Use Newton's method to approximate a root of the equation (2 x^3 + 4 x + 4 =0) as follows.
Let (x_1 = -1\) be the initial approximation.
The second approximation (x_2) is ?
and the third approximation (x_3) is ?
First of all, why not reduce it to
x^3 + 2x + 2 = 0
Your iteration formula should be
xnew = (2x^3 - 2)/(3x^2 + 2)
so for a start of x=-1
we get
x1 = -1.333333
x2 = -.91919
x3 = -.78357
BTW, x5 = -.770917
which is correct to 7 decimal places.
looks like I forgot to close my subscript code
The end should read:
Your iteration formula should be
xnew = (2x^3 - 2)/(3x^2 + 2)
so for a start of x=-1
we get
x1 = -1.333333
x2 = -.91919
x3 = -.78357
BTW, x5 = -.770917
which is correct to 7 decimal places.
using by newton's method an approximate value the positive root
To approximate the root using Newton's method, we follow these steps:
Step 1: Given the equation 2x^3 + 4x + 4 = 0, we need to find the derivative of the function f(x) = 2x^3 + 4x + 4. Differentiating f(x) with respect to x, we get f'(x) = 6x^2 + 4.
Step 2: Choose an initial approximation, x₁. In this case, x₁ = -1.
Step 3: Now, we need to use the formula for Newton's method to find the next approximation, x₂. The formula is given by:
x₂ = x₁ - (f(x₁) / f'(x₁))
Step 4: Plug in the values of x₁, and calculate f(x₁) and f'(x₁).
f(x₁) = 2(-1)^3 + 4(-1) + 4 = -2 - 4 + 4 = -2
f'(x₁) = 6(-1)^2 + 4 = 6 + 4 = 10
Step 5: Substitute the values of f(x₁) and f'(x₁) into the formula to find x₂:
x₂ = x₁ - (f(x₁) / f'(x₁))
= -1 - (-2 / 10)
= -1 + 0.2
= -0.8
So, the second approximation, x₂, is -0.8.
Step 6: Repeat steps 3 to 5 to find the third approximation, x₃.
f(x₂) = 2(-0.8)^3 + 4(-0.8) + 4
= -1.024 - 3.2 + 4
= -0.224
f'(x₂) = 6(-0.8)^2 + 4
= 3.84 + 4
= 7.84
x₃ = x₂ - (f(x₂) / f'(x₂))
= -0.8 - (-0.224 / 7.84)
= -0.8 + 0.02857
= -0.77143
Therefore, the third approximation, x₃, is approximately -0.77143.