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

Question

cos(x^2+3)=x^3 as follows . Let x_1=1 be the initial approximation. x_2?

Answer 1

Given equation :

x³ - cos(x²+3) = 0

Let f(x) = x³ - cos(x²+3)

x₁ = 1

Find x₂ = ?

f '(x) = 3x^3-1 - (- sin(x²+3)) (2x)

f '(x) = 3x² + 2xsin(x²+3)

According to Newton Method 

x₂ = x₁ – [ f(x₁) / f’(x₁) ]

f(x₁) = (x₁)³ - cos(x₁²+3)

f(x₁) = (1)³ - cos(1²+3)

=  1 - cos4

=  1 - 0.99756

= 0.00244

f '(x₁) =  3(x₁)² + 2x₁sin(x₁²+3)

=  3(1)² + 2*1*sin(1²+3)

=  3 + 2sin(4)

=  3 + 0.1395

=  3.1395

x₂ = 1 – [ 0.00244 / 3.1395 ]

x₂ = 1 – 0.0007772

x₂ = 0.999223

Solution :

x₂ = 0.999223