Mathematics, 21.07.2019 02:00 jocelynfray16
How do you find absolute extrema of this function? f(x)= (x)/ (x^2+2); interval is [0,4]
Answers: 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}](/tpl/images/0113/8245/26e97.png)
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How do you find absolute extrema of this function? f(x)= (x)/ (x^2+2); interval is [0,4]...
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