# Inflection point of $\,f(t) = \frac{1}{1+e^{(-t)}}$

I am trying to calculate the inflection point of the logistic function $f(t) = \dfrac{1}{1+e^{(-t)}}$. According to the definition given in Wikipedia,

"A differentiable function has an inflection point at $(x, f(x))$ if and only if its first derivative, $f′$, has an isolated extremum at $x$". using the definition, I try to differentiate the logistic function and equate it to $0$. so then I get this formulation $\dfrac{-e^{-t}}{(1+e^{-t})^2} =0$. Removing the denominator, I get $-e^{-t}=0$.

It looks like this equation has no solution, since there is no value of $t$ that can fit this equation. Does that mean the logistic function has no inflection point??

But I guess that is not correct., since my intuition tells that there must be an inflection point at $t=0$. since the curve changes from being concave to convex at that point.

Could someone please clarify?

• Follow Wikipedia's instructions and look for the extremum of $f'$, not its zero. – Yves Daoust Nov 23 '14 at 13:03

You are searching a critical point of $f'$, i.e., a $t$ with $f''(t)=0$.

As you say, you need to find an isolated, local maximum of $f'$. This might or might not happen when $f''(x)=0$ and solving that equation doesn't change the fact that you need to analyze the behavior of the function at that point. However, $$f'(x) = \frac{e^{-t}}{\left(1+e^{-t}\right)^2}$$ and it's not too terribly hard to show that this expression attains a maximum of $1/4$ at the origin. Indeed, $f'(0)=1/4$ by direct computation. Furthermore, $$0\leq(1-e^{-t})^2 = 1-2e^{-t}+e^{-2t}.$$ Adding $4e^{-t}$ to both sides we see that $$4e^{-t} \leq 1+2e^{-t}+e^{-2t} = (1+e^{-t})^2$$ which implies that $$\frac{e^{-t}}{(1+e^{-t})^2} \leq \frac{1}{4}.$$

Of course, the second derivative allows you to easily see that there are no other inflection points but, again, the fact that $f''(x)=0$ does not immediately imply that that the point is an inflection point.

• " it's trivial to see... ": can't agree with that. – Yves Daoust Nov 23 '14 at 13:04
• I disagree to say that this is trivial, especially to a junior OP. – Yves Daoust Nov 23 '14 at 13:18
• @YvesDaoust Edited. The main point that I was trying to get across, though, was that we have an inflection point because $f'$ has an isolated local extreme at zero - not because $f''$ is zero at zero, which is a different thing. – Mark McClure Nov 23 '14 at 13:36
• This is a very elegant proof for the inflection point, without using derivatives. – user76170 Nov 27 '14 at 9:45
• @user76170 Thanks! Though, the vote tally currently stands at -1; the standard way to express approval is via an upvote. – Mark McClure Nov 27 '14 at 17:29

Potential extrema (the critical points of a function $f(t)$) are found where $f'(t)= 0$.

Potential inflection points (the critical points of a function $f'(t)$) are found where where $f''(x)=0$.

The definition is referring to the latter, so you need to differentiate once again, and solve $f''(t) = 0$.

So, to find $f''(t),$ we find \begin{align} f''(t) = \dfrac{d}{dt}\left(\dfrac{-e^{-t}}{(1+e^{-t})^2}\right) & = \dfrac{e^{-t}(1+e^{-t})^2 + e^{-t}(2\cdot (1+e^{-t})(-e^{-t}))}{(1 + e^{-t})^4}\\ \\ & = \dfrac{e^{-t}(1+e^{-t})\Big((1+e^{-t}) -2e^{-t}\Big)}{(1 + e^{-t})^4} \\ \\ &= \dfrac{e^{-t}(1-e^{-t})}{(1 + e^{-t})^3} \end{align}

Now, $f''(t) = 0$ if and only if $$1-e^{-t} = 0 \iff e^{-t} = 1 \iff t=0$$

Now we need only confirm, as you have visually, that when $t = 0$, we indeed have an inflection point.