Newtons Method (1 Viewer)

[Lukybear]

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]

I have generated a plot of the function e^x - 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, e^x - 3x² = 0
The derivative does have a root between x = -1 and x = 0

 

[Lukybear]

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

+rep uncle

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

 

[Dragonmaster262]

How did you draw the graph?

 

[Uncle]

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

+rep uncle

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

maths in focus.
very typical as a dodgy textbook.

How did you draw the graph?

Maple 13
Symbolic mathematics software used by mathematics and engineering students.