Inflection Point of a Logistic Function

preview_player
Показать описание

Рекомендации по теме
Комментарии
Автор

Thank you so much! This video is the only thing that actually helped me.

alexisowen
Автор

I've noticed something: If we write the derivative of P(t) = M/(1+Ae^(-MKt)) as dP/dt = kP(M-P), where M is the carrying capacity, the inflection point occurs when the P value is half of the carrying capacity. You can see this by taking the second derivative, which is K(MKP(M-P)-2KP^2(M-P)). Set that to zero and we can get MKP(M-P)-2KP^2(M-P) = 0. From this, we get M/2P(M-P)-P^2(M-P) = 0. This means that M/2P(M-P) = P^2(M-P). We can cancel out the M-P term, leaving us with M/2P=P^2, and therefore P=M/2. Then, just plug in the P value into the original equation to solve for t. Do you think that this method is valid?

potatoman
Автор

or you can just set the function equal to half of its limiting value, in this case 50, and solve for x. It is much simpler that way.

DumbleDork