When a logistic law loses his first bend? I doing a regression of an experimental curve, it best fits  with a logistic law (best becauseI already compare  it to some polynomial regressions with different degrees).
The logistic law is defined by 
$$ f(z)= D+ \frac{A-D}{ 1+\Big( \frac{z}{C}\Big) ^B} $$
Usually the curve of a logistic law have a sigmoid shape defined as follows:
With $A$, $B$, $C$, and $D$ are the paramters of the  logistic law.
While my experimental curve is fited by a logistic law, it does not have two dendings: 

For example, for the first curve appearing in  the second graph, the values of the logistic parameters that appear after fitting are: A=-11.39962,  B= 0.988052231,  C=0.01338638, and  D=-1.493300048.  hope this helps  @ClaudeLeibovici  .  I plotted it here
My question is when a logistic law loses his first bending? 
I tried to derive some formulation for this interms of the variation of  the parameters  A, B, C , and D (for example if $B<1 $ the first bending disappear, or if  $C$ is near zero ) but unfortunately I failed as this appears not to be valid  graphically.
Can some one give me some tips in this manner.
Thank you in advance.
 A: Your curve is defined by the equation $$f(z) = D + \frac{A-D}{1 + (\frac{z}{C})^B} $$
The interesting part of it is the  second term (ignoring the inconsequential, for our purposes, constants at the numerator), 
$$\frac{1}{ 1 + (\frac{z}{C})^B}$$
Let us rescale the $z$ variable for simplicity, $\frac{z}{C} \to z$, and now look at the second derivative of $   \frac{1}{ 1 + z^B}$, which reads
$$ \frac{\mathrm{d}^2}{\mathrm{d}x^2} \Big( \frac{1}{ 1 + z^B}\Big) =\frac{Bx^{B-2} (x^B + B(x^B -1) +1)}{(x^B+1)^3} $$
For $B=1$ the second derivative is easily checked to have a constant sign. For $0 \leq B \leq 1$, as in your case, this is also true.
Indeed, the denominator is always positive, for $x>0$. The first term on the numerator s positive.
One is then left with te term 
$$ x^B + B(x^B -1) +1$$ 
whose derivative $B(B+1)x^{B-1}$ is positive and whose value for $x=0$ equals $ 1-B$, so positive for $B < 1$. 
It seems to me that only for $B>1$ one gets a sigmoid shape, i.e. the second derivative changes sign. 
