# What are the local extrema of f(x)= xe^-x?

Oct 19, 2016

$\left(1 , {e}^{-} 1\right)$

#### Explanation:

We need to use the product rule: $\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

$\therefore f ' \left(x\right) = x \frac{d}{\mathrm{dx}} \left({e}^{-} x\right) + {e}^{-} x \frac{d}{\mathrm{dx}} \left(x\right)$
$\therefore f ' \left(x\right) = x \left(- {e}^{-} x\right) + {e}^{-} x \left(1\right)$
$\therefore f ' \left(x\right) = {e}^{-} x - x {e}^{-} x$

At a min/max $f ' \left(x\right) = 0$
$f ' \left(x\right) = 0 \implies {e}^{-} x \left(1 - x\right) = 0$
Now, ${e}^{x} > 0 \forall x \in \mathbb{R}$
$\therefore f ' \left(x\right) = 0 \implies \left(1 - x\right) = 0 \implies x = 1$

$x = 1 \implies f \left(1\right) = 1 {e}^{-} 1 = {e}^{-} 1$

Hence, there a single turning point at $\left(1 , {e}^{-} 1\right)$

graph{xe^-x [-10, 10, -5, 5]}