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Newtons Method (1 Viewer)

Lukybear

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The root of the equation e^x-x^3 = 0 lies between x=-1 amd x=0. Use newton's method once to find an approximation to the root correct to 2 decimla places.

This question has me perplexed. Using x=-0.5, and e^x-3x^2 = f'(x)

i get a= -0.5 - (f(-0.5)/f'(-0.5))

= 4... something

Help?
 

Uncle

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I have generated a plot of the function ex - x3 = 0
You can see that there is no root between x = -1 and x = 0, there is something wrong with the question.
The function and its derivative is continuous, what you are doing is correct though.



They must have confused it with its derivative, ex - 3x2 = 0
The derivative does have a root between x = -1 and x = 0
 

Lukybear

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o thx goodness... its math in focus... lolz

+rep uncle

EDIT: You must spread some Reputation around before giving it to Uncle again.
 

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