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Task1 5

Content type
User Generated
Subject
Calculus
School
American Public University System
Type
Homework
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Solution:
Given: g(x) = x^3 h(x) = -3x+6
Using Newton’s Methods x(n+1) = x(n) f(x(n))/f’(x(n))
Since we need the intersection point of these 2 graphs, so:
g(x) = h(x)
x^3 = -3x+6
x^3 + 3x 6 = 0 f(x)
f’(x) = 3x^2 + 3
f (1) = (1) ^ 3 +3(1) 6 = -2
1. X1 = 1
x (1) = 1 f (1)/f’ (1) f ‘(1) = 3(1)^2 + 3 = 6
= 1 (-2)/6 = 4/3 = 1.333
f(1.333) = (1.333)^3 + 3(1.3333) 6 = 0.3701
2. X2 = n+1 = 1+1 = 2
x (2) = x(1) f(x(1))/f’(x(1))
= 1.333 f(1.333)/f’ (1.333)
= 1.333 - 0.3701/8.3331
= 1.2889
f’(1.333) = 3(1.333)^2 + 3 = 8.3331
3. X3 = 2+1 = 3
X(3) = x(2) f(x(2))/f’(x(2))
= 1.2889 f(1.2889)/f’(1.2889)
= 1.2889 0.0079/7.9838
= 1.2879
So, approximately the intersection point of these 2 graphs is x = 1.2879

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Solution: Given: g(x) = x^3 Using Newton’s Methods h(x) = -3x+6 x(n+1) = x(n) – f(x(n))/f’(x(n)) Since we need the intersection point of these 2 graphs, so: g(x) = h(x) x^3 = -3x+6 x^3 + 3x – 6 = 0 f(x) f’(x) = 3x^2 + 3 f (1) = (1) ^ 3 +3(1) – 6 = -2 1. X1 = 1 x (1) = 1 – f (1)/f’ (1) = 1 – (-2)/6 = 4/3 = 1.333 f ‘(1) = 3(1)^2 + 3 = 6 f(1.333) = (1.333)^3 + 3(1.3333) – 6 = 0.3701 2. X2 = n+1 = 1+1 = 2 x (2) = x(1) – f(x(1))/f’(x(1)) = 1.333 – f(1.333)/f’ (1.333) = 1.333 - 0.3701/8.3331 = 1.2889 f’(1.333) = 3(1.333)^2 + 3 = 8.3331 3. X3 = 2+1 = 3 X(3) = x(2) ...
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