For the curve \(y = xe^{2x}\)
minimum occurs at \(x = -\dfrac{1}{2}\)
For \(y = xe^{2x}\), \(y' = e^{2x}(1+2x)\), which vanishes at \(x=-\dfrac{1}{2}\). Then \(y'' = 4e^{2x}(x+1)\), and at \(x=-\dfrac{1}{2}\), \(y'' = 4e^{-1}\left(\dfrac{1}{2}\right) = \dfrac{2}{e} > 0\), so the curve has a minimum at \(x = -\dfrac{1}{2}\).
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