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Sagot :
Lets first find the derivative of f:
[tex]\frac{d}{dx}(e^{2x}+e^{-x})=e^{-x}(2e^{3x}-1)[/tex]Now lets set the derivative equal to zero to find the minimum value:
[tex]e^{-x}(2e^{3x}-1)=0[/tex][tex](2e^{3x}-1)=0[/tex][tex]x=\frac{-ln(2)}{3}[/tex]And replacing that value of x, we have:
[tex]e^{2(\frac{-\ln(2)}{3})}+e^{-(\frac{-\ln(2)}{3})}[/tex][tex]e^{\frac{-2\ln (2)}{3}}+e^{(\frac{\ln (2)}{3})}[/tex][tex]\frac{3\sqrt[3]{2}}{2}[/tex]The local minimum of f is at:
[tex](\frac{-ln(2)}{3},\frac{3\sqrt[3]{2}}{2})[/tex]
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