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

Jun 1, 2018

This function has no real critical points

#### Explanation:

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

$f \left(x\right) = 1 - \frac{x}{3 - {x}^{2}}$, so, by the quotient rule, $f ' \left(x\right) = - \frac{3 + {x}^{2}}{3 - {x}^{2}} ^ 2$.

Note that this function blows up to infinity at $x = \pm \sqrt{3}$, when the denominator is zero.

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