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 + 1 = 0, x_1 = −3
To use Newton's method to find the third approximation, x3, to the root of the equation 1/3x^3 + 1/2x^2 + 1 = 0, we need to follow these steps:
Step 1: Find the derivative of the function f(x).
The derivative of the function f(x) = 1/3x^3 + 1/2x^2 + 1 is f'(x) = x^2 + x.
Step 2: For each approximation, xn, calculate f(xn) and f'(xn).
For the initial approximation, x1 = -3:
- Calculate f(x1): f(-3) = (1/3*(-3)^3) + (1/2*(-3)^2) + 1 = -16.5
- Calculate f'(x1): f'(-3) = (-3)^2 + (-3) = 12
Step 3: Use the formula xn+1 = xn - f(xn) / f'(xn) to find the next approximation, x2.
Using the values calculated in step 2:
x2 = x1 - f(x1) / f'(x1) = -3 - (-16.5) / 12 = -3 + 1.375 = -1.625
Step 4: Repeat step 2 for the new approximation, x2.
- Calculate f(x2): f(-1.625) = (1/3*(-1.625)^3) + (1/2*(-1.625)^2) + 1 = 0.4468
- Calculate f'(x2): f'(-1.625) = (-1.625)^2 + (-1.625) = 1.625
Step 5: Use the formula xn+1 = xn - f(xn) / f'(xn) to find the third approximation, x3.
Using the values calculated in step 4:
x3 = x2 - f(x2) / f'(x2) = -1.625 - 0.4468 / 1.625 = -1.8935
Therefore, the third approximation, x3, to the root of the equation is -1.8935 (rounded to four decimal places).