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