**g(x) = 1/2 x**^{3} - 2x + 3x + 2 on the domain [-1, 1]
I have to perform iterations of Newton's method based on this theorem:

**Theorem.** Let [tex]f\, :\, [a,\, b]\, \rightarrow\, \mathbb{R}[/tex] have the following properties:
. . . . .[tex]\mbox{1. }\, f(a)\, f(b)\, <\, 0,[/tex]

. . . . .[tex]\mbox{2. }\, f\, \mbox{ has no critical points in }\, (a,\, b),\, \mbox{ and}[/tex]

. . . . .[tex]\mbox{3. }\, f''\, \mbox{ exists, is continuous, and fulfills }\, f''\, \geq\, 0\, \mbox{ or }\, f''\, \leq\, 0\, \mbox{ on whole }\, (a,\, b).[/tex]

Then [tex]f(x)\, =\, 0[/tex] has exactly one solution [tex]x^*.[/tex] The Newton sequence [tex]x_n[/tex] converges always towards [tex]x^*[/tex] as [tex]n\, \rightarrow\, \infty[/tex] if the initial guess [tex]x_0[/tex] is chosen according to
. . . . .[tex]\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \leq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \geq\, 0\, \mbox{ then }\, x_0\, \in\, [a,\, x^*],\, \mbox{e.g.}\, x_0\, =\, a[/tex]

. . . . .[tex]\bullet\, \mbox{ if }\, f(a)\, <\, 0,\, f''\, \geq\, 0,\, \mbox{ or }\, f(a)\, >\, 0,\, f''\, \leq\, 0\, \mbox{ then }\, x_0\, \in\, [x^*,\, b],\, \mbox{e.g.}\, x_0\, =\, b [/tex]

In both cases we have for all iterates [tex]x_n[/tex] the estimate
. . . . .[tex]\displaystyle \big| x_n\, -\, x^{*} \big| \, \leq\, \dfrac{\big|f(x_n)\big|}{\min\limits_{a \leq x \leq b}\, \big|f'(x)\big|}[/tex]

a) I have to show that the Conditions 1-3 hold, I proved that.

b) Compute the

denominator of that expression in the end, min| g'(x) |, I completed this one also, the value is 1/2

c) choose an initial guess according to the theorem and perform iterations until the error |Xn - X*| lies below 10^-4

I do not understand this part how can I do the interations, if for example my Xo = 1, then how much is my X1, can someone do or explain just one or two interations, so then I am able to continue?

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