How do you find the critical points for $xe^x$ ?

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Answer 1 · 2017-01-13T13:32:11

$x=-1$ is a local minimum for $xe^x$

Critical points for a function are identified equating its first derivative to zero.

As:

$d/(dx) (xe^x) = (x+1)e^x$

its critical points are the roots of the equation:

$(x+1)e^x = 0$

or, as $e^x > 0$ for every $x$ :

$x+1=0$

that is:

$x=-1$

We can easily see that:

$d/(dx) (xe^x) < 0$ for $x<-1$

$d/(dx) (xe^x) > 0$ for $x> -1$

which means that $x=-1$ is a local minimum.

How do you find the critical points for xe^xxe^x?