Function with certain characteristics I am wanting a function $f(x)$ that is continuous and differentiable on the open interval $(0..1)$ that has the following characteristics, for some given constant $c$, also between $(0..1)$
$$  \lim\limits_{x \to 0^+}{f(x)} = 0$$
$$ \lim\limits_{x \to 1^-}{f(x)} = 1$$
$$ \lim\limits_{x \to 0^+}{f'(x)} = \infty$$
$$ \lim\limits_{x \to 1^-}{f'(x)} = \infty$$
$$ f(0.5) = c$$
Also, the second derivative of $f(x)$ should cross the x axis only once.
Are there any tractable closed-form functions that satisfy these criteria?
I was able to contrive one for the special case where $c = 0.5$, but I am not sure how to generalize it.
How I found the special case was I took a function that satisfies the first two criteria as well as the 3rd: $f_1(x)=\sqrt{x}$, and then came up with a second function: $f_2(x)=1-\sqrt{1-x}$, which is essentially $f_1$ mirrored twice, once horizontally around $x=0.5$ and once vertically around $y=0.5$.  This causes $f_2$ to satisfy the first two criteria as well as criteria 4.  I then interpolated between the two functions as a linear function of x, so my final $f(x)= \sqrt{x}(1-x)+(1-\sqrt{1-x})x$.  Because $f_1$ is concave down, while $f_2$ is concave up, and because interpolating between them in this way works out to taking the average of $f_1$ and $f_2$ at exactly $x=0.5$, the single inflection point in $f$ shows up there, where the second derivative crosses the x axis.
However, as I said, this function is fixed only for $c=0.5$... I would like a more general solution for an arbitrary $c \in (0..1)$.
Also, if there is a technique by which I could generalize the method that I used to come up with it to work for different values of $c$, that would also be satisfactory.
EDIT:
My bad, and most sincere apologies... I thought that the above would be sufficient for what I am looking for, but I forgot one also very important characteristic, which is that $f(x)$ must be strictly increasing from left to right (its first derivative may contain a zero, but no more than one). 
 A: How's this?
$$f\left(x\right)=\frac{1}{2}\sqrt[3]{x}+\frac{1}{2}\sqrt[3]{x-1}+\frac{1}{2}+2\left(c-\frac{1}{2}\right)\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}$$
I have to go, but I will post my technique later. :)
A: I found one. 
$$f(x) = \left( \sqrt {x} \left( c\sqrt {2} \right) ^{2-2\,x} \left( 1-x
 \right) + \left( 1-\sqrt {1-x} \left(  \left( 1-c \right) \sqrt {2}
 \right) ^{2\,x} \right) x-c \right)  \left( 2\, \left( x-1/2 \right) 
^{2}+1/2 \right) +c
$$
How I came up with this was as follows.   I started with a function that I knew was defined at 0, but whose first derivative at 0 was $\infty$:  $\sqrt{x}$
Now this equals 1 at x=1, but equals $\sqrt{1/2}$ at x=1/2... so what I then needed was another function to multiply this by, which had the characteristics such at x=1/2, it equaled $c/\sqrt{1/2}$, but which was still equal to 1 at x=1, and which was always positive for x=0..1.   
An exponential function was what I needed, and the one that turned out to meet the criteria was: $\left( c\sqrt {2} \right) ^{2-2\,x}$
So multiplying them, I get the following:
$$f_1(x) = \sqrt {x} \left( c\sqrt {2} \right) ^{2-2\,x}.$$  
This function rises vertically at x=0, continues to rise through the point $(1/2,c)$, and then continues to rise until it gets to (1,1)
The resulting function looks like this for c=1/2 vs c=2/3:


I then flipped this function upside-down to get a second function that passes through $(1/2,c)$ but rises vertically at (1,1): 
$$f_2(x) = 1-\sqrt {1-x} \left(  \left( 1-c \right) \sqrt {2} \right) ^{2\,x}.$$
Which looks like this (again for c=1/2 vs c=2/3):


Now, with two functions that both intersect (1/2,c), I can interpolate between the two functions using a linear function of x, ie $f_1(x)(1-x) + f_2(x)(x)$.  Since the two functions have the same value at x=1/2, the resulting function still intersects (1/2,c), and the combined function is:
$$\sqrt {x} \left( c\sqrt {2} \right) ^{2-2\,x} \left( 1-x \right) +
 \left( 1-\sqrt {1-x} \left(  \left( 1-c \right) \sqrt {2} \right) ^{2
\,x} \right) x.$$
Here again, we show the resulting function for c=1/2 vs c=2/3:


Now at this point, the function meets all of the requirements that I stated above, but I noted that this function does not have a very nice shape.  It was very "straight" in the middle, and I was actually hoping to find a function that had a nice logistic equation-like "S" curve to it.
What I did, therefore, was subtract c from the above equation and multiplied the result by a parabolic equation $\left( 2\, x - 1 \right) ^{2}/2 + 1/2$, and then added c back.  This gave me a nice smooth S-curve that I was after, $f(x)$, again shown here for c=1/2 vs c=2/3:


