If Newton's Method is used to locate a root of the equation f(x)=0f(x)=0 and the initial approximation is x_1=2x1=2, find the second approximation x_2x2?

Full question below


"Suppose that the tangent line to the curve y=f(x)y=f(x) at the point (3,8)(3,8) has the equation y=5-3xy=53x. If Newton's Method is used to locate a root of the equation f(x)=0f(x)=0 and the initial approximation is x_1=2x1=2, find the second approximation x_2x2?"

PLEASE APPLY CALCULUS I METHODS.
I have solved the equation with my own efforts, check my answer please?

1 Answer
Nov 29, 2016

x_2=5/3x2=53

Explanation:

set information given

f(x)=y=5-3xf(x)=y=53x and f'(x)=(5-3x)'=-3(1)=-3

  • f(x)=5-3x, f'(x)=-3

Newton's Method formula x_(n+1)=x_n-(f(x))/(f'(x))

  • translate with f(x),f'(x) \rArrx_(n+1)=x_n-(5-3x_n)/(-3)
  • x_2=x_1-(5-3x_1)/(-3)

calculations

x_2=x_1-(5-3x_1)/(-3)=2-(5-3(2))/(-3)=2-(5-6)/(-3)
=2-((-1)/-3)=2-1/3=6/3-1/3=5/3

therefore the answer is x_2=5/3