Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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<style>
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h1 {
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<div id="container" style=" padding: 6px;
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<i class="fa fa-envelope fa-2x"></i>
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<i class="fa fa-camera fa-2x"></i>
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<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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<div><img class="left"/><div class="alert gray">
For the function f(x)=4x^2-6x, Find the formula for f'(x).
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<div><div class="alert blue">

To begin this problem we first need the limit definition equation which is $f'\left(x\right)=\lim _{h\to 0\:}\frac{f\left(x+h\right)-\left(x\right)}{h}$.
The derivative f'(x) is the slope of the functine and f(x) is at a point.
The dereivative of a function f'(x) allows us to the change of f(x) over a time interval.
We can alos find the instantaneous rate of change where $x=a$
($a$ is a real number)

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

I see, so using the limit definition equation $f'\left(x\right)=\lim _{h\to 0\:}\frac{f\left(x+h\right)-\left(x\right)}{h}$, we can find the derivative

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ok,What does $h$ represent?

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Yes, we will use the limit definition equation to solve the problem.
The $h$ value in the equation represents the change in ($x$)

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

ok, so how would I use $f(x)=4x^2-6x$ and $f'\left(x\right)=\lim _{h\to 0\:}\frac{f\left(x+h\right)-\left(x\right)}{h}$ to solve for the derivative?

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Yes, so what do you think we have to do first?

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

We can plug in $(x+h)$ for every (x) value in $4x^2-6x$ and then subtract it by $4x^2-6x$

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

Yes that is correct and dont forget to divide everything by $h$
$f'\left(x\right)=\lim _{h\to 0\:}\frac{f\left(x+h\right)-\left(x\right)}{h}$

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<div><img class="left"/><div class="alert gray">
so the equation should look like this

$f'\left(x\right)=\lim _{h\to 0\:}\frac{4\left(x+h\right)^2-6\left(x+h\right)-\left(4x^2-6x\right)}{h}$

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That looks great now the next step would be to foil but be careful becasue this is where alot of mistakes can be made. Work it and send me your results.

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After working the equation out this is what I ened up with :
$f'\left(x\right)=\lim _{h\to 0\:}\frac{4\left(x^2+2xh+h^2\right)-\left(6x+6h\right)-\left(4x^2-6x\right)}{h}$

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Is this correct?

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

Yes that is correct now remember to multiply 4 into $(x^2+2xh+h^2)$ and apply the negative sign (-1) to $(4x^2-6x)$

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

My equation now looks like
$f'\left(x\right)=\lim _{h\to 0\:}\frac{4x^2+8xh+4h^2-6x-6h-4x^2+6x}{h}$
I do have some like terms do I cancel them out?

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Yes, eliminate any like terms that cancel each other out such as $4x^2$ and $-4x^2$. They have oppostie signs

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After eliminating the like terms I am left with
$f'\left(x\right)=\lim _{h\to 0\:}\frac{8xh+4h^2-6h}{h}$

I know I have to simplify but im not sure how to eliminate the $h^2$

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

Ok to simplify the equation you must factor out the $h$ to cancel the $h$ from the denominator and numerator. After factoring $h^2$ you should be left with $h$.

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

The equation I have now look like this
$f\:'\left(x\right)=\lim _{h\to 0\:}\frac{h\left(8x+4h-6\right)}{h}$

Then after the elimination I am left with
$f\:'\left(x\right)=\lim _{h\to 0\:}\left(8x+4h-6\right)$

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

Looks great, now since we are looking for the limit as $h$ aproaches 0, plug in 0 for every $h$ remaining in the equation.
Then that would leave us with the exact derivative formula for f'(x)

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After I plug the 0's, I am left with

$f'(x)=8x+4(0)-6$
$f'(x)=8x-6$

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

Yes your final answer would be $f'(x)=8x-6$

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