construction of a curve connecting two points Let $a,b,c$ be positive reals numbers. Assume $a<b$. I'm trying to construct a $C^1$ function (meaning a function with continuous derivative) $f$ with the following properties:


*

*$f$ is increasing and supported on $[a,b]$;

*$f(a)=0$ and $f(b)=c$;

*$f'(a)=0$ and $f'(b)=1$;


Geometrically $f$ is a curve connecting $(a,0)$ and $(b,c)$ with additional derivative requirements at two end points.
I think such a function is very useful in application but I can't find any book containing methods of constructions. I guess the construction may be tedious. Do you know any reference providing constructions of lots of special functions? 
 A: You can look for a polynomial of degree 4 of the form
$$
p(x)=(x-a)^2(A\,x+B)
$$
where the constants $A$ and $B$ are chosen to satisfy the conditionsconditions at the point $b$.
The special case $a=0$, $b=c=1$ gives $p(x)=x^2(2-x)$, with $p'(x)=x(4-3\,x)\ge0$ if $0\le x\le1$.The general case is then
$$
c\,p\Bigl(\frac{x-a}{b-a}\Bigr),
$$
which is increasing on $[a,b]$.
A: Just write down the formula for a degree 4 polynomial or higher and you should be able to solve for the coefficients to satisfy the constraints.
A: Using linear interpolation one reduces to the same problem with $a=b=c=0$, replacing the condición $f'(b)=1$ by the more general $f'(1)=\lambda>0$. Now the conditions mean that the tangents to the graph are horizontal at $t=0$ and have slope $\lambda$ at $t=1$. There are different ways to construct such a smooth function. One is through the so-called bump functions, as found typically in many Differential Calculus texts. But one can also use splines. Try one of them
$$
f(t)=n(1-\tfrac{1}{n}\lambda)t^{n-1}(1-t)+t^n
$$
with $n>\lambda$. What is important here is that the function is strictly increasing, hence a difeo, which makes necessary a possibly very high degree $n$.
In general you can produce anything form $[a,b]$ to $[c,d]$ with any chosen slopes at $a$ and $b$.
A: Not very different from the other answers, but maybe easier to understand and use:
$$
f(x) = \frac{(a-x)^2 \left( (a-b+2c)x -ab +ac -3bc + b^2   \right)}{(a-b)^3}
$$
