Mathematics, 21.07.2019 02:00 dthompson365
How do you find the absolute extrema of this function? f(x) = x+3x^(2/3); interval is [-10,1]
Answers: 1
![\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}} \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x} \\\\\\ (-2)^3=x\implies \boxed{-8=x}\\\\ -------------------------------\\\\ 0=\sqrt[3]{x}\implies \boxed{0=x}](/tpl/images/0113/8255/af2cc.png)
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How do you find the absolute extrema of this function? f(x) = x+3x^(2/3); interval is [-10,1]...
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