Question

What are the critical numbers of `f(x) = 1-x/(3-x^2)`?

Answer 1

This function has no real critical points

Explanation:

I'll assume you mean "critical points". These are the points on the curve where `f'(x)=0`.

`f(x)=1-x/(3-x^2)`, so, by the quotient rule, `f'(x)=-(3+x^2)/(3-x^2)^2`.

Note that this function blows up to infinity at `x=+-sqrt(3)`, when the denominator is zero.

`f'(x)=0` implies `3+x^2=0`, which has no real solutions - so this functions has no critical points in the real domain.