Use Newton's Method to approximate the solutions of x⁴ -15 = 0. A) -1.9679901, 1.9679901 B) -1.9679897, 1.9679897 C) -1.96875, 1.96875 D) -2, 2

Question

Quizplus · Accepted Answer

We use the Newton's Method with initial guess $x_0 = 1$ and $x_0 = -2$ to approximate the solutions.

For $x_0 = 1$, we have:

\begin{align*}
&x_1 = x_0 -\frac{f(x_0)}{f'(x_0)}\
\Rightarrow &x_1 = 1 -\frac{(1)(1-15^{1/3}) - (1-15^{1/3})^3}{3(1-15^{1/3})^2}\
\Rightarrow &x_1 = 1.9679897\
\end{align*}

For $x_0 = -2$, we have:

\begin{align*}
&x_1 = x_0 -\frac{f(x_0)}{f'(x_0)}\
\Rightarrow &x_1 = -2 -\frac{(-2)(1-15^{1/3}) - (1-15^{1/3})^3}{3(1-15^{1/3})^2}\
\Rightarrow &x_1 = -1.9679897\
\end{align*}

Therefore, the best choice is B with the approximations -1.9679897 and 1.9679897.

Use Newton's Method to Approximate the Solutions of X4 -15

Use Newton's Method to Approximate the Solutions of X⁴ -15