Choosing the largest interval around a root that Converges with Newton's Method with every Xo guess

[Bluewolf1986]

Here's a question from my Calculus 1: Differential Calculus course on the Newton's Method Unit:

Find the largest interval around each of the roots x−x3=0 such that Newton's method converges to that root for every initial guess x0 in that interval.

(Enter answer as an interval (a, b). Enter infty for ∞ and -infty for −∞. Use round parentheses. You may type sqrt(2) for 2–√ etc. Type ∗ for multiplication, / for division, and ∧ for exponentiation.)

1) I got the first two answers right that the root around x=1 the interval is (sqrt(1/3), infinity) and around x=-1 the interval= (-infinity,-sqrt(1/3)

2) I need help with the root around x=0, I came up with (-sqrt(1/3), 0) but it was incorrect. I am stuck on how to contrive this interval. Please provide hints or methods to help me find this interval.

Thank you!

 

[Deleted member 4993]

Here's a question from my Calculus 1: Differential Calculus course on the Newton's Method Unit:

Find the largest interval around each of the roots x−x3=0 such that Newton's method converges to that root for every initial guess x0 in that interval.

(Enter answer as an interval (a, b). Enter infty for ∞ and -infty for −∞. Use round parentheses. You may type sqrt(2) for 2–√ etc. Type ∗ for multiplication, / for division, and ∧ for exponentiation.)

1) I got the first two answers right that the root around x=1 the interval is (sqrt(1/3), infinity) and around x=-1 the interval= (-infinity,-sqrt(1/3)

2) I need help with the root around x=0, I came up with (-sqrt(1/3), 0) but it was incorrect. I am stuck on how to contrive this interval. Please provide hints or methods to help me find this interval.

Thank you!

Is your function:

f(x) = x - x^3

You said:

I got the first two answers right that the root around x=1 the interval is (sqrt(1/3), infinity) and around x=-1 the interval= (-infinity,-sqrt(1/3)​

Please share your work regarding these two roots.

 

[tkhunny]

Were you tempted to include [math]\dfrac{1}{\sqrt{3}}\;or\;-\dfrac{1}{\sqrt{3}}[/math] in any of your intervals?

Why, specifically, might one care?

 

[Bluewolf1986]

Is your function:

f(x) = x - x^3

You said:

I got the first two answers right that the root around x=1 the interval is (sqrt(1/3), infinity) and around x=-1 the interval= (-infinity,-sqrt(1/3)​

Please share your work regarding these two roots.

I found the derivative of the function and set it to zero to find the positive and negative sqrt(1/3) value. Then I graphed the function noticing that the function converged at negative one and one near the minima and maxima of the function and the convergence was infinite since both the left and right hand side of the cubic function converged infinitely.

I actually figured out how to calculate the other root interval correctly which was to solve f(-x)= -f(x), which makes X2=Xo and X1=-Xo. Then the formula that I used solved this is X1=−Xo=Xo−Xo−x^3/1−3x^2o=−2x^3o/1−3x^2o. This then gave me both values of -1/sqrt(5) and 1/sqrt(5) which is of course the interval that converges on the root zero at the graph's origin.

 

[Bluewolf1986]

Thank you all for your responses, they were very helpful.