# Find the absolute maxmium of the function $f(x) = x \cdot {e}^{-x}$

How can I find the absolute maximum of this exponential function?

$f(x) = x \cdot {e}^{-x}$

I know that the first step is to take the derivative of the function, like so:

${f}^{\prime}(x) = x \cdot {e}^{-x}(-1) + {e}^{-x}(1)$

${f}^{\prime}(x) = {e}^{-x}(1 - x)$

Then the next step is to set it equal to zero and find the critical points for the first derivative test. How can I do that with an exponential function?

Could someone please show how they have done that?

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$e^{-x}$ is never 0; but that's good here, because you can "throw it away": $$e^{-x}(1-x)=0\iff 1-x=0\iff x=1.$$
(A product is 0 if and only if one of the factors is 0; so, since $e^{-x}$ is never 0, the first equation above holds if and only if the second one does. Alternatively, dividing both sides of the first equation above by $e^{-x}$ is justified.)
ohhhh... You divide through by ${e}^{-x}$. Can't believe I didn't see that. Thank you! Points up and answer accepted! – Oliver Spryn Jan 26 '12 at 1:24