How do you use the first and second derivatives to sketch #f(x)=e^(-x^2)#?

1 Answer
Dec 12, 2016

#f(x)# has a maximum in #x=0#, is strictly increasing for #x<0# and strictly decreasing for #x>0#, is concave down in the interval #(-1/sqrt(2), 1/sqrt(2))#, concave up outside the interval and has two inflection points for #x=+-1/sqrt(2)#

Explanation:

Using the chain rule we can calculate:

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

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

#f''(x) = (-2+4x^2)e^(-x^2)#

So we can see that #f(x)# is strictly increasing for #x<0# and strictly decreasing for #x>0# and hat in #x=0# it has a critical point.

The second derivative is positive for:

#(4x^2-2)>0# or #|x|>1/sqrt(2)#

Considering the sign of the second derivative this critical point is a maximum, and #f(x)# is concave down in the interval #(-1/sqrt(2), 1/sqrt(2))#, concave up outside the interval and has two inflection points for #x=+-1/sqrt(2)#

graph{e^-(x^2) [-1.293, 1.207, -0.15, 1.1]}