Question
Find the intervals in which $f(x) = (x + 2)e^{-x}$ is increasing or decreasing.
✓
Answer
$f(x) = (x + 2)e^{-x}$
$f'(x) = -e^{-x}(x + 2) + e^{-x}$
$= -xe^{-x} - 2e^{-x} + e^{-x}$
$= -xe^{-x} - e^{-x}$
$= e^{-x}(-x - 1)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$\Rightarrow e^{-x}(-x - 1) > 0$
$\Rightarrow -x - 1 > 0$
$\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
$\Rightarrow -x > 1$
$\Rightarrow x < -1$
$\Rightarrow\text{x}\in(-\infty,-1)$
So, $f(x)$ is increasing on $(-\infty,-1).$
For $f(x)$ to be decreasing, we must have
$f'(x) < 0$
$\Rightarrow e^{-x}(-x - 1) < 0$
$\Rightarrow -x - 1 < 0$
$\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
$\Rightarrow -x < 1$
$\Rightarrow x < -1$
$\Rightarrow\text{x}\in(-1,\infty)$
So,$ f(x)$ is decreasing on $(-1,\infty).$
$f'(x) = -e^{-x}(x + 2) + e^{-x}$
$= -xe^{-x} - 2e^{-x} + e^{-x}$
$= -xe^{-x} - e^{-x}$
$= e^{-x}(-x - 1)$
For $f(x)$ to be increasing, we must have
$f'(x) > 0$
$\Rightarrow e^{-x}(-x - 1) > 0$
$\Rightarrow -x - 1 > 0$
$\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
$\Rightarrow -x > 1$
$\Rightarrow x < -1$
$\Rightarrow\text{x}\in(-\infty,-1)$
So, $f(x)$ is increasing on $(-\infty,-1).$
For $f(x)$ to be decreasing, we must have
$f'(x) < 0$
$\Rightarrow e^{-x}(-x - 1) < 0$
$\Rightarrow -x - 1 < 0$
$\big[\because\ \text{e}^{-\text{x}}>0,\forall\ \text{x}\in\text{R}\big]$
$\Rightarrow -x < 1$
$\Rightarrow x < -1$
$\Rightarrow\text{x}\in(-1,\infty)$
So,$ f(x)$ is decreasing on $(-1,\infty).$
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