algebra-precalculusfunctions
Here's the question: Give three examples of input-output relationships in real life that cannot have negative values in the practical range? Explain why their range cannot have negative values?
It's a confusing question because how can a relationship have both and input and output?
Could time be a input-output relationship?
For fun, I like to liven-up the "black box"/machine view of a function by putting a monkey into the box. (I got pretty good at chalkboard-sketching a monkey that looked a little bit like Curious George, but with a tail.)
Give the Function Monkey an input and he'll cheerfully give you an output. The Function Monkey is smart enough to read and follow rules, and make computations, but he's not qualified to make decisions: his rules must provide for exactly one output for a given input. (Never let a Monkey choose!)
You can continue the metaphor by discussing the monkey's "domain" as the inputs he understands (what he can control); giving him an input outside his domain just confuses and frightens him ... or, depending upon the nature of the audience, kills him. (What? You gave the Reciprocal Monkey a Zero? You killed the Function Monkey!) Of course, it's probably more appropriate to say that the Function Monkey simply ignores such inputs, but students seem to like the drama. (As warnings go, "Don't kill the Function Monkey!" gets more attention than "Don't bore the Function Monkey!")
The Function Monkey comes in handy later when you start graphing functions: imagine that the x-axis is covered with coconuts (one coconut per "x" value). The Function Monkey strolls along the axis, picks up a "x" coconut, computes the associated "y" value (because that's what he does), and then throws the coconut up (or down) the appropriate height above (or below) the axis, where it magically sticks (or hovers or whatever). So, if you ever want to plot a function, just "Be a Function Monkey and throw some coconuts around". (Warning: Students may insist that that's not a coconut the Monkey is throwing.)
Further on, you can make the case that we're smarter than monkeys (at least, we should strive to be): We don't always have to mindlessly plot points to know what the graph of an equation looks like; we can sometimes anticipate the outcome by studying the equation. This motivates manipulating an equation to tease out clues about the shape of its graph, explaining, for instance, our interest in the slope-intercept form of a line equation (and the almost-never-taught intercept-intercept form, which I personally like a lot), the special forms of conic section equations (which aren't all functions, of course), and all that stuff related to translations and scaling.
Parametric equations can be presented as a way to let the Function Monkey plot elaborate curves ... both in the plane and in space (and beyond).
All in all, I find that the Function Monkey can make the course material more engaging without dumbing it down; he provides a fun way to interpret the definitions and behaviors of functions, not a way to avoid them. Now, is the Function Monkey too cutesy for a College Algebra class? My high school students loved him, even at the Calculus level. One former student told me that he would often invoke the Function Monkey when tutoring his college peers. If it's clear to the students that the instructor isn't trying to patronize them, the Function Monkey may prove quite helpful.
Note that $y=x^2$ is an equation, to be more precisely a statement form. It states: “The second coordinate is the square of the first one.” For the coordinates of a point that statement may be true -- for $(-2,4)$, e.g.-- or false for, say $(7,50)$. The set of points for which the statement form yields a true statement is called the solution set of the statement form. The statement form $y-x^2=0$ states: “The difference of the first coordinate and the square of the second equals zero.” is a different statement form, but has the same solution set as the first one.
The second one is given in a so called implicit form, wheras the first is called an explicit form. Why bother about the to forms? Well, if you want to determine the second coordinate of a point in the solution set, given the first coordinate is $7$, let's do it using the imlicit form. We have $$y-7^2=0\iff y=49,$$ so you have to solve an equation. (That's the reason for why we call it implicit.). Using the explicit form we get immediately $y=7^2=49$ without solving any equation.
Your first example $y=1/x$: “The second coordinate is the reciprocal of the first one.” reads in implicit version $xy=1$: “The product of both coordinates equals one.” Another explicit form is $x=1/y$. So if you are able to isolate one coordinate in an implicit form, this coordinate is called dependent, the other independent. In $x=1/y$ the coordinate $x$ is dependent and $y$ independent whereas in $y=1/y$ it's vice versa.
Now in $y-x^2=0$ you can't isolate $x$ in a closed form, here we have $\sqrt{y}=|x|$. So if for example $y=25$ we have $\sqrt{25}=|x|\iff5=|x|\iff x=5\lor x=-5$, so we have to solve an equation anyway. You may write $x=\pm\sqrt{y}$ and see that for all positive $y$ you'll get two values of $x$, so that explicit form is not a function, which mathematicians prefer.
In the example $x^2+y^2=1$ given by James there doesn't exist any explicit form which is a function, for an obvious reason: the solution set is the unit circle, centered at the origin.
Facit: if a statement form allows an explicit form of one of the coordinates and that explicit form is a function, you are free to call the isolated coordinate dependent and the other independent. I never use those attributes, because they are rather useless, won't gain any knowledge and are causing a lot of trouble as we see here.
Best Answer
To give an example of an input-output relationship, just think about a function.
For instance, suppose $f$ is a function that tracks revenue at a lemonade stand. It takes "number of cups sold" as an input and gives "dollars made" as an output. It might look like this:
$f(x)=.25x$
Assuming that each cup of lemonade costs .25 cents. So if I sell $5$ cups, then I make:
$f(5)=.25(5)=1.25$ dollars. If I sell 20 cups, I make:
$f(10)=.25(10)=2.50$ dollars. In this case, we call $10$ the input and $2.50$ the output. Does that make sense?
Also note: the domain of $f$ in this example is $[0,\infty).$ That is, the input value must be positive, since we can't sell a negative number of cups. The range is also positive, since $.25*(a\,non-negative\,number)$ is always non-negative.
Let me know if you have any questions!