How can I derive monotonically increasing polynomial functions with steep curves from [0,1] to [0,1]? For computer graphics reasons, I need a Taylor polynomial function on the interval [0,1] like the classic smoothstep function 3x^2-2x^3 which outputs a monotonically increasing value on the interval [0, 1] but that has a much steeper curve. It should have only one inflection point (not counting the beginning or the ends.) The beginning should be convex and the end should be concave.
I would also like to be able to tweak the function to my needs and make it steeper or not so steep with some parameters if possible.
How can I derive steep polynomial functions from [0,1] to [0,1]?
 A: "Taylor series" is the wrong terminology to use.
What you're probably
looking for is a polynomial satisfying some properties. Let $f(x)=\sum_{k=0}^{n}c_{k}x^{k}$.
We need to satisfy:
\begin{align*}
f(0) & =0;\\
f(1) & =1;\\
f^{\prime}(x) & =\sum_{k=1}^{n}c_{k}kx^{k-1}\geq0 & \text{for }0\leq x\leq1.
\end{align*}
The first two equations are self-explanatory. The third ensures that
the polynomial is nondecreasing. Save from the degenerate case of
a line ($f(x)=x$), you should have enough freedom to fool around
with the first derivative to get "steeper" functions.
For example, if $k=2$, then you'll find that the only polynomials
that satisfy the above are $cx+(1-c)x^{2}$ parameterized by $0\leq c\leq2$.
The parameter $c$ can be used to control the steepness:

Another example are the polynomials
$x^k$ and $1-(1-x)^k$.
These also satisfy the above. The larger the value of $k$,
the steeper the result:


However, you might not want to use high degree polynomials as these are more expensive (in terms of floating point operations) to compute.
A: I want a polynomial, probably low-degree.
So let's try a + bx + cx^2 + dx^3 + e^4.
I want the point at zero to be zero so a must be zero.
I want the point at one to be one so c + d + e must be one.
I want the start to be convex and curve up so the first derivative at zero must be zero and hence b must be zero.
I want the end to be concave and curve down so the first derivative at one must be zero and hence 2c + 3d + 4e = 0.
Then d = 4 - 2 c and e = 1 - (c + d) and c is left free.
I am unsure of how to determine if c is two low or too high. I know that values lower than -1 don't work and values higher than around 7 don't work but I don't know why.
