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

Aug 25, 2014

The answer is $\left(\frac{\ln A}{k} , \frac{K}{2}\right)$, where $K$ is the carrying capacity and $A = \frac{K - {P}_{0}}{P} _ 0$.

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

$P \left(t\right) = \frac{K}{1 + A {e}^{- k t}}$
$= K {\left(1 + A {e}^{- k t}\right)}^{- 1}$
$P ' \left(t\right) = - K {\left(1 + A {e}^{- k t}\right)}^{- 2} \left(- A k {e}^{- k t}\right)$ power chain rule
$P ' ' \left(t\right) = 2 K {\left(1 + A {e}^{- k t}\right)}^{- 3} {\left(- A k {e}^{- k t}\right)}^{2} - K {\left(1 + A {e}^{- k t}\right)}^{- 2} \left(A {k}^{2} {e}^{- k t}\right)$ product and chain rule

Now we solve:

$2 K {\left(1 + A {e}^{- k t}\right)}^{- 3} {\left(- A k {e}^{- k t}\right)}^{2} - K {\left(1 + A {e}^{- k t}\right)}^{- 2} \left(A {k}^{2} {e}^{- k t}\right) = 0$
$2 {\left(1 + A {e}^{- k t}\right)}^{- 1} {\left(- A k {e}^{- k t}\right)}^{2} - \left(A {k}^{2} {e}^{- k t}\right) = 0$ cancel
$2 {\left(1 + A {e}^{- k t}\right)}^{- 1} {\left(A k {e}^{- k t}\right)}^{2} - k \left(A k {e}^{- k t}\right) = 0$ factor out
$2 {\left(1 + A {e}^{- k t}\right)}^{- 1} \left(A k {e}^{- k t}\right) - k = 0$ cancel
$2 {\left(1 + A {e}^{- k t}\right)}^{- 1} \left(A k {e}^{- k t}\right) = k$
$2 A k {e}^{- k t} = k \left(1 + A {e}^{- k t}\right)$ cancel
$2 A {e}^{- k t} = 1 + A {e}^{- k t}$
$2 A {e}^{- k t} - A {e}^{- k t} = 1$
$A {e}^{- k t} = 1$
${e}^{- k t} = \frac{1}{A}$
$- k t = - \ln A$ log rules
$t = \frac{\ln A}{k}$

This gives us $t$ which we can substitute into $P \left(t\right)$:

$P \left(\frac{\ln A}{k}\right) = \frac{K}{1 + A {e}^{- k \frac{\ln A}{k}}}$
$= \frac{K}{1 + A {e}^{- \left(\ln A\right)}}$ log rules
$= \frac{K}{1 + \frac{A}{A}}$
$= \frac{K}{1 + 1}$
$= \frac{K}{2}$

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