Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.)
1/3x^3 + 1/2x^2 + 8 = 0, x1 = −3
I got -3.4808, but it's wrong. Help?
1/3x3+1/2x2+11=0,x1=−3
To use Newton's method to find the third approximation, x3, to the root of the equation 1/3x^3 + 1/2x^2 + 8 = 0 with an initial approximation x1 = -3, follow these steps:
Step 1: Calculate f(x1)
To find f(x), substitute x = x1 into the equation:
f(x1) = 1/3(x1)^3 + 1/2(x1)^2 + 8
f(-3) = 1/3(-3)^3 + 1/2(-3)^2 + 8
f(-3) = 1/3(-27) + 1/2(9) + 8
f(-3) = -9 + 4.5 + 8
f(-3) = 3.5
Step 2: Calculate f'(x1)
To find f'(x), differentiate the equation with respect to x:
f'(x) = d/dx(1/3x^3 + 1/2x^2 + 8)
f'(x) = x^2 + x
Now, substitute x = x1 into the equation:
f'(-3) = (-3)^2 + (-3)
f'(-3) = 9 - 3
f'(-3) = 6
Step 3: Calculate x2
Use the formula for the next approximation, x2, which is given by:
x2 = x1 - f(x1)/f'(x1)
Substitute the values calculated in steps 1 and 2 into the formula:
x2 = -3 - 3.5/6
x2 = -3 - 0.5833
x2 ≈ -3.5833
Step 4: Calculate f(x2)
To find f(x), substitute x = x2 into the equation:
f(x2) = 1/3(x2)^3 + 1/2(x2)^2 + 8
f(-3.5833) = 1/3(-3.5833)^3 + 1/2(-3.5833)^2 + 8
f(-3.5833) = 1/3(-45.625) + 1/2(12.8544) + 8
f(-3.5833) = -15.2083 + 6.4272 + 8
f(-3.5833) = -0.7811
Step 5: Calculate f'(x2)
To find f'(x), differentiate the equation with respect to x:
f'(x) = d/dx(1/3x^3 + 1/2x^2 + 8)
f'(x) = x^2 + x
Now, substitute x = x2 into the equation:
f'(-3.5833) = (-3.5833)^2 + (-3.5833)
f'(-3.5833) = 12.8301 - 3.5833
f'(-3.5833) = 9.2468
Step 6: Calculate x3
Use the formula for the next approximation, x3, which is given by:
x3 = x2 - f(x2)/f'(x2)
Substitute the values calculated in steps 4 and 5 into the formula:
x3 = -3.5833 - (-0.7811)/9.2468
x3 = -3.5833 + 0.0845
x3 ≈ -3.4988
Therefore, the third approximation to the root of the equation 1/3x^3 + 1/2x^2 + 8 = 0 with the initial approximation x1 = -3 is approximately -3.4988 (rounded to four decimal places).
To find the third approximation, x3, using Newton's method, we need to iterate the process using the formula:
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)
Given the equation:
(1/3)x³ + (1/2)x² + 8 = 0
We need to find the derivative, f'(x), which will be used in the above formula.
f'(x) = d/dx[(1/3)x³ + (1/2)x² + 8]
= (1/3)(3x²) + (1/2)(2x)
= x² + x
Now, let's calculate the values using Newton's method:
1. Calculate f(x):
f(x) = (1/3)x³ + (1/2)x² + 8
f(-3) = (1/3)(-3)³ + (1/2)(-3)² + 8
= (-1/3)(27) + (1/2)(9) + 8
= -9 + 4.5 + 8
= 3.5
2. Calculate f'(x):
f'(x) = x² + x
f'(-3) = (-3)² + (-3)
= 9 - 3
= 6
3. Apply Newton's method:
x₃ = x₂ - f(x₂) / f'(x₂)
We are given x₁ = -3, which is our initial approximation.
We need to find x₂ first and then use it to find x₃.
4. Calculate x₂:
x₂ = x₁ - f(x₁) / f'(x₁)
x₂ = -3 - 3.5 / 6
= -3 - 0.58333 (rounded to 5 decimal places)
5. Calculate x₃:
x₃ = x₂ - f(x₂) / f'(x₂)
Now, substitute the value of x₂ and calculate:
x₃ = (-3.58333) - f(-3.58333) / f'(-3.58333)
Repeat steps 1 to 5 as needed until you reach the desired accuracy.
By following these steps, you should be able to find the third approximation, x₃, to the root of the equation.