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Find the extreme value(s) of the function f(x)=1/(1-x^2).

I tried, and for the first derivative I got -1/(2*sqrt((1-x^2)^3)). I then found critical values of the function to be -1 and 1.
However, the textbook says that the minimum value is 1 at f(0), and I have absolutely no idea where they came up with that answer... Help please!

  • Calculus - ,

    I think there's posibly a piece of information missing in the question as you've stated it, which is that you're only supposed to be looking at the function between the values X = -1 and X = +1, because the function is discontinuous at both of those values, and also takes smaller values than f(X) = 1 outside that range: try X = -2 or +2, and you'll get f(X) = -1/3, which is certainly less than 1.
    Secondly, I think you've got the differentiation wrong: I get df/dx = 2x/(1-x^2)^2, which if you set it to zero gives you just one real turning point at x = 0, for which f(x) = 1 as stated in the textbook.

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