Given the function: $\displaystyle f(x)=5e^{-x}-29x$ When, $\displaystyle x_{0}=\ln(6)$ $\displaystyle x_{1}=?$ I know... $\displaystyle x_{n+1}=x_{n}-\frac{5e^{-x}-29x}{-5e^{-x}-29}$
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You will have: $\displaystyle x_1=\ln(6)-\frac{5e^{-\ln(6)}-29\ln(6)}{-5e^{-\ln(6)}-29}$ Now, use the two identities: $\displaystyle -\ln(y)=\ln\left(\frac{1}{y} \right)$ and $\displaystyle e^{\ln(y)}=y$ to simplify.
Thanks, I think it was the ln's that threw me... I hate those things...
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.
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.
I get: $\displaystyle x_1=\ln(6)-\frac{\frac{5}{6}-29\ln(6)}{-\frac{5}{6}-29}\approx0.077982108079$
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...
I enter the entire expression. Exactly what are you entering?
I believe that it was just misplaced parentheses.
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