Question

The curve $y=x{e}^x$ has minimum value equal to

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

Answer 1

The correct option is A

$-\frac{1}{e}$

Explanation for correct answer:

Finding the minimum value:

Given that the equation of the curve is $y=x{e}^x$

Differentiating the curve w.r.t. '$x$'

$$\begin{gathered} \frac{dy}{dx}=x.e^x+e^x \\ =e^x\left(x+1\right) \end{gathered}$$

For minimum value ,$\frac{dy}{dx}=0$

$$\begin{gathered} \therefore e^x\left(x+1\right)=0 \\ \Rightarrow x=-1 \end{gathered}$$

Now, differentiating it again we get

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=e^x\left(x+1\right)+e^x \\ =e^x\left(x+2\right) \end{gathered}$$

At $x=-1$,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=e^{-1}\left(-1+2\right) \\ =\frac{1}{e} \\ >0 \end{gathered}$$

At $x=-1$,$\Rightarrow \frac{d^{2}y}{d{x}^2}>0$

Therefore, $y$ is minimum at $x=-1$.and

$$\begin{gathered} y_{min}=-1.e^{-1} \\ =-\frac{1}{e} \end{gathered}$$

Hence, option (A) is the correct answer