How Do You Determine Which Values to Use in Linear Approximation?

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SUMMARY

The discussion focuses on the application of linear approximation using the formula L(x) = f(a) + f'(a)(x-a) to estimate values such as sqrt(25.1) and 2.001^5. The key insight is to identify the appropriate values for 'a' and 'x' in the context of the problem. For sqrt(25.1), 'a' is 25 and 'x' is 25.1, while for 2.001^5, 'a' is 2 and 'x' is 2.001. Understanding the derivative at 'a' and how to apply it in the linear approximation is crucial for accurate estimations.

PREREQUISITES
  • Understanding of derivatives and their notation
  • Familiarity with the concept of Taylor series
  • Basic knowledge of functions such as square root and polynomial functions
  • Ability to perform simple algebraic calculations
NEXT STEPS
  • Study the application of Taylor series for function approximation
  • Learn how to calculate derivatives of common functions
  • Practice linear approximation problems with various functions
  • Explore the concept of error estimation in linear approximations
USEFUL FOR

Students preparing for calculus exams, educators teaching linear approximation, and anyone seeking to improve their understanding of function estimation techniques.

Rizzice
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So I have this as the last thing I don't understand before tomorrow's test.. I have tried reading in the book and online, but it's just not clicking for me!

There are so many numbers, and it they seem to just plug them in from nowhere.

Like some example problems would be like sqrt(25.1) and 2.001^5.

Supposedly I use the equation L(x) = f(a) - f'(a)(x-a)? But I have no idea which numbers I plug in where.. is this easier than I'm making it to be?
 
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It would be easier if you read it as:

L(a+e)= f(a) +f'(a)e

that is the value at 'a' plus a little bit (e) is approximately linearly related to the value at 'a' using the derivative at 'a'.

so the square root of 25.1 is approximately the square root at 25 plus the derivative at 25 (of the square root function) times 0.1

All you're doing is using the first two terms of the Taylor/MacLaurin series to estimate the whole sum. For small 'e' this is approximately good enough.
 
Rizzice said:
So I have this as the last thing I don't understand before tomorrow's test.. I have tried reading in the book and online, but it's just not clicking for me!

There are so many numbers, and it they seem to just plug them in from nowhere.

Like some example problems would be like sqrt(25.1) and 2.001^5.

Supposedly I use the equation L(x) = f(a) - f'(a)(x-a)? But I have no idea which numbers I plug in where.. is this easier than I'm making it to be?
I think it might help you to read the problem correctly! You say "Like some example problems would be like sqrt(25.1) and 2.001^5."

I hope you understand that "sqrt(25.1)" and "2.001^5" are numbers, not problems! State (at least to yourself) clearly what you are supposed to do with those numbers!

I suspect that the problem really is something like this:
"Use the tangent line approximation to give a simple approximation to sqrt(25.1)" or "approximate sqrt(25.1) without using a calculator".

Then you should think to yourself, "Alright, I know that sqrt(25)= 5 so I don't need a calculator for that! I also know that the tangent line to f(x) at x= a is given y(x)= f(a)+ f'(a)(x-a). Since I am asked to find sqrt(25.1), I had better take x= 25.1! Since I need to be able to find f(a)= sqrt(a) (with a as close to the given point as possible) , I had better take a= 25.

Similarly with "Use the tangent line approximation to give a simple approximation to 2.0015. Since you are specifically asked to find f(2.001) (approximately), you should take x= 2.001. Since you need to be able to find f(a)= a5 easily, and have a close to 2.001, you should take a= 2.
 

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