Question

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?

Answer 1

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]}