Question

Use Newton's method with initial approximation x1 = −2 to find x2, the second approximation to the root of the equation x3 + x + 7 = 0. (Round your answer to four decimal places.)

Answer 1

Answer:

$x_2 \approx -1.769$

Step-by-step explanation:

Let $f(x)=x^3+x+7$

So $f'(x)=3x^2+1$

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

Let $x_1=-2$

We are going to find $x_2$

So we are evaluating $-2-\frac{f(-2)}{f'(-2)}$

First step find f(-2)

Second step find f'(-2)

Third step plug in those values and apply PEMDAS!

$f(-2)=(-2)^3+(-2)+7=-8-2+7=-10+7=-3$

$f'(-2)=3(-2)^2+1=3(4)+1=12+1=13$

So

$x_2=-2-\frac{-3}{13} \\\\ x_2=\frac{-26+3}{13} \\\\ x_2=\frac{-23}{13} \\\\ x_2 \approx -1.769$