anonymous
  • anonymous
Help with linear approximation please! I need help in part a) my calculation is as follows f(2)+(2.2-2)(f(2)) which then gave me 40+(0.2)(40) which gave me 48 but that is not the correct answer =/ .. Help please!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
@Luigi0210
Luigi0210
  • Luigi0210
@Jhannybean @ganeshie8 @primeralph Sorry, I threw in the towel awhile ago

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Luigi0210
  • Luigi0210
@johnweldon1993
anonymous
  • anonymous
@e.mccormick help please? =(
e.mccormick
  • e.mccormick
hmmm... do you know the slope at 2?
e.mccormick
  • e.mccormick
Cause that graph is a little small for me to read. Hehe.
anonymous
  • anonymous
I will get it enlarged for ya just one sec
anonymous
  • anonymous
https://webwork.elearning.ubc.ca/webwork2_files/tmp/MATH184-922_2013S1//gif/6TGFHE2ZLS05-1297-setAssignment_10prob3image1.png this help?
anonymous
  • anonymous
slope of 4 at x=2
anonymous
  • anonymous
4*.2 +40 im guessing here
anonymous
  • anonymous
oh what 40.8 is right what did I do wrong? @_@
e.mccormick
  • e.mccormick
Yah. That changes things. For linear approximation you use this: \(f(x) \approx f(a)+f'(a)(x-a)\) That pic was the \(f'(a)\) part.
anonymous
  • anonymous
and @e.mccormick I thought I did that in my calculations? oO
anonymous
  • anonymous
f'(a) is the first derivative of f(a) instead of using f ' (a), you ended up using f(a) instead
anonymous
  • anonymous
oo so can someone show me the actual calculation then with numbers please? =/
e.mccormick
  • e.mccormick
\(f(x)=40+(4)(2.2-2)\)
anonymous
  • anonymous
ahh darnit ok thanks so much!
anonymous
  • anonymous
4 was determined using the graph because a=2, we needed to look at the graph for a=2 in order to determine f'(2)
anonymous
  • anonymous
yeah that was cleared up thanks to you guys! Thanks again! The help is much appreciated!
e.mccormick
  • e.mccormick
Yep, small mistake. The graph makes you think that is the function, but it was stated as being the prime. Makes it an easy one to get wrong.

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