Question

The point for the curve $y=x{e}^x$

• A. $x=–1$is minimum
• B. $x=0$ is minimum
• C. $x=–1$is maximum
• D. $x=0$ is maximum

Answer 1

The correct option is A

$x=–1$is minimum

Find the maximum or minimum point satisfying the condition

Given: curve is $y=x{e}^x$

On differentiating w.r.t to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=x{e}^x+e^x \\ \frac{dy}{dx}=e^x(1+x) \end{gathered}$$

Again differentiating ,we get
$\frac{d^{2}y}{d{x}^2}=(x+2)e^x$

For maximum or minimum, put $\frac{dy}{dx}=0$
$$\begin{gathered} e^x\left(1+x\right)=0 \\ \Rightarrow x=-1\left[\because e^x\ne 0\right] \end{gathered}$$

Thus ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-1}=+ve$ value
Therefore, the curve is minimum at $x=-1$

Hence, option (A) is the correct answer.