Asked by Bob
                Find the points of intersection of the curves y=2sin(x-3) and y=-4x^2 + 2
            
            
        Answers
                    Answered by
            Reiny
            
    tough question!
I used Newton's Method
and I let
y = 2sin(x-3) + 4x^2 - 2
then dy/dx = 2cos(x-3) + 8x
x<sub>new</sub> = x - (2sin(x-3) + 4x^2 - 2)/(2cos(x-3) + 8x)
I started with x = 1
and my first x<sub>new</sub> was.974691
make that your current x and find the next x<sub>new</sub>
I quickly converged to
x = 0.974237
set your calculator to radians and test my answer, it works making both equations equal.
(error was .000000418 between LS and RS
    
I used Newton's Method
and I let
y = 2sin(x-3) + 4x^2 - 2
then dy/dx = 2cos(x-3) + 8x
x<sub>new</sub> = x - (2sin(x-3) + 4x^2 - 2)/(2cos(x-3) + 8x)
I started with x = 1
and my first x<sub>new</sub> was.974691
make that your current x and find the next x<sub>new</sub>
I quickly converged to
x = 0.974237
set your calculator to radians and test my answer, it works making both equations equal.
(error was .000000418 between LS and RS
                    Answered by
            Mgraph
            
    Reiny found a positive root but there is a negative root x=-0.55 (approximately)
We can started with x=-1
    
We can started with x=-1
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.