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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#?

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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

Scarlett Allison

#x_2=5/3#

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

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

#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#

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Answer 2

Aurora Alvarado

To find the second approximation (x_2), we use Newton's Method formula:

[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ]

Given the initial approximation (x_1 = 2), we first need to find (f(x_1)) and (f'(x_1)). If (f(x) = 0), then (f'(x)) is the derivative of (f(x)).

If you provide the function (f(x)), I can help you calculate (f(x_1)) and (f'(x_1)), and then we can find (x_2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.