Newton's method to approximate a root

Question

use newtons law to approximate the root of 2x^7+3x^4+4=0 let x1=3 find second and third apporximation of

carry atleast 4 decimal places in your calculations

Answer 1

x_n+1=x_n-f(x_n)/f'(x_n) = x_n-(2x⁷+3x⁴+4)/(14x⁶+12x³)

x₁ = 3

x₂ = 2.56115

x₃ = 2.18099

Actually, it takes around 50 iterations to settle to the root of -1.3

Answer 2

Newton's method to iterate towards a root is x_n+1=x_n - f(x_n)/f'(x_n) with x₁ = 3 (apparently that is the first approximation, although I would call it x₀ . This is a silly starting guess as the zero must be a negative number (about -1.3) and Newton's method is sensitive to initial guess. This may also fail as f'(0) = 0 which can lead to NM not working) - Maybe the equation is copied wrong.

Work out general expression for f'(x) = 14x⁶+12x³

Plug into the formula after subbing values for f(3) and f'(3):

x₂ = 3 - (2(3)⁷+3(3)⁴+4)/(14(3)⁶+12x³) = (I get 2.56116)

The next step is to do the same process for x₃ from x₂

Please consider a tutor. Take care.