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.
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 understoodposted by bobpursley
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