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.