Answer:
- 0.510914462964
- 2.93017792012
Step-by-step explanation:
A graphing calculator is an extremely useful tool for problems like this. Not only does it give you an initial value to work with, it can also do the iteration for you.
You start by defining a function that is zero at the desired values of x. It usually works well to subtract one side of the equation, effectively looking for the condition that the difference between one side of the equal sign and the other is zero. Here, we have used ...
[tex]f(x)=\ln{(x)}-\dfrac{1}{x-2}[/tex]
The Newton's method iterator is ...
new x = x - f(x)/f'(x)
We have defined that function to be g(x). Then entering an approximate value as the argument of g(x) gives the iterated value as a result. Since the approximate values shown on the graph are accurate to 3 decimal places, one iteration is really all that is required to get 6 decimal place accuracy.*
That is, ...
g(0.511) = 0.510914
g(2.93) = 2.930178 . . . . rounded to 6 d.p.
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The attached interactive graphing calculator gives the iterated value as quickly as you can type in the "guess". If you watch what it is showing you, you can type in that iterated value to extend the solution to full calculator precision very quickly.
The attachment shows the two solutions to be ...
- x = 0.510914462964
- x = 2.93017792012
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* Newton's method effectively doubles the number of accurate decimal places with each iteration.
