Question

The curve $y = xe^x$ has minimum value equal to

• A. $- \frac{1}{e}$
• B. $\frac{1}{e}$
• C. #NAME?
• D. e

Answer 1

Answer: (A)

$- \frac{1}{e}$

Answer 2

Let $y = xe^x$. Differentiate both side w.r.t. �$x$�. $\Rightarrow \frac{dy}{dx} =e^{x} + xe^{x}=e^{x} \left(1+x\right)$ Put $\frac{dy}{dx} = 0$ $\Rightarrow e^{x}\left(1+x\right)=0$ $\Rightarrow x=-1$ Now, $\frac{d^{2}y}{dx^{2}} = e^{x} + e^{x} \left(1+x\right)=e^{x} \left(x+2\right)$ $\left(\frac{d^{2}y}{dx^{2}}\right) _{\left(x=-1\right)} = \frac{1}{e} + 0>0$ Hence, $y = xe^x$ is minimum function and $y_{\min} = - \frac{1}{e}$