2x^4-3x^2-7x+1=0
Estimate what the real solution will be using Newton's method. Answers must be 4 decimal places. Let the first guess be x=1.75
let y = 2x^4 - 3x^2 - 7x + 1
y ' = 8x^3 - 6x - 7
newx = x - y/y'
= x - (2x^4 - 3x^2 - 7x + 1)/(8x^3 - 6x - 7)
=(6x^4- 3x^2 - 1)/(8x^3 - 6x - 7)
x = 1.75
newx = 1.8162 ---> x
newx = 1.8111 ---> x
newx = 1.811102
I would call 1.8111 a good estimate
To estimate the real solution of the equation 2x^4 - 3x^2 - 7x + 1 = 0 using Newton's method, you can follow these steps:
Step 1: Calculate the derivative of the function.
Let's find the derivative of the function f(x) = 2x^4 - 3x^2 - 7x + 1. Taking the derivative, we get:
f'(x) = 8x^3 - 6x - 7.
Step 2: Set up the recursive formula for Newton's method.
Newton's method uses the formula:
x(n+1) = x(n) - f(x(n)) / f'(x(n)),
where x(n) represents the nth guess and x(n+1) is the next guess.
Step 3: Start with the first guess, x(0) = 1.75, and iterate using the formula from Step 2.
Using the given first guess, let's calculate the next guess, x(1).
x(1) = x(0) - f(x(0)) / f'(x(0)).
Substituting the values, we have:
x(1) = 1.75 - (2(1.75)^4 - 3(1.75)^2 - 7(1.75) + 1) / (8(1.75)^3 - 6(1.75) - 7).
Simplifying the equation leads us to:
x(1) ≈ 1.75 - (3.324) / (8.27 - 10.5 - 7).
Calculating further:
x(1) ≈ 1.75 - (-2.324) / (-9.23).
The result is:
x(1) ≈ 1.75 + 0.2515 ≈ 2.0015 (approximated to 4 decimal places).
Therefore, the estimated real solution using Newton's method with the first guess x=1.75 is approximately 2.0015.