Question

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

• 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

Given curve is $y=x{e}^x$
$\Rightarrow \frac{dy}{dx}=x{e}^x+e^x=e^x(1+x)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=(x+2)e^x$
For maximum or minimu, put $\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=0\Rightarrow x=-1(\because e^x\ne 0)$
Thus ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=-1}=+ve$
Hence, $f\left(x\right)$ is minimum at $x=-1$