The second approximation x2 is and the third approximation x3 is

Question

Use Newton's method to approximate a root of the equation .

Let  be the initial approximation.

The second approximation  is   

and the third approximation  is

Answer 1

f(x) = e^{0.6x^2}+x-2

f '(x) = 1.2xe^{0.6x^2} + 1

x_n+1 = x_{n - (}e^{0.6x^2}+x-2)/1.2xe^{0.6x^2} + 1

x₁ = 2

x₂ = 2- 11.023/27.4556 = 1.5985

x₃ = 1.5985 - 4.23111/12.1182 = 1.1705

Note: x eventually settles to 0.6801 after 8 iterations.