Answer:
Part A.
Let f(x) = 0;
suppose x= a+h
such that f(x) =f(a+h) = 0
By second order Taylor approximation, we get
f(a) + hf'±(a) +  f''(a) = 0
f''(a) = 0
  ±
 ± ![\frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B%5D%7B%28f%27%28a%29%29%5E%7B2%7D-2f%28a%29f%27%27%28a%29%20%7D%20%7D%7Bf%27%27%28a%29%7D)
So, we get the succeeding equation for Newton's method as
 ±
 ± ![\sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]](https://tex.z-dn.net/?f=%5Csqrt%7Bf%28x_%7Bi%7D%29%5E%7B2%7D-2fx_%7Bi%7Df%27%27x_%7Bi%7D%20%7D%20%5D)
Part B.
It is evident that Newton's method fails in two cases, as:
1.  if f''(x) = 0
2. if f'(x)² is less than 2f(x)f''(x)    
Part C.
In case   is close to
 is close to  , the choice that shouldbe made instead of ± in part A is:
, the choice that shouldbe made instead of ± in part A is:
f'(x) =  ⇔
  ⇔ 
Part D.
As given  =
 =  = h
 = h
or                 h =  -
 - 
We get, 
f(a) + hf'(a) +(h²/2)f''(a) = 0
or h² = -hf(a)/f'(a)
Also,             ( -
- )² = -(
)² = -( -
- )(f(
)(f( )/f'(
)/f'( ))
))
So,                f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0
It becomes   h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]
Also,              =
 =  -f(
 -f( )/f'(
)/f'( ) + [(
) + [( -
 -  )f''(
)f''( )f(
)f( )]/[2(f'(
)]/[2(f'( ))²]
))²]