hi!
this is the function I'm looking at:
f(x)= {(x^3-3x) /(1-x^2)}
i got to f'(x)= (-x^4-3)/(1-x^2)
and so I'm having a hard time finding the max and min
(f'(x)=0)
What am I'm doing wrong?
thanks!
Hi,
there is a small typo in your derivative:
$\displaystyle f'(x)=\frac{-x^4-3}{(1-x^2)^2}$
A quotient equals zero is the denominator is not zero and the nurmerator equals zero. Therefore you only have to solve for x:
$\displaystyle -x^4-3=0~\iff~x^4=-3$ . This equation doesn't have any real root and that means your function hasn't any extreme values.