For the curve y = x e^x,the point: | vedclass

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(A) Given the function $y = f(x) = x e^x$.First,find the first derivative:$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) = 1 \

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$x = -1$ is a minimum

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(A) Given the function $y = f(x) = x e^x$.First,find the first derivative:$\frac{dy}{dx} = \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) = 1 \cdot e^x + x e^x = e^x(1 + x)$.For critical points,set $\frac{dy}{dx} = 0$:$e^x(1 + x) = 0$.Since $e^x$ is never zero,we have $1 + x = 0$,which gives $x = -1$.Next,find the second derivative:$\frac{d^2y}{dx^2} = \frac{d}{dx}(e^x + x e^x) = e^x + (1 \cdot e^x + x e^x) = e^x(2 + x)$.Evaluate the second derivative at $x = -1$:$f''(-1) = e^{-1}(2 - 1) = e^{-1}(1) = \frac{1}{e}$.Since $\frac{1}{e} > 0$,the function has a local minimum at $x = -1$.

For the curve $y = x e^x$,the point:

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