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anonymous

  • one year ago

Approximate a critical number of y = x^3 + e^−x using Newton’s method

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  1. anonymous
    • one year ago
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    Critical numbers are where the derivative is 0 or doesn't exist. Did you find the derivative?

  2. anonymous
    • one year ago
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    Ohh okay got it! I was a little confused on what it was xD

  3. anonymous
    • one year ago
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    ok

  4. anonymous
    • one year ago
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    Wait so the derivative is \[3x^2 - e^x\] right?

  5. anonymous
    • one year ago
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    \[3x^2-e ^{-x}\]

  6. anonymous
    • one year ago
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    hmm okay! thanks :)

  7. anonymous
    • one year ago
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    I think I'm doing this completely wrong lol... I misunderstood the question!

  8. anonymous
    • one year ago
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    Approximate a critical number of y = x^3 + e^−x using Newton’s method

  9. anonymous
    • one year ago
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    That's the question

  10. anonymous
    • one year ago
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    Not really sure how many roots the derivative has, but I guess start with xn = 0 |dw:1435192530939:dw|

  11. anonymous
    • one year ago
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    |dw:1435192671369:dw| Then keep going until the numbers don't change much

  12. anonymous
    • one year ago
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    Ohh So the same process is just repeated over an over?

  13. anonymous
    • one year ago
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    and*

  14. anonymous
    • one year ago
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    |dw:1435192838109:dw| wrong equation. Thanks to @jim_thompson5910 for pointing it out

  15. anonymous
    • one year ago
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    Ahh I see

  16. anonymous
    • one year ago
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    What is this equation called?

  17. jim_thompson5910
    • one year ago
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    Newton's method

  18. anonymous
    • one year ago
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    Ohh so every time I have to use newton's method I just apply this?

  19. jim_thompson5910
    • one year ago
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    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

  20. jim_thompson5910
    • one year ago
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    x1 is generated from x0 x2 is generated from x1 etc etc the further you go, the more accurate the approximation

  21. anonymous
    • one year ago
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    Oh I understand! Thanks so much!

  22. jim_thompson5910
    • one year ago
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    no problem

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