Question

Find the absolute extrema if they exist as well as all values of x where they occur for the function f(x)=-x/x^2+5

Answer 1

$\bf f(x)=\cfrac{x}{x^2+5} \\\\\\ \cfrac{df}{dx}=\cfrac{(x^2+5)-2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5-x^2}{(x^2+5)^2}\impliedby \begin{array}{llll} using\ the\\ quotient\ rule \end{array}\\\\ -------------------------------\\\\ 0=\cfrac{5-x^2}{(x^2+5)^2}\implies 0=5-x^2\implies x^2=5\implies x=\pm\sqrt{5} \\\\\\ f(\sqrt{5})\approx 0.2236\impliedby \textit{only maximum, thus absolute maximum} \\\\\\ f(-\sqrt{5})\approx -0.2236\impliedby \textit{only minimum, thus absolute minimum}$

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.