How do you find the critical points of: #f(x)=(x)(e^(-x^2))#?

1 Answer
Oct 21, 2015

Answer:

See the explanation section, below.

Explanation:

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

#f'(x)=(1)(e^(-x^2)) + (x)(e^(-x^2)(-2x))# (product and chain rules)

# = e^(-x^2)(1-2x^2) #

#f'(x)# is never undefined and is #0# at #x = +-1/sqrt2#.

Both #1/sqrt2# ans #-1/sqrt2# are in the domain of #f# (Domain is #RR#), so both are critical numbers for #f#.

Note

Some treatments have "critical points" being points in 2-space, so we also find #y = f(x)# at these critical numbers. I do not use that terminology, but if your teacher does, then you should find #(1/sqrt2 , 1/(sqrt2root4e))# and #(-1/sqrt2 , -1/(sqrt2root4e))# (or whatever form is expected).