# Is f(x)=(x-2)e^x  increasing or decreasing at x=-2 ?

Oct 12, 2016

decreasing.

#### Explanation:

You need to remember these rules:

• If $f ' \left(x\right) > 0$ for every $x$ on some interval I, then $f \left(x\right)$ is increasing on
the interval.
• If f((x)<0 for every $x$ on some interval I, then $f \left(x\right)$ is
decreasing on the interval.
• If f((x)=0 for every $x$ on some interval I, then $f \left(x\right)$ is constant on the interval

So $f \left(x\right) = \left(x - 2\right) {e}^{x}$; to differentiate we need to use the product rule:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$

Applying this rule gives us:

$f ' \left(x\right) = \left(x - 2\right) \left({e}^{x}\right) + \left(1\right) \left({e}^{x}\right)$
$\therefore f ' \left(x\right) = x {e}^{x} - 2 {e}^{x} + {e}^{x}$
$\therefore f ' \left(x\right) = x {e}^{x} - {e}^{x}$
$\therefore f ' \left(x\right) = \left(x - 1\right) {e}^{x}$

When $x = - 2 \implies f ' \left(- 2\right) = \left(- 2 - 1\right) {e}^{-} 2$
When $\therefore f ' \left(- 2\right) = - 3 {e}^{-} 2$

Remember that ${A}^{x} > 0 \forall x \in \mathbb{R} \implies {e}^{-} 2 > 0$

Hence, $f ' \left(- 2\right) = - 3 {e}^{-} 2 < 0$ so we can conclude that when $x = - 2$ then $f \left(x\right)$ is decreasing .

A graph always helps to visualise and confirm these solutions:
graph{(x-2)e^x [-10.114, 7.666, -4.41, 4.48]}