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

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Answer 1 · 2015-11-22T18:28:59

$P(1/3, 1/(3e))$

First we derivate this, so

$y = xe^(-3x)$

$ln(y) = ln(x) -3x$

$1/y*dy/dx = 1/x - 3$

$dy/dx = e^(-3x) -3xe^(-3x)$

Critical points are those where the derivative is 0, so

$0 = e^(-3x) - 3xe^(-3x)$

Or

$3xe^(-3x) = e^(-3x)$

Because the exponential is never 0 you can divide it out

$3x = 1$
$x = 1/3$

The function has a critical point, in $P(1/3, 1/(3e))$

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