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

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

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.1We 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

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

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