# 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 ##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.