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kirbykirby
 2 years ago
Best 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?

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

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.0Yes. The right limit is zero and the left limit is zero. Nevertheless, it is not continuous.
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