Building a function out of a given domain and range How does one build a function while given domain and range ? For example, the domain $(0,5]$ and range $[0, \infty)$
 A: To start  we can think of the square root of a quadratic function with zeros $0$ and $5$ and positive between them, e.g. $$\sqrt{x(5-x)}.$$
Its domain $[0,5]$ is "quite perfect" (but not the range).   
We have to exclude $0$ from the domain. This will be done if we divide by $x.$ Here we realize that the signs of $\;x(5-x)\;$ and  $\;\dfrac{5-x}{x}\;$ are equal.
Thus consider the function $$f(x)=\sqrt{\frac {5-x}{x}}.$$
Here we successfully reduced the domain to $(0,5].$ Moreover, the range is $[0,\infty)$ as required.
A: Hint: When you see square brackets in your domain/range, you know that your function must contain the point with the $x$ and $y$ values that have the square brackets. In your example, the function must contain the point $(5,0)$. 
In terms of defining particular functions that will satisfy these conditions, you may simply define a function with infinite range over a specific domain, for example: $$\frac{1}{e^{\left(x-5\right)}}-1 \ \ \ \ \ 0< x\leq5$$However, if you want to define a function that will satisfy these conditions for all $x$ values, I recommend using $y = \arcsin(x)$ or $y = \arccos(x)$ for finite domain and infinite range, and $y = \sin(x)$ or $y = cos(x)$ for finite range with infinite domain. For your example, something like: $$\left|\frac{1}{\arcsin\left(\frac{2x}{5}-1\right)x}\right|$$ should do the trick.
A: something like 
$f(x)= \frac 1x - \frac 15$ seems like it would fit the bill.
or $g(x) = -\ln x + \ln 5$
I am looking to map the open end in the domain with the open end in the range, and the closed end of the domain with the closed end of the range.  It is not entirely necessary, but if we don't it seems to me that we have to start looking for discontinuous functions.
$f(x)$ approaches infinity as x approaches 0.  And $f(5) = 0$
Any function with asymptotic behavior around $0$ should be a good place to start.  Then we do what we need to to fit the endpoint the other endpoint.  Functions that are strictly decreasing will be nice, to take away the possibility that they might dip below 0.  And continuous functions are nice, as they will go through every point between the extreme values of the range.
