Ease-in function I need a function for ease-in effect, exactly like this ease-in-out function, but without the end (infinite).
I've tried two variants that I don't like too much for a while.
The first one never becomes really straight. I'd prefer to have it absolutely or almost absolutely straight after some certain $x_1$ value.
$$
f(x)=x*(1-\frac{1}{x+1})
$$
Second one is a combination of two functions which works well (for $x_1=\frac{π}{2}$), but I don't like the fact that I have two functions instead of one.
$$
f_1(x)=1-cos(x), x<\frac{π}{2}
$$
$$
f_2(x)=x-\frac{π}{2}+1, x>=\frac{π}{2}
$$
Is there a better variant?
 A: Given any ease-inout function $f(x)$ with $f(\frac{1}{2})=\frac{1}{2}$ and domain and range [0, 1], such as  the 
$$f(x,a)=\frac{x^a}{x^a+(1-x)^a}$$ suggestion from the other thread, you convert to ease-in only by "zooming" on the first half, i.e. use 2 * f(x/2).  Similarly, for an ease-out function zoom on the second half using 2 * f(1/2 + x/2) - 1.
Easy.  :-)
(You will lose the property that $f(\frac{1}{2})=\frac{1}{2}$, but I think that's not important for one-sided easing.)
Edit:  As described in my comment, you can patch in the line $y=a(x-1)+1$ at $x=1$ as shown in the following figure for $a=2$.  If you specifically need a gradient of 1, transform $x\rightarrow\frac{x}{a}$.

A: Just use an IF-THEN block:
Pseudo-Code:
IF x > 10 THEN
 return value
 ELSE:
 return f(x)
 ENDIF

A: Your first example is a good thought.  It can be written as $f(x)=x-\frac x{x+1}=x-1+\frac 1{x+1}=\frac{x^2}{x+1}$  This has an asymptote at $f(x)=x-1$ but doesn't approach it very quickly.  It seems like you want to start at $(0,0)$.  Your second example has the derivative at $0$ equal to zero. Do you want that? How sharp a corner is allowed?  There are functions that will satisfy any answer you give to these questions.  You might try $f(x)=x-\left (\frac {x}{x+k}\right)^n$ for some moderately large $n,k$ and see how you like it.
