The graphs of f(x) = x^3 and g(x) = 3X+5 have one intersection point. Determine the x-value for this intersection point using Newton's Method with an initial estimate of X1=2. You should use this method three times in order to obtain estimates X2, X3, and X4.
To find the intersection point, set
x^3 = 3x+5
x^3-3x-5 = 0
let y = x^3 - 3x -+ 5
dy/dx = 3x^2 - 3
newx = x - y/y' = x - (x^3 - 3x - 5)/(3x^2 - 3)
= (2x^3 + 5)/(3x^3-3)
Using my calculator
x1 = 21/9 = 2.333333
x2 = 2.28055555
x3 = 2.27902
x4 = 2.27902 --- correct to 5 decimal places !!!!
What is this newx and how did you get the equation that says x-y/y'
Your problem says to use Newton's Method
I stated the formula for Newton's Method.
Are you in over your head for this question?
To determine the x-value for the intersection point of the graphs of f(x) = x^3 and g(x) = 3x + 5 using Newton's Method, we will first find the derivative of f(x) to obtain f'(x). Then, we will use the formula
X(n+1) = X(n) - f(X(n))/f'(X(n))
where X(n) represents the nth estimate, X(n+1) is the (n+1)th estimate, f(X(n)) is the value of the function at X(n), and f'(X(n)) is the value of the derivative at X(n).
Let's begin the calculations:
1. Finding f'(x):
Taking the derivative of f(x) = x^3, we have:
f'(x) = 3x^2
2. Applying Newton's Method:
Given initial estimate X1 = 2, let's calculate X2, X3, and X4 using the formula:
a. Iteration 1 (n = 1):
X2 = X1 - f(X1)/f'(X1)
To calculate f(X1):
f(X1) = X1^3 = 2^3 = 8
To calculate f'(X1):
f'(X1) = 3X1^2 = 3(2^2) = 12
Substituting these values into the formula, we get:
X2 = 2 - 8/12 = 2 - 2/3 = 4/3
Therefore, X2 = 4/3.
b. Iteration 2 (n = 2):
X3 = X2 - f(X2)/f'(X2)
To calculate f(X2):
f(X2) = (4/3)^3 = 64/27
To calculate f'(X2):
f'(X2) = 3(4/3)^2 = 3(16/9) = 16/3
Substituting these values into the formula, we get:
X3 = 4/3 - (64/27)/(16/3) = 4/3 - (64/27)*(3/16) = 4/3 - 2/9 = 38/27
Therefore, X3 = 38/27.
c. Iteration 3 (n = 3):
X4 = X3 - f(X3)/f'(X3)
To calculate f(X3):
f(X3) = (38/27)^3 = 54872/19683
To calculate f'(X3):
f'(X3) = 3(38/27)^2 = 3(1444/729) = 4332/729
Substituting these values into the formula, we get:
X4 = 38/27 - (54872/19683)/(4332/729) = 38/27 - (54872/19683)*(729/4332) ≈ 1.201
Therefore, X4 ≈ 1.201.
After these three iterations, the estimated x-value for the intersection point obtained using Newton's Method with an initial estimate of 2 is approximately 1.201.