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
Jim H · mason m
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.)

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