Using Newton's method, a root of x^3+3x+7=0 correct to one decimal place is: A. x=1.6 B. x=1.5 C. x=1.4 D. x=1.3 Help please?

1 Answer
Oct 7, 2017

The answer is option C, i.e =-1.4

Explanation:

The Newton's method : To obtain successive approximations of x_1, x_2, x_3,......, start with a guess x_0 and use

x_(n+1)=x_n-f(x_n)/(f'(x_n))

Here,

f(x)=x^3+3x+7

f'(x)=3x^2+3

Let x_0=0

Therefore,

x_1=0-f(0)/(f'(0))=-7/3

x_2=-7/3-(f(-7/3))/(f'(-7/3))=-2.3333+12.704/19.333=-1.676

x_3=-1.676+2.74/11.4=-1.44

x_4=-1.44+0.27/0.19=-1.41

x_5=-1.41+0.0038/8.94=-1.41

graph{x^3+3x+7 [-7.56, 4.92, -2.27, 3.97]}