I will look on the graph of function y=(x) and y=((6—3-1)/4)0. 2. Below I show you that graph. The x-coordinates of the three points where these graphs intersect are the roots of the original equation. I will perform the iteration using a starting value of x0 =-2 and attempt to find the root in the interval [-2,-1]. We must also remember this graph is an example of a Staircase convergence. And I have found the first root, which lies in that interval.

I must say that that process will take a bit of time if somebody would like to work in Excel. But Ive found the root at x= -1. 273 (3d. p) Im, trying to find now next root which lies in [1,2]. For my starting value I choose x0 =1. The staircase diagram goes towards the root. Because gradient at x=1 is less than 1. To show that method could also failure. If I would like to find where exactly the root lies, we now that in an interval [0,1].

But where is it exactly? I will try two starting point, one at x=0, and next one at x=0. 7. Look what Ive done. At x=0. At x=0. 7. This is a little bit weird. Why they not come to this root which I want. Answer is very simple and depends of the gradient of g(x). Gradient is in this method responsible for finding the roots. A rearrangement will find only root at x if the gradient of g(x) at this point x is between -1 and 1. So algebraically, if -1

I must say that that process will take a bit of time if somebody would like to work in Excel. But Ive found the root at x= -1. 273 (3d. p) Im, trying to find now next root which lies in [1,2]. For my starting value I choose x0 =1. The staircase diagram goes towards the root. Because gradient at x=1 is less than 1. To show that method could also failure. If I would like to find where exactly the root lies, we now that in an interval [0,1].

But where is it exactly? I will try two starting point, one at x=0, and next one at x=0. 7. Look what Ive done. At x=0. At x=0. 7. This is a little bit weird. Why they not come to this root which I want. Answer is very simple and depends of the gradient of g(x). Gradient is in this method responsible for finding the roots. A rearrangement will find only root at x if the gradient of g(x) at this point x is between -1 and 1. So algebraically, if -1

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