# What are the critical values, if any, of f(x)=xe^(-3x)?

Nov 22, 2015

$P \left(\frac{1}{3} , \frac{1}{3 e}\right)$

#### Explanation:

First we derivate this, so

$y = x {e}^{- 3 x}$

$\ln \left(y\right) = \ln \left(x\right) - 3 x$

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x} - 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{- 3 x} - 3 x {e}^{- 3 x}$

Critical points are those where the derivative is 0, so

$0 = {e}^{- 3 x} - 3 x {e}^{- 3 x}$

Or

$3 x {e}^{- 3 x} = {e}^{- 3 x}$

Because the exponential is never 0 you can divide it out

$3 x = 1$
$x = \frac{1}{3}$

The function has a critical point, in $P \left(\frac{1}{3} , \frac{1}{3 e}\right)$