Question

If Newton's Method is used to locate a root of the equation `f(x)=0` and the initial approximation is `x_1=2`, find the second approximation `x_2`?

Full question below

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"Suppose that the tangent line to the curve `y=f(x)` at the point `(3,8)` has the equation `y=5-3x`. If Newton's Method is used to locate a root of the equation `f(x)=0` and the initial approximation is `x_1=2`, find the second approximation `x_2`?"

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

Answer 1

`x_2=5/3`

Explanation:

set information given

`f(x)=y=5-3x` 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`