Question

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.) 2x3 − 3x2 + 2 = 0, x1 = −1

Answer 1

Answer:

The third approximation is x_3=-0.6825.

The value of f(x) for x_3 is f(x_3)=-0.0332.

Step-by-step explanation:

The Newton's method allow to approximate the root of a function as:

$x_n=x_{n+1}-\frac{f(x_n)}{f'(x_n)}$

For this function, it is:

$x_1=-1\\\\f(x)=2x^3-3x^2+2\\\\f'(x)=6x^2-6x$

For x1=-1

$f(x_1)=2x_1^3-3x_1^2+2=2*(-1)^3-3(-1)^2+2=-2-3+2=-3\\\\f'(x_1)=6(-1)^2-6(-1)=6+6=12\\\\\\x_2=x_1+\frac{f(x_1)}{f'(x_1)} =(-1)-\frac{-3}{12} =-1+0.25=-0.75$

Then, for x2=-0.75

$f(x_2)=2x_2^3-3x_2^2+2=-0.53125\\\\f'(x_1)=6x_2^2-6x_2=7.875\\\\\\x_3=x_2+\frac{f(x_2)}{f'(x_2)} =(-0.75)-\frac{-0.53125}{7.875}=-0.75+ 0.0675=-0.6825$

Then, the third approximation is x_3=-0.6825.

The value of f(x) for x_3 is f(x_3)=-0.0332.