zeesbrat3
  • zeesbrat3
If f(1) = 5 and f prime of x equals the quotient of 6 and the quantity x squared plus 2, which of the following is the best approximation for f(1.03) using local linearization?
Mathematics
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SOLVED
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chestercat
  • chestercat
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zeesbrat3
  • zeesbrat3
\[f(1) = 5\] \[f'(x) = \frac{ 6 }{ x^2 + 2}\]
zeesbrat3
  • zeesbrat3
@ganeshie8
SolomonZelman
  • SolomonZelman
you first need to integrate the 6/(x²+2)

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SolomonZelman
  • SolomonZelman
you would need a trig substitution....
zeesbrat3
  • zeesbrat3
I haven't gotten to integration yet..
SolomonZelman
  • SolomonZelman
oh, then just recongize the derivative I guess
ganeshie8
  • ganeshie8
\[f(1+0.03) = f(1) + f'(1)(1.03-1)\]
SolomonZelman
  • SolomonZelman
I feel stupid now:)
SolomonZelman
  • SolomonZelman
you don't even need to find f(x).
zeesbrat3
  • zeesbrat3
There is no defined f'(1) I thought?
ganeshie8
  • ganeshie8
Since it is an approximation, it makes more sense to use \(\approx\) \[f(1+0.03) \approx f(1) + f'(1)(1.03-1)\] simplify
SolomonZelman
  • SolomonZelman
for f`(1) plug in 1 for x into the f'(x)
SolomonZelman
  • SolomonZelman
(and it is always defined, because x²+6 is never=0, it is at least 6)
zeesbrat3
  • zeesbrat3
So \[5 + 2(.03)\]
ganeshie8
  • ganeshie8
Looks good! it simplifies further tho
zeesbrat3
  • zeesbrat3
To 5.06
ganeshie8
  • ganeshie8
Yep
zeesbrat3
  • zeesbrat3
Thank you!
SolomonZelman
  • SolomonZelman
I wish one day I would be that good at math too.... :D f(1+0.03) was quite awesome, I will remeber this trick.
ganeshie8
  • ganeshie8
yw
SolomonZelman
  • SolomonZelman
tnx ganesh.
ganeshie8
  • ganeshie8
np

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