For the curve \(y=xe^{x}\), the point
Answer & explanation
Correct answer: option 1
\(\begin{aligned}\frac{dy}{Dx}&=e^{x}(x+1)\\ \text{so }\frac{dy}{Dx}&=0\Rightarrow x=-1\\ \frac{d^{2}y}{dx^{2}}&=e^{x}(x+1)+e^{x}\\ \frac{d^{2}y}{dx^{2}}\left|_{x=-1}\right.&=e^{-1}>0\end{aligned}\)