What are the critical points and also inflection points of #3e^(-2x(^2))#?

1 Answer
Aug 7, 2015

Assuming that the function is: #f(x) = 3e^(-2x^2)#, the only critical number is: #0#, and the inflection points are: #(1/sqrt12, 3e^(-1/6))# and #(-1/sqrt12, 3e^(-1/6))#

Explanation:

#f(x) = 3e^(-2x^2)#

#f'(x) = 3e^(-2x^2) * (-4x) = -12xe^(-2x^2)#

#f'(x)# is never undefined and is #0# when #x=0#.

The only critical number is #0#.

#f''(x) = -12e^(-2x^2) -12x(-12xe^(-2x^2))#

# = -12e^(-2x^2)(1-12x^2)#

#f''(x) is never undefined, so the only chance it has to change signs is when it is zero.

Solving: #-12e^(-2x^2)(1-12x^2) = 0#, gives us

#x = +-1/sqrt12#

Observe that the sign of #f''(x)# is the opposite of the sign of #1-12x^2# which does, indeed, change sign at both #+1/sqrt12# and #-1/sqrt12#

#f# is an even function so we calculate: #f( +-1/sqrt12) = 3e^(-2(1/12)) = 3e^(-1/6)#

The inflection points are: #(1/sqrt12, 3e^(-1/6))# and #(-1/sqrt12, 3e^(-1/6))#

(Simplify and rationalize if you prefer.)