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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Critical numbers are where the derivative is 0 or doesn't exist. Did you find the derivative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh okay got it! I was a little confused on what it was xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait so the derivative is \[3x^2  e^x\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I'm doing this completely wrong lol... I misunderstood the question!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Approximate a critical number of y = x^3 + e^−x using Newton’s method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not really sure how many roots the derivative has, but I guess start with xn = 0 dw:1435192530939:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435192671369:dw Then keep going until the numbers don't change much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh So the same process is just repeated over an over?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435192838109:dw wrong equation. Thanks to @jim_thompson5910 for pointing it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is this equation called?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Newton's method

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh so every time I have to use newton's method I just apply this?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes 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
 one year ago
Best ResponseYou've already chosen the best response.1x1 is generated from x0 x2 is generated from x1 etc etc the further you go, the more accurate the approximation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I understand! Thanks so much!
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