Find the zeroes of the function using Newtons method.
f(x)=(1/4)x^3-3x^2+(3/4)x-2
I find that when x=11.803 y=0
Then I do Newtons method ((xn))-((f(x)n))/((f'(x)n))
((x))-((1/4x^3-3x^2+3/4x-2))/((3/4x^2-6x+3/4))
I find that when x=11.803 y=11.803
So Did I do this correctly?
What do you mean by that? Newton's method does not give you "y" it gives you , the next number in a sequence hopefully converging to a zero of the function. If you mean you put 11.803 into the Newton's method formula, for , and got back , yes, of course! If [tex]f(x_n)= 0, then the formula is which is just .
That's essentially how you tell when you have arrived at a solution- when the iteration starts giving you the same number again and again.
I am not sure what you did! You say found that "when x= 11.803, f(x)= 0". How did you find that value of x? And once you have found it why then put that value into the Newton's method formula?So Did I do this correctly?
I ran a quick Newton's method starting with
and got , , , , , , ,
Since we are now getting the same result, we can say that one solution, to three decimal places, is 11.803.