Use newton's method with initial approximation x1 = 1 to find x2, the second approximation to the root of the equation x4 ? x ? 6 = 0. x2 =

Next time please correctly write the equation. I believe in this case it would be:

x^4 – x – 6 = 0

Correct me if I am wrong though.

Now let us go back in answering this problem. Using newton’s method of approximation, the value of x2 would be:

x2 = x1 – [f(x1) / f’(x1)]

Since x1 = 1, therefore:

f(x1) = 1^4 – 1 – 6 = - 6

And the 1st derivative of f(x) is f’(x) so:

f’(x) = 4 x^3 – 1

f’(x1) = 4 (1)^3 – 1 = 3

Calculating for x2 now:

x2 = 1 – [- 6 / 3]

x2 = 1 – [- 2]

x2 = 3