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Math 181 Miniproject 3: Texting Lesson.md
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My lesson Topic
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Hello, Im a calculus student and I've been having trouble understanding the concept and importance for linear approximation.  

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Hello, the importance for why linear approximation is used is when the function of $f(x)$ is to complicated to solve to find the value of an exact point. By solving the linear approximation you are able to find the tangent line, a much easier function to solve for. 

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So, how is linear approximation used?

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Linear approximation is a method of estimating the value of funciton, $f(x)$, near a point; $x=a$, using this formula:
$$L(x)=f(a)+f'(x)(x-a)$$


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Can we work through an example?

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Yes, so for instance a question would be asked like: Use the derivative formula $f'(x)= 4x^3$ to find the linear approximation of $f(x)=x^4$ at $x=2$.

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Okay, so I know the formula is $L(x)=f(a)+f'(x)(x-a)$. So since $f'(x)=4x^3$ and $f(x)=x^4$ and $a=2$. It would be:
$L(x)=(2)^4+ 4(2)^3(x-a)$
$=16+32(x-2)$

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Correct, $15+32(x-2)$ is the tangent line that we will use to eastimate any value instead of using the function for $f(x)$.

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So now use the linear approximation to estimate the value of $2.05^4$.

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Okay, that is the part I'm having trouble with. Where do I plug in $2.05^4$ to estimate the value?

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Okay, so we now have $15+32(x-2)$, which is easier to solve for than $f(x)$.
$2.05^4=f(2.05)$
$\sim L\left(2.05\right)$
$L(2.05)=12+ 32+(2.05-2)$

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Oh okay, so $2.05$ is the $x$ of $$L(x)=f(a)+f'(x)(x-a)$$
So,now that $L(2.05)=12+ 32+(2.05-2)$
$=16 +32(0.05)$
$=16+1.6$
$=17.6$
The estimate of the value $2.05^2$ is 17.6

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Correct! For exams I suggest making your own practice problems of all the concepts and explainging those problems to another person or going over them to yourself outloud. 

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Thank you! I will make my own cheat sheat and make up my own problems to study for the exam.

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