Answer:
There is only one real zero and it is located at x = 1.359
Step-by-step explanation:
After the 4th iteration the solution was repeating the first 3 decimal places. The formula for Newton's Method is
[tex]x_{n}-\frac{f(x_{n}) }{f'(x_{n}) }[/tex]
If our function is
[tex]f(x)=x^5+x-6[/tex]
then the first derivative is
[tex]f'(x)=5x^4+1[/tex]
I graphed this on my calculator to see where the zero(s) looked like they might be, and saw there was only one real one, somewhere between 1 and 2. I started with my first guess being x = 1.
When I plugged in a 1 for x, I got a zero of 5/3.
Plugging in 5/3 and completing the process again gave me 997/687
Plugging in 997/687 and completing the process again gave me 1.36976
Plugging in 1.36976 and completing the process again gave me 1.359454
Plugging in 1.359454 and completing the process again gave me 1.359304
Since we are looking for accuracy to 3 decimal places, there was no need to go further.
Checking the zeros on the calculator graphing program gave me a zero of 1.3593041 which is exactly the same as my 5th iteration!
Newton's Method is absolutely amazing!!!
