1. ## Newton's Method

Given the function:

$f(x)=5e^{-x}-29x$

When, $x_{0}=\ln(6)$

$x_{1}=?$

I know...

$x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}$

2. ## Re: Newton's Method

You will have:

$x_1=\ln(6)-\frac{5e^{-\ln(6)}-29\ln(6)}{-5e^{-\ln(6)}-29}$

Now, use the two identities:

$-\ln(y)=\ln\left(\frac{1}{y} \right)$ and $e^{\ln(y)}=y$ to simplify.

3. ## Re: Newton's Method

Thanks, I think it was the ln's that threw me... I hate those things...

4. ## Re: Newton's Method

I advise you to learn to love the natural log function, as you will get to know it well during your study of the calculus.

5. ## Re: Newton's Method

Could you give me an answer. I have plugged it in two different times and gotten.
-80.169265 and 23.752784.
I have to go to 6 decimal places.

6. ## Re: Newton's Method

I get:

$x_1=\ln(6)-\frac{\frac{5}{6}-29\ln(6)}{-\frac{5}{6}-29}\approx0.077982108079$

7. ## Re: Newton's Method

When you put it into your calculator, do you put it in all at once? Or do you break it up into segments to put in your calculator. Because I plugged the same thing you have there and I got a different answer...

8. ## Re: Newton's Method

I enter the entire expression. Exactly what are you entering?

9. ## Re: Newton's Method

I believe that it was just misplaced parentheses.