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kirbykirby

  • 2 years ago

x-intercepts of: y=x^2*ln|x|+ex^2

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  1. kirbykirby
    • 2 years ago
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    \[y=x^2\ln|x|+ex^2\], yes that is the constant "e" times x, not the exponential e^x. I end up with \[x^2(\ln|x|+e)=0 \implies x^2=0, \ln|x|=e^{-e}\] \[\implies x=0, x=\pm e^{-e}\] 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=x-intercepts+x%5E2ln%7Cx%7C%2Bex%5E2 where one zero is (0,inderteminate)... what does this mean?

  2. tkhunny
    • 2 years ago
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    You have it right. x = 0 is NOT in the Domain. What is the limit of the function as x approaches zero?

  3. kirbykirby
    • 2 years ago
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    it would be 0?

  4. tkhunny
    • 2 years ago
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    Yes. The right limit is zero and the left limit is zero. Nevertheless, it is not continuous.

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