# 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##.

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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
##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.

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