Thread: Newton-Raphson Approximation

A student, asked to find a root of the equation

did not notice the solution x=8 but chose, instead, to use Newton's Method taking x=7.2 as the first approximation. He then calculated correctly, f(7.2)=-7.712, f'(7.2)=2.02, and deduced (again correctly) the second approximation 9.84. By means of a graph, or otherwise, explain why Newton's Method failed to give a better approximation in this case. Prove that using Newton's Method, a first approximation , for a value of in the interval , would give a second approximation which is closer to the root x=8 provided that and deduce that any value of in the above range exceeding would in fact give improvement.
[You may, if you wish, assume without proof that f'(x) is positive and increasing forx>7.2; also that .]

For the first part, I found


There are three reasons given in the examples of this book for why the method doesn't work.
1) h is too large
2) f'(a) is too small
3) f''(a) is too large
I thought it would be that f''(a) is too large, f''(7.2)=15.2, but in the next portion, f''(7.3), say, is 15.8, so that can't be the case.
f'(a)= 2.92, which isn't too small, the other explanation then is that h=0.8 is too large, the problem is I'm given in the book that 0<h<1.
This is where I'm stuck
Thanks!

Look at the graph of and you should be able to see immediately what the problem is.
Notice that has a minimum at and that 7.2 is just to the right of this.

so what exactly is the reason, the one given sounds a bit vague to me. is it because its only a little to the right, hence its gradient is too small?

Yes, that's pretty much it. The graph provides you with an explanation of why you might be kicked further away from the root than you started. In this case it isn't a problem, the iteration will converge quite nicely from the RHS, (starting from 9.84).
If you are looking for something analytical, check out the value of the expression

that you see later in the question.