[tex]\bf f(x)=\cfrac{x}{x^2+5}
\\\\\\
\cfrac{df}{dx}=\cfrac{(x^2+5)-2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5-x^2}{(x^2+5)^2}\impliedby
\begin{array}{llll}
using\ the\\
quotient\ rule
\end{array}\\\\
-------------------------------\\\\
0=\cfrac{5-x^2}{(x^2+5)^2}\implies 0=5-x^2\implies x^2=5\implies x=\pm\sqrt{5}
\\\\\\
f(\sqrt{5})\approx 0.2236\impliedby \textit{only maximum, thus absolute maximum}
\\\\\\
f(-\sqrt{5})\approx -0.2236\impliedby \textit{only minimum, thus absolute minimum}[/tex]
we also get critical points when the denominator is 0, namely (x²+5)² = 0
however, this denominator, doesn't give us any critical points
critical points when the denominator is 0, are usually asymptotic or "cusps", where the derivative is not continuous, but has an extrema.
