Question

Use Newton's method with initial approximation
x1 = −2
to find x2, the second approximation to the root of the equation
x3 + x + 3 = 0.

x2 =

I got -7/13 = -.5384615385 but wrong answer and i need help with this one

Use Newton's method to find all roots of the equation correct to six decimal places. (Enter your answers as a comma-separated list.)
6 cos x = x + 1

Answer 1

Newton's method formula goes with\[x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]
So you've got to find f'(x) first by differentaiting f(x), then use newton's method to find the second approximation \(x_2\),
\[f(x)=x^3 + x + 3 \\ \\ f'(x)=3x^2+1\]
\[x_2 \approx -2-\frac{(-2)^3+(-2)+3}{3(-2)^2+1} \\ \\ x_2 \approx -\frac{19}{13} \approx -1.461538462\]

Answer 2

so you use -2- instead of -1- because i got 7/13