How do you find the critical numbers of #e^(-x^2)#?

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Answer 1 · 2017-05-15T14:15:34

Domain: #(-oo, +oo)#
Range: #(0, 1]#

#f(x) = e^(-x^2)#

#lim_"x->+-oo" f(x) =0#

and

#f(x)>=0 forall x in RR#

#f'(x) = e^(-x^2) * (-2x)# [Chain rule]

#:. f'(x) = 0# at #x=0#

Hence: #f_max = f(0) = e^0 = 1#

Thus the domain of #f(x)# is #(-oo,+oo)# and
the range of #f(x)# is #(0,1]#

This can be seen by the graph of #f(x)# below:

graph{e^(-x^2) [-2.433, 2.435, -1.217, 1.216]}