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

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Answer 1 · 2018-06-01T20:49:25

This function has no real critical points

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.

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