Need help with a nooby question

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In summary, the conversation discusses using the NR method to find the intersection of a function and its derivative. There is a question about the use of a constant and the process for finding the second approximation. It is noted that ln(4x) and ln(x) have the same derivative because ln(4x)= ln(x)+ ln(4). The correct second approximation is 0.16.
  • #1
semc
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5
alright there is this question which uses NR method to find intersect but i kept getting the wrong ans can anybody help me check which part i went wrong?:biggrin:

f(x)=2x + ln4x , f'(x)=2+ 1/x

btw,differentiate ln4x and lnx get the same result but the arbitary constant is different right?so do we consider the constant when finding the intersect?
 
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  • #2
If you integrate 1/x, you get lnx + C. ln4x = ln4 + lnx, so C = ln4 in this case.
 
  • #3
hmm...i see now so the constant changes although there's 1/x in both cases.

regarding the question, if i were to find the 2nd approx. , i am suppose to use 0.1 as 1st approx. however, 0.1-f(0.1)/f'(0.1) is not the ans:confused:
 
  • #4
Why not? It won't be the exact solution to the equation, of course, but it certainly is a "second approximation" with first approximation 0.1. I get 0.16 as the second approximation.

By the way, the reason ln(4x) and ln(x) have the same derivative (1/x) is that ln(4x)= ln(x)+ ln(4).
 
  • #5
hmm...i also get 0.16 guess the ans is wrong :biggrin: Thanks
 

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