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kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
\[y=x^2\lnx+ex^2\], yes that is the constant "e" times x, not the exponential e^x. I end up with \[x^2(\lnx+e)=0 \implies x^2=0, \lnx=e^{e}\] \[\implies x=0, x=\pm e^{e}\] However, shouldn't the function not be defined at x=0, since ln(0) is undefined? Wolfram gives me this: http://www.wolframalpha.com/input/?i=xintercepts+x%5E2ln%7Cx%7C%2Bex%5E2 where one zero is (0,inderteminate)... what does this mean?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
You have it right. x = 0 is NOT in the Domain. What is the limit of the function as x approaches zero?
 one year ago

kirbykirby Group TitleBest ResponseYou've already chosen the best response.1
it would be 0?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Yes. The right limit is zero and the left limit is zero. Nevertheless, it is not continuous.
 one year ago
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