find the extreme values of the function and where they occur y=1/(sqrt 1-x^2).

Question

So far I have y'=x/((1-x^2)(1-x^2)^(1/2)) but I am stuck.

Answer 1

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.

Answer 2

Thanks i got it I think I just hiccuped there