A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

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
 one year 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.
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.