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?
$e^{-x}$ 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.)