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December 18, 2014

December 18, 2014

Posted by **Sam** on Thursday, March 17, 2011 at 2:43am.

- Calculus! -
**Reiny**, Thursday, March 17, 2011 at 9:23amso you are solving

e^(2x) + 3x - 25 = 0

let y = e^(2x) + 3x - 25

y' = 2e^(2x) + 3

newtons's formula says

X = x - f(x)/f'(x)

where x is your starting x value and X is the new value.

Hopefully your X will approach x, when that happens you have the solution.

X = x - (e^(2x) + 3x - 25)/(2e^(2x) + 3)

= (2xe^(2x) - e^(2x) + 25)/(2e^2(2x) + 3)

I made a rough sketch of y = e^(2x) + 3x - 25

and noticed that there was an x-intercept, which would be your solution, at appr x = 1.5

So I will make that my starting x

X ---- x

-------- 1.5

1.5096 ---1.50915695

1.50951695 -- 1.509515688

1.509515688 -- 1.50951688

Wow, got the answer correct to 9 decimal places after only 3 iterations.

The key thing is that you start with an initial guess close to the real number.

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