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
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