Question

For the equation f(x)=x/(x-1), the equilibria are at x=2 and x=0, but at each of these points, the slope is exactly -1. How can I tell if these equilibria are stable or unstable?

Thanks!!

Answer 1

check f'(2) from left and right side unstable is when away from pt both sides <---- . ----> and stable is when ----> . <--- approaching pt from both sides on the graph i.e if at F'(1.9)=- and F'(2.1) = + then its unstable as the arrows wud be going away from F'(2)

Answer 2

theres also semistable where its like ----> . ----> at the point or <----- . <-----

Answer 3

so what happens if f'(1.9) and f'(2.1) are both negative? Is that what you mean by semistable?

Answer 4

yes semistable

Answer 5

u shud just do theres minigraphs like this

down (-)
pt 2----------
down (-)

pt 0 --------

Answer 6

lemme send u a pic i saw online

Answer 7

Ok, thanks.

Answer 8

http://home.comcast.net/~sharov/PopEcol/lec9/equilib.html

Answer 9

hmmm.... it just gives me an error.

Answer 10

ok its like this you see a parabola say like ---> U
at the bottom where its derivative = 0 is an equalibria pt
so u check the derivate of that function from left before it hits that 0 and from right
u will see that the f' from left will be decreasing to increasing at right so that means its going -----> . -----> which wud make it an unstable

Answer 11

also if u want a visual understanding of this concept like imagine a ferris wheel at the top it wud be unstable and bottom wud be stable because when its going down the gravity is pulling and when its going up from the bottom most point the gravity is again pulling like ----> btm pt <---- but on the way up the gravity is pulling it away from the top most point and when its top and slightly down the gravity is once again pulling it away from the top point <---- top pt ----> voila

Answer 12

Ok. Thanks.