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title: Linear Approximation

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

Hello Professor, I need help on the practice exam probelm 3. It asks: use linear approximation to estimate $\sqrt{9.3}$. Let $f(x)=\sqrt{x}$ and $x=9.$.  

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

Well first, it is asking to find the linear approximation. What equation do we use that relates to linear approximation?  

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The linear approximation equation is $$L(x) = f(a) + f'(a)(x-a)$$

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

Correct! By understanding the formula, we can now have a visualization of the slope of the tangent line. So if you look at the graph at y=L(x) the linear function is able to tell you information of the graph at y=f(x). In other words, it is the equation of the tangent line to the graph of y=f(x).     

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Ah I see, if I look at y=L(x) for a function, we can see the original function's value and the slope at the tangency.   

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

Correct! 


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

Ok, so I understand the formula but how do we use the formula of linear appoximation? How do I apply it?   

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

Well, theres a few different ways to go about this. In my opinion the first think you should think about is the forumla and what you know. 

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

Ok, so I know 
$$L(x)=f(a)+f'(a)(x-a)$$ 

$$L(x)=f\left(\sqrt{9}\right)+f'\left(\sqrt{9}\right)\left(x-9\right)$$

Next, I know that $$f\left(\sqrt{9}\right)=3$$

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

Right so now, you need to solve of $$f'\left(\sqrt{9}\right)$$ Do you remember how you could solve this? 

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Hmm, oh yes I use the limit definition. 

$$f'(x) = $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

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

Correct! But it might be helpful for you to solve it with what was given to you in the question. Which was $f(x)=\sqrt{x}$.

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

Oh okay so it'll look like this.

$$f'(\sqrt{x}) = \lim_{h \to 0}\frac{f(\sqrt{x+h})-f(\sqrt{x})}{h}$$
$$= \lim_{h \to 0}\frac{f(\sqrt{x+h})-f(\sqrt{x})}{h}\cdot\ \frac{\left(\sqrt{x+h}\ +\ \sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)}$$
$$ = \lim_{h \to 0}\frac{\left(x+h\right)-x}{h\sqrt{x+h}+\ \sqrt{x}}$$
$$ = \lim_{h \to 0}\frac{1}{\sqrt{x+h}+\ \sqrt{x}}$$
$$ = \frac{1}{\sqrt{x+0}+\ \sqrt{x}}$$
$$ = \frac{1}{2\sqrt{x}}$$

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

Correct! I appreciate that you show all your work! So what would you do next?

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$$L(x)=f\left(\sqrt{9}\right)+f'\left(\sqrt{9}\right)\left(x-9\right)$$
$$ = 3 + f'(\frac{1}{2\sqrt{9}})(x-9)$$
$$ = 3 + 1/6 (x-9)$$

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

Yes! So now you have linear approximation in slope-point form. So we can use this formula $\sqrt{9.3}$ to approximate f(9.3) and L(9.3)

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

Oh yea so it can help us estimate. 
$\sqrt{9.3} = f(9.3)$
$= L(9.3)$
$= 3 + 1/6 (9.3-9)$
$=3+1/6(.3)$
$=3.05$

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

I get it now! Thank you for helping me solve this!

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

Of course! I am glad this helped.

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