Question

for the function (f(x) = -2x^3 + 3x + 2), write newtons formula, (n(x) = x - \frac{f(x)}{f(x)}), for solving (f(x) = 0).

(n(x) = )

Explanation:

Find the derivative of the function

To construct Newton's formula, we first need the derivative of the given function \(f(x) = -2x^3 + 3x + 2\). We apply the power rule for differentiation:

$$ f'(x) = \frac{d}{dx}(-2x^3 + 3x + 2) = -6x^2 + 3 $$

Substitute into Newton's formula

Newton's formula is defined as:

$$ N(x) = x - \frac{f(x)}{f'(x)} $$

Substituting \(f(x) = -2x^3 + 3x + 2\) and \(f'(x) = -6x^2 + 3\) into the formula gives:

$$ N(x) = x - \frac{-2x^3 + 3x + 2}{-6x^2 + 3} $$

Simplify the expression

We can simplify the negative signs in the fraction by factoring out \(-1\) from both the numerator and the denominator, or by rewriting the denominator as \(3 - 6x^2\):

$$ N(x) = x - \frac{-2x^3 + 3x + 2}{3 - 6x^2} $$

Alternatively, factoring out \(-1\) from the fraction:

$$ N(x) = x + \frac{-2x^3 + 3x + 2}{6x^2 - 3} $$

Or keeping it in its direct substitution form:

$$ N(x) = x - \frac{-2x^3 + 3x + 2}{-6x^2 + 3} $$

Both forms are mathematically equivalent and standard for Newton's method.

Answer:

For the function \(f(x) = -2x^3 + 3x + 2\), write Newton's formula, \(N(x) = x - \frac{f(x)}{f'(x)}\), for solving \(f(x) = 0\).

\(N(x) =\) <blank>\(x - \frac{-2x^3 + 3x + 2}{-6x^2 + 3}\)</blank>