For the curve y = xe^x,which of the following | vedclass

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(C) Given the function $y = xe^x$.First,find the first derivative:$\frac{dy}{dx} = x \cdot e^x + e^x \cdot 1 = e^x(x + 1)$.To find the critical points

Vedclass · Accepted Answer

$x = -1$ is a point of minimum.

Vedclass · Answer

(C) Given the function $y = xe^x$.First,find the first derivative:$\frac{dy}{dx} = x \cdot e^x + e^x \cdot 1 = e^x(x + 1)$.To find the critical points,set $\frac{dy}{dx} = 0$:$e^x(x + 1) = 0$.Since $e^x > 0$ for all $x$,we have $x + 1 = 0$,which implies $x = -1$.Now,find the second derivative:$\frac{d^2y}{dx^2} = \frac{d}{dx}[e^x(x + 1)] = e^x(1) + (x + 1)e^x = e^x(x + 2)$.Evaluate the second derivative at $x = -1$:$\left. \frac{d^2y}{dx^2} \right|_{x=-1} = e^{-1}(-1 + 2) = e^{-1}(1) = \frac{1}{e}$.Since $\frac{1}{e} > 0$,the function has a local minimum at $x = -1$.

For the curve $y = xe^x$,which of the following is true?

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