Given f(x)=1-x+sinx ; Approximate the zero(s) of the function. Use Newton’s Method where x1=2 and continue the process until two successive approximations differ by less than 0.001.

Question

Given f(x)=1-x+sinx ; Approximate the zero(s) of the function. Use Newton’s Method where x1=2  and continue the process until two successive approximations differ by less than 0.001.

anonymous · Answer

Sorry not to actually be answering your question, but dang, it looks like you stole my cat! XD

ganeshie8 · Answer

$$x_{n+1} =   x_n - \dfrac{f(x_n)}{f'(x_n)}$$

anonymous · Answer

@ganeshie8 ok so whats next? is n equal to 0.001?

ganeshie8 · Answer

your first guess is $x_1 = 2$

ganeshie8 · Answer

newton's method gives u the next good guess :

$$x_{1+1} =   x_1 - \dfrac{f(x_1)}{f'(x_1)}   $$

ganeshie8 · Answer

evaluate $f(x_1)$ and  $f'(x_1)$  ,
plug them above ^

anonymous · Answer

i have to go but ill be on tomorrow to review :( thanks!

ganeshie8 · Answer

okay sure :)