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AERONIK Group TitleBest ResponseYou've already chosen the best response.0
at x=2 function is not defind.
 2 years ago

AERONIK Group TitleBest ResponseYou've already chosen the best response.0
so the function is discontinuous at that point.....
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We have discontinuities whenever the DENOMINATOR is zero. So let's set the denominator equal to 0, and solve for x! Those values of X are the trouble makers! :O \[\large (x^22x)=0\]Factoring an x out of each term gives us: \[\large x(x2)=0\] Know what to do from here to solve for x? :D Need to take advantage of the Zero Factor Property! :D
 2 years ago

arifelgon94 Group TitleBest ResponseYou've already chosen the best response.0
thnk you... but how did you solve it?
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Aeron mentioned one of the values, but there is also another one!! Make sure you don't miss that trouble maker! :3
 2 years ago
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