Another examples of strictly monotone surjective function from an interval to another interval Let $[a,b],[c,d]$ are two intervals in $\mathbb{R}$. I know two strictly monotone surjective functions from $[a,b]$ onto $[c,d]$:
$$f(x)=c+\frac{x-a}{b-a} (d-c)$$
and
$$f(x)=d-\frac{x-a}{b-a} (d-c)$$
Could you give more examples?
 A: Notice that any interval $[a,b]$ can be mapped bijectively onto $[0,1]$ through
$$
    \lambda(x) = \frac{x-a}{b-a}
$$
Notice that $\lambda$ is increasing.
Similarly, we can bijectively map $[c,d]$ onto $[0,1]$ through
$$
    \mu(x) = \frac{x-c}{d-c}
$$
Note that $\mu^{-1}(y) = c + (d-c)y$ for any $y$.
Now if we take any monotone surjective function $g$ from $[0,1]$ to itself, we can form the composition $f = \mu^{-1} \circ g \circ \lambda$:
$$
    f(x) = c + (d-c) g\left(\frac{x-a}{b-a}\right)
$$
It's not too hard to show that $f$ is monotone and surjective.  
The function $g(x) = x$ gives you your first $f$, because 
$$
    \mu^{-1}(\lambda(x)) = c+(d-c)\frac{x-a}{b-a}
$$
The function $g(x) = 1-x$ gives you your second $f$, because
$$
    \mu^{-1}(1-\lambda(x)) = c + (d-c)\left(1-\frac{x-a}{b-a}\right)
    = d - \frac{d-c}{b-a}(x-a)
$$
The function $g(x) = x^2$ gives you Siong Thye Goh's answer:
$$
    \mu^{-1}(\lambda(x)^2) = c + (d-c)\left(\frac{x-a}{b-a}\right)^2
$$
but any positive exponent will work in place of $2$.
You can also use $g(x) = \frac{t^x -1}{t-1}$ for any $t>0$.  We get
$$
    f(x) = \mu^{-1}\left(\frac{t^{\lambda(x)}-1}{t-1}\right)
    = c + \frac{d-c}{t-1}\left(t^{(x-a)/(b-a)}-1\right)
$$
For various $t$ this produces the exponential examples from Mohammad Riezi-Kermani.
A: Let $(a,c)$ be the vertex of a quadratic equation and $(b,d)$ be another point on the quadratic curve.
then $y=A(x-a)^2+c$
substitute $(b,d)$ inside, we have 
$$d=A(b-a)^2+c$$
$$A=\frac{d-c}{(b-a)^2}$$
Hence, $$f(x) = (d-c) \left( \frac{x-a}{b-a}\right)^2 + c$$
