The function $f(x)=(x+2) e^{-x}$ is
The function $f(x)=(x+2) e^{-x}$ is
- decreasing in $(-\infty,-1)$ and increasing in $(-1, \infty)$
- decreasing for all $x$
- increasing in $(-\infty,-1)$ and decreasing in $(-1, \infty)$
- increasing for all $x$
Solution
Given
$\begin{array}{l}
f(x)=(x+2) e^{-x} \\
\begin{aligned}
\therefore f^{\prime}(x) &=(x+2)\left(e^{-x}\right)(-1)+e^{-x}(1) \\
&=e^{-x}[1-(x+2)]=e^{-x}(-x-1) \\
&=-e^{-x}(x+1)
\end{aligned}
\end{array}$
Here $e^{-x}$ is always positive.
$\begin{array}{l}
\text { Now }-(x+1)>0 \Rightarrow-x>1 \Rightarrow x < -1 \\
\text { Also }-(x+1) < 0 \Rightarrow-x < 1 \Rightarrow x>-1
\end{array}$
Thus $\mathrm{f}(\mathrm{x})$ increases in $(-\infty,-1)$ and decreases in $(-1, \infty)$
Asked in: MHT CET 2020 (19 Oct Shift 1)
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