calculus

f'(x) < 0 if |x| < 1

And apparently this means f is decreasing on (-1, 1).

I don't understand this. I know f'(x) < 0 means the function is decreasing. But I don't understand the absolute value. It means I can put -1 or 1 inside, but wouldn't it still be 1 = 1 and not 1 < 1? I don't understand.

asked by Kelly
  1. Consider f(x)=x^2 -1

    f(x) <0 if abs(x)<1
    However, f(x) is not decreasing over that entire range (-1 to 1)

    So your meaning of the problem statement is not understood

    posted by bobpursley
  2. Sorry, I don't understand your reply...

    This question has a prime after the f by the way. It's f prime, aka the derivative. That's why the function is decreasing.

    posted by Kelly

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