I have the answer can somone please check me
f(x)=(500x)/(x^2+1)
has a relative maximum at x=? and a relative minimum at x=?

Question

I have the answer can somone please check me
f(x)=(500x)/(x^2+1)
has a relative maximum at x=? and a relative minimum at x=?

agent0smith · Answer

anonymous · Answer

so then you got 0 for the maximum and minimum right

agent0smith · Answer

x will not be zero.

anonymous · Answer

that is what I got so that is why I am questioning yall

anonymous · Answer

$$f'(x)={500\over x^2+1}-{1000x^2\over (x^2+1)^2}=0\
500(x^2+1)-1000x^2=0\
1-x^2=0\
\boxed{x=1}\qquad\boxed{x=-1}
$$

anonymous · Answer

one of them is the maxima, the other a minima.

anonymous · Answer

oh ok I understand what I did wrong so then the maximum is 1  mininmum is -1

anonymous · Answer

yep

anonymous · Answer

ok thanks