Question

F(x)= -x over x^2+5 on the domain [0,4]

Answer 1

If you're asking for extrema, like the previous posting

well

$\bf f(x)=\cfrac{-x}{x^2+5} \\\\\\ \cfrac{df}{dx}=\cfrac{(x^2+5)+2x^2}{(x^2+5)^2}\implies \cfrac{df}{dx}=\cfrac{5+3x^2}{(x^2+5)^2}\impliedby \begin{array}{llll} using\ the\\ quotient\ rule \end{array}$

like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points

and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4

f(0) = 0        <---- only maximum, and thus absolute maximum

f(4) ≈ - 0.19  <----  only minimum, and thus absolute minimum