Question

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

Answer 1

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.