Question

Maximum value of ${(x-1)}^2{e}^x$

• A. $4.$
• B. $4e.$
• C. $4/e.$
• D. $e.$

Answer 1

Answer: (C)

$4/e.$

$y={(x-1)}^2{e}^x$
$\frac{dy}{dx}={(x-1)}^2.e^x+2(x-1)e^x$
$=e^x(x^2-2x+1+2x-2)$
$\frac{dy}{dx}=e^x(x^2-1)$
For maxima or minima,
$\frac{dy}{dx}=0$
$\Rightarrow e^x(x^2-1)=0$
$\Rightarrow x=1,-1$ ($\because e^x$ cannot be 0.)
Now, $\frac{d^{2}y}{d{x}^2}=e^{(}x)2x+(x^2-1)e^x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=e^x(x^2+2x-1)$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-2}{e}<0$ at $x=-1$
Hence, $x=-1$ is point of maxima.
Maximum value $=\frac{4}{e}$