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Zarrin [17]
3 years ago
11

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

Mathematics
1 answer:
DIA [1.3K]3 years ago
4 0
\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.
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Step-by-step explanation:

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