Respuesta :
[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.
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.
