort of confusing. So when trying to move forward with the problem it'll look like this then?
$\lim_{h \to 0} \frac{\left(18x+18h-x^{2}+h^{2}-18x-x^{2}\right)}{h}$
and then I would simplify right?

</div></div>

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

So then I just got 18 after simplifying, so that's the final answer right?

</div></div>

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

So you're not exactly wrong in your thinking of simplfying the formula after we've plugged everything in, do you see the $(x+h)^2$? It actually has to be foiled out because $(x+h)^2$ is just$(x+h)*(x+h)$ So the first part of your partly solved equation is actually correct, it'll be $18x+18h$ Good job on figuring that part out! Like I said these types of problems are tricky. So what did you get when you foiled the $(x+h)^2$?

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

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

Okay! That does make sense, so i got:
$\frac{\left(18x+18h)-(x^{2}+2xh+h^{2})-(18x-x^{2}\right)}{h}$
I put in what I already knew as well, the first part of the equation and then putting the function in place of the x at the end. Once I started simplifying the whole thing I got:
$\frac{\left(18h-2hx-h^{2}\right)}{h}$
And then my final equation would be:
$18h-2hx-h$. And with that I finsihed the problem right?

</div></div>

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

You're exactly right! Great job! However, don't forget to continue labelling your work using the $\lim_{h \to 0}$ in front of the function you're solving, it's important so you don't forget what you're doing! Also, you're not quite finished yet! I know it seems like you are but do you see that h you have at the end of the equation and attached to the 2x? You actually have to plug in 0 because we are finding the limit as h approaches 0, so h must equal 0. (This is why it's important to keep notating the $\lim_{h \to 0}$). So you'd plug that in and get $f'(x)= (18-2x)$ We did't use the $\lim_{h \to 0}$ in front of this last part because we already input the 0 for h and it asked us to find the exact formula for $f'(x)$!

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

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

Geez, this was so difficult! Why is this even important? it's like he's tying to torture us with this!

</div></div>

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

Oh don't be dramatic! THese problems are extremely important if you want to understand all aspects of the derivative. At the same time it is important to understand how limits work even if it is just approaching 0 because limits are used to figure out continuity and derivatives which is what we are learning now!

I know it can be difficult now but the homework has helped me immensely in studying for the exam! Praticing these harder problems on a sheet of paper with different numbers and different expressions can help you prepare for these types of problems on the exam. I know you can do it! You essentially solved the whole problem yourself, just some little mistakes here or there. You got this!

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

---

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