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

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:

For this function, it is:

For x1=-1

Then, for x2=-0.75

Then, the third approximation is x_3=-0.6825.

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

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step-by-step explanation:

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step-by-step explanation: