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dy/dx = e^x^2 fin… - QuestionCove
OpenStudy (anonymous):

dy/dx = e^x^2 find coordinates of any turning points??

4 years ago
OpenStudy (debbieg):

Turning points would mean that the graph of the function goes from increasing to decreasing or vice-versa, which means that there would be a local min or max, which means that the 1st derivative would = 0. so..... for what value of x would \(\large e^{x^2}=0\)?

4 years ago
OpenStudy (anonymous):

and then how do I get the point(s)?

4 years ago
OpenStudy (debbieg):

Well, if/when you find values of x that solve that equation, you would need the anti-derivative. But don't worry about that just yet. What do you think about the solutions to the equation, \(\large e^{x^2}=0\)? Hmmmm?

4 years ago
OpenStudy (isaiah.feynman):

That equation doesn't seem to have real solutions.

4 years ago
OpenStudy (debbieg):

That's correct. :) Because x^2 is always positive, and when you take a positive number (e) to a postive power, you won't ever get 0.

4 years ago
OpenStudy (anonymous):

the answer is supposed to be 18

4 years ago
OpenStudy (debbieg):

How so? That isn't coordinates of a point. Are you sure that you're reading the problem correctly?

4 years ago
OpenStudy (debbieg):

I mean, you said "find coordinates of any turning points".... so how is "18" the coordinates of a turning point?

4 years ago
OpenStudy (isaiah.feynman):

|dw:1380226383842:dw| @yaysocks

4 years ago
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