Asked by Clair
find the extreme values of the function and where they occur y=1/(sqrt 1-x^2).
So far I have y'=x/((1-x^2)(1-x^2)^(1/2)) but I am stuck.
So far I have y'=x/((1-x^2)(1-x^2)^(1/2)) but I am stuck.
Answers
Answered by
Steve
Look at that fraction. The domain is (-1,1). As long as x ≠ ±1, the denominator is not zero.
So, the fraction is zero when x=0. On either side of that, y increases, so (0,1) is a minimum.
To check, see the graph at
http://www.wolframalpha.com/input/?i=1%2F%E2%88%9A%281-x^2%29
Lots of times the derivative is a very messy fraction, but to find the extremes, all you need to work with is the numerator.
So, the fraction is zero when x=0. On either side of that, y increases, so (0,1) is a minimum.
To check, see the graph at
http://www.wolframalpha.com/input/?i=1%2F%E2%88%9A%281-x^2%29
Lots of times the derivative is a very messy fraction, but to find the extremes, all you need to work with is the numerator.
Answered by
Clair
Thanks i got it I think I just hiccuped there
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.