Practicing Success
For the curve \(y=xe^{x}\), the point |
\(x=-1\) is a point of minimum \(x=0\) is a point of minimum \(x=-1\) is a point of maximum \(x=0\) is a point of maximum |
\(x=-1\) is a point of minimum |
\(\begin{aligned}\frac{dy}{Dx}&=e^{x}(x+1)\\ \text{so }\frac{dy}{Dx}&=0\Rightarrow x=-1\\ \frac{d^{2}y}{dx^{2}}&=e^{x}(x+1)+e^{x}\\ \frac{d^{2}y}{dx^{2}}\left|_{x=-1}\right.&=e^{-1}>0\end{aligned}\) |