Question

How do you find absolute extrema of this function? f(x)= (x)/ (x^2+2); interval is [0,4]

Answer 1

$\bf f(x)=\cfrac{x}{x^2+2}\implies \cfrac{dy}{dx}=\cfrac{1(x^2+2)-(x)(2x)}{(x^2+2)^2}\\\\\\ \cfrac{dy}{dx}=\cfrac{x^2+2-2x^2}{(x^2+2)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{2-x^2}{(x^2+2)^2}\implies 0=2-x^2\implies x=\pm\sqrt{2} \\\\\\ \textit{now }-\sqrt{2}\textit{ is outside the range of }[0,4],\textit{ so is only }\sqrt{2}$

the denominator yields no critical points, so is only that one, which IS within the range of [0, 4].

f(0) = 0    and f(4) is about 0.2222...   whilst  f(√2) is about 0.3536

now, doing a first-derivative test, the √2 points is a maximum, and and the 0 and 4 are both minima, from which the 0 is lowest than the 4, notice f(0) = 0 and f(4) is up above that.

so the absolute minimum is f(0), and the absolute maximum is f(√2).