00:01
For this problem, we are told that newton's method will also fail when the difference of successive approximations does not decrease as k increases.
00:09
We're asked to, in part a, show that this happens for the function f of x equals the cubed root of x minus 2 when we choose x0 equals 1.
00:17
So to begin, we would have that x1.
00:21
Oh, actually, let me back up one second here.
00:24
So we have f prime of x will be equal to 1 over 3 times.
00:30
X minus 2 to the power of negative 2 over 3.
00:34
So we'll start off by having f of 1 be equal to the cubed root of negative 2.
00:44
So that is just, yeah, so that is just going to be negative 2 to the power of 1 over 3.
00:51
And we have f prime of 1...