Newton-Raphson Approximation

A student, asked to find a root of the equation

$\displaystyle f(x)\equiv x^3-14x^2+49x-8=0$

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 $\displaystyle \alpha$, for a value of $\displaystyle \alpha$ in the interval $\displaystyle 7.2<\alpha<8$, would give a second approximation which is closer to the root x=8 provided that $\displaystyle 2(8-\alpha)f'(\alpha)+f(\alpha)>0$ and deduce that any value of $\displaystyle \alpha$ in the above range exceeding $\displaystyle 5+\sqrt{5.6}$ 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 $\displaystyle 2(8-\alpha)f'(\alpha)+f(\alpha)=(8-\alpha)(5\alpha^2-50\alpha+97)$.]

For the first part, I found

$\displaystyle f'(x)=3x^2-28x+49$

$\displaystyle f''(x)=6x-28$

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!