anonymous
  • anonymous
Approximate a critical number of y = x^3 + e^−x using Newton’s method
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Critical numbers are where the derivative is 0 or doesn't exist. Did you find the derivative?
anonymous
  • anonymous
Ohh okay got it! I was a little confused on what it was xD
anonymous
  • anonymous
ok

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anonymous
  • anonymous
Wait so the derivative is \[3x^2 - e^x\] right?
anonymous
  • anonymous
\[3x^2-e ^{-x}\]
anonymous
  • anonymous
hmm okay! thanks :)
anonymous
  • anonymous
I think I'm doing this completely wrong lol... I misunderstood the question!
anonymous
  • anonymous
Approximate a critical number of y = x^3 + e^−x using Newton’s method
anonymous
  • anonymous
That's the question
anonymous
  • anonymous
Not really sure how many roots the derivative has, but I guess start with xn = 0 |dw:1435192530939:dw|
anonymous
  • anonymous
|dw:1435192671369:dw| Then keep going until the numbers don't change much
anonymous
  • anonymous
Ohh So the same process is just repeated over an over?
anonymous
  • anonymous
and*
anonymous
  • anonymous
|dw:1435192838109:dw| wrong equation. Thanks to @jim_thompson5910 for pointing it out
anonymous
  • anonymous
Ahh I see
anonymous
  • anonymous
What is this equation called?
jim_thompson5910
  • jim_thompson5910
Newton's method
anonymous
  • anonymous
Ohh so every time I have to use newton's method I just apply this?
jim_thompson5910
  • jim_thompson5910
yes where the first term \(\Large x_0\) is the initial guess. You can randomly pick a number (say 0) or look at the graph and make an educated guess where the root is
jim_thompson5910
  • jim_thompson5910
x1 is generated from x0 x2 is generated from x1 etc etc the further you go, the more accurate the approximation
anonymous
  • anonymous
Oh I understand! Thanks so much!
jim_thompson5910
  • jim_thompson5910
no problem

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