Question

Question part points submissions used 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

Set $f(x)=2x^3-3x^2+2$. Find the tangent line $\ell_1(x)$ to $f(x)$ at the point when $x=x_1$:

$f'(x)=6x^2-6x\implies f'(x_1)=12$ (slope of $\ell_1$)

$\implies\ell_1(x)=12(x-x_1)+f(x_1)=12(x+1)-3=12x+9$

Set $x_2=-\dfrac9{12}$, the root of $\ell_1(x)$. The tangent line $\ell_2(x)$ to $f(x)$ at $x=x_2$ has slope and thus equation

$f'(x_2)=\dfrac{63}8\implies\ell_2(x)=\dfrac{63}8\left(x+\dfrac9{12}\right)-\dfrac{17}{32}=7x+\dfrac{151}{32}$

which has its root at $x_3=-\dfrac{151}{224}\approx-0.6741$.

(The actual value of this root is about -0.6777)