How do you find the inflection point of a logistic function?

The answer is #((ln A)/k, K/2)#, where #K# is the carrying capacity and #A=(K-P_0)/P_0#.

To solve this, we solve it like any other inflection point; we find where the second derivative is zero.

#P(t)=K/(1+Ae^(-kt))#
#=K(1+Ae^(-kt))^(-1)#
#P'(t)=-K(1+Ae^(-kt))^(-2)(-Ake^(-kt))# power chain rule
#P''(t)=2K(1+Ae^(-kt))^(-3)(-Ake^(-kt))^2-K(1+Ae^(-kt))^(-2)(Ak^2e^(-kt))# product and chain rule

Now we solve:

#2K(1+Ae^(-kt))^(-3)(-Ake^(-kt))^2-K(1+Ae^(-kt))^(-2)(Ak^2e^(-kt))=0#
#2(1+Ae^(-kt))^(-1)(-Ake^(-kt))^2-(Ak^2e^(-kt))=0# cancel
#2(1+Ae^(-kt))^(-1)(Ake^(-kt))^2-k(Ake^(-kt))=0# factor out
#2(1+Ae^(-kt))^(-1)(Ake^(-kt))-k=0# cancel
#2(1+Ae^(-kt))^(-1)(Ake^(-kt))=k#
#2Ake^(-kt)=k(1+Ae^(-kt))# cancel
#2Ae^(-kt)=1+Ae^(-kt)#
#2Ae^(-kt)-Ae^(-kt)=1#
#Ae^(-kt)=1#
#e^(-kt)=1/A#
#-kt=-lnA# log rules
#t=(lnA)/k#

This gives us #t# which we can substitute into #P(t)#:

#P((lnA)/k)=K/(1+Ae^(-k(lnA)/k))#
#=K/(1+Ae^(-(lnA)))# log rules
#=K/(1+A/A)#
#=K/(1+1)#
#=K/2#

It's a lot of algebra, so be very careful with factoring, cancelling, and negative signs.