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then you'll need to know the max/min points. use differentiation to find them and plot the points.

so like two and 1? also how do you know or like get if it has a 0 or not?

\(y=2x(x^2-1)^{-1}\)
\(y'=2(x^2-1)^{-1} -2x(x^2-1)^{-2} \)

how did you get the two and 1?

umm cuz there's a 2 and a 1? i was just trying but i knew it probably wasn't that easy

at the max point, |dw:1355369628774:dw|

wait. is this a precalculus question?

okay so far so good next?

i.e. the graph at those lines will go into infinity. (as shown in the pic on top)

OH! okay! i understand!

OH so if x=0 then 0^2-1 would =-1 and that's where a vertical asyptote in your graph is!!!

okay so now what?

so, what did you get when x=-0.1 and 0.1?

x=.01and .2