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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<div><img class="left"/><div class="alert gray">

So... what exactly is this assignment? It looks tricky.

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

Which one?  

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

Number 3. The one that says, 
"(a) Use the formula $f'(x)=\frac{1}{2\sqrt{x}}$ to find the linear approximation of $f(x)=\sqrt[2]{x}$ at $x=49$.
(b) Use the linear approximation to estimate the value of $f(50)$."

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

Is that the one that helps us estimate the value of a number on the function with the help of a tangent line?  

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

Yeah, that one! I'm having trouble understanding it. HELP ME PLZ!

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

Okay, okay, I'll help you. So basically all we are trying to find is an equation that we can use to estimate the value of $f(x)$ at a certain point. Especially when the numbers get more and more specific and decimals become a little too much to handle.

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

In this case we will be using the equation $L(X)=f(a)+f'(a)(x-a)$.
L(x) is simply the value we are trying to find, while x is the value we are plugging in.
f(a) is our original equation $f(x)=\sqrt[2]{x}$.
While f'(a) is just the derivative of f(a) or $f'(x)=\frac{1}{2\sqrt{x}}$.

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

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

Alright, that seems simple enough. Now how do we find each of those values?

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

Since we can easily find the value of $f(49)$ we can use that number as a reference or as the a in the equation.
In other words we just find the values of $f(49)$and $f'(49)$ and plug those in to the equation $L(x)$.
A little hint tho, it might be easier to plug in the whole equations into those values.

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

Okay, give me a minute to try it out.

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

So like this?
$L(x)=f(49)+f'(49)(x-49)$
$L(x)=\sqrt[2]{49}+\frac{1}{2\sqrt{49}}(x-49)$
$L(x)=7+\frac{1}{14}(x-49)$
This is as far as I got.

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

That's perfect! Now all you need to do is plug in the value we are trying to find, in this case $x=50$. So plug in $50$ for every $x$ in the equation.

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

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

Alright!
$L(x)=f(49)+f'(49)(x-49)$
$L(x)=\sqrt[2]{49}+\frac{1}{2\sqrt{49}}(x-49)$
$L(x)=7+\frac{1}{14}(x-49)$

So I continue it like this, right?
$L(50)=7+\frac{1}{14}(50-49)$
$L(50)=7+\frac{1}{14}(1)$
$L(50)=\frac{98}{14}+\frac{1}{14}$
$L(50)=\frac{99}{14}$
or
$L(50)=7.071428571$

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

Yes, exactly!
Now another hint, in order to check your answer and to see if it is at least somewhat accurate to the actual number just use a calculator.
Plug in f(50) in to the calculator and see if that number is equal to your estimate.
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<div><img class="left"/><div class="alert gray">

My estimate was $L(50)=\frac{99}{14}$ or $L(50)=7.071428571$
If I plug it in the calculator then I get... $f(50)=7.07106781187$
Wow, I was pretty close!
Is that all?

</div></div> 

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

Yep, that's all.
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<div><img class="left"/><div class="alert gray">

Really? Omg, thank you so much for your help!

</div></div> 

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

No problem! See you in class on Monday!
</div><img class="right"/></div>



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