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
± ![\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 
, the choice that shouldbe made instead of ± in part A is:
f'(x) = 
⇔ 
Part D.
As given = = h
or h = -
We get,
f(a) + hf'(a) +(h²/2)f''(a) = 0
or h² = -hf(a)/f'(a)
Also, (-)² = -(-)(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()]/[2(f'())²]
Answer:
$22.50
Step-by-step explanation:
900 x 2.5%
=22.50
Answer:
Step-by-step explanation:
as given j=h and k=m then which expression for g is
as from attached image , we can see that its look like two right angle triangle
so from lower triangle we apply the pythagorean theorem .
which state that sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse.
so small triangle ΔEBD
h² +k²= g² ---------------(1)
now from big triangle ΔABC
(j + h)² +(m + h)² =f²---------(2)
given j=h and k=m so from equation-2 we can find ,
(h + h)² +(k+ k)² =f²
(2h)² +(2k)²=f²
4h²+ 4k²= f²
4(h²+k²)= f² ----------(3)
now from equation 1 and 3 we can obtained .
4×g² =f²
take square root both side we get,
2×g=f
so g=f/2----- Answer
Answer:
a = -48
Step-by-step explanation:
Answer:
x=64
y=40
Step-by-step explanation:
2x+x-12=180
3x=192
x=64
3y+y+20=180
4y=160
y=40
