Question #db8e2

Jul 10, 2016

For increasing you have to find out where the first derivative is positive, for decreasing where first derivative is negative

Explanation:

Let's calculate the first derivative of the function $f \left(x\right)$:

$\left[\left(x + 2\right) \cdot {e}^{-} x\right] '$ = $\left(x + 2\right) ' \cdot {e}^{-} x + \left(x + 2\right) \cdot \left({e}^{-} x\right) ' = 1 \cdot {e}^{-} x + \left(x + 2\right) \left(- {e}^{-} x\right) = {e}^{-} x - x {e}^{-} x - 2 {e}^{-} x = - x {e}^{-} x - {e}^{-} x = {e}^{-} x \left(- x - 1\right)$

Now, regardless of the value of $x$, ${e}^{-} x$ is always positive, so the sign of the first derivative depends on whether $\left(- x - 1\right)$ is positive or negative. That is:

• If $\left(- x - 1\right) > 0$, the first derivative is positive and the function is increasing. Thus, if $- 1 > x$ the function is increasing
• Conversely, if $\left(- x - 1\right) < 0$, the first derivative is negative and the function is decreasing. Thus, if $- 1 < x$ the function is decreasing