Consider the function #f(x)=e^(-x^(2))#?

Find the following:

  • domain; found: #(-\infty,\infty)#
  • x-intercepts; found: #-x^2=-\infty# thus there is NO x-intercept
  • y-intercept; found: #y=1#
  • symmetry
  • vertical asymptotes; found: NONE
  • horizontal asymptotes; found: #x=0# (exponential function rule?)
  • intervals of increase and decrease
  • local maxima and minima
  • intervals of concavity
  • inflection points

If any characteristics are not present in the function, state "NONE". Then graph the function

I apologize for the confusion regarding the parts I can understand/answer! I have worked out which ones I actually need help with now.

1 Answer
Nov 23, 2016

See below.

Explanation:

Symmetry

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

So the function is even and the graph of the function is symmetric with respect to the #y#-axis.

Increase/Decrease and Extrema

#f'(x) = -2xe^(-x^2)#

Since #e^"real number"# is always positive, the sign of #f'# is the opposite of the sign of #x#.

So #f# is increasing on #(-oo,0)#, has local maximum #f(0) = 1# and then #f# decreases on #(0,oo)#

Concavity and Inflection

#f''(x) = (-2)(e^(-x^2)) + (-2x)(-2xe^(-x^2))#

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

#f''# is defined everywhere and is #0# only at #+-sqrt2/2#.

Analysis of the sign of #f''# shows that the sign of #f''# agrees with that of #2x^2-1#, so:

on #(-oo,-sqrt2/2)#, #f''(x) >0# and the graph of #f# is concave up
on #(-sqrt2/2,sqrt2/2)#, #f''(x) < 0# and the graph of #f# is concave down
on #(sqrt2/2,oo)#, #f''(x) >0# and the graph of #f# is concave up

Therefore, there are two inflection points: #(-sqrt2/2,e^(-1/2))# and #(sqrt2/2,e^(-1/2))#.

(Note that #e^(-1/2) = 1/sqrte# if you prefer to write it that way.)