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Functions with multiple outputs are often written as vector-valued functions.
How is it possible for a function to give multiple outputs?
[Math] How to a function have multiple outputs
functionsvectors
Related Solutions
The first step is to look at the simple $XY$ plot of the data:
Whoa. What if we take the logarithm of $Y$?
This looks like a power of $X$ with a lot of noise. Averaging the ratio of $\ln\ln Y/\ln X$ gives $0.42$. Indeed, $\ln Y\approx X^{0.42}$ is a reasonable "smoothed" approximation.
But I don't think that curve-fitting is the way to go here. The nature of the function, and the fact that the argument runs from $1$ to $256$ suggest that some binary operations might be involved. And look, we have spikes at
113 = 0b01110001
159 = 0b10011111
242 = 0b11110010
Actually, I suspect we should subtract $1$ from $X$ to make it run from $0$ to $255$. Then these three spikes are at
112 = 0b01110000
158 = 0b10011110
241 = 0b11110001
Must be something about these chains of 0s and 1s.
What you are getting confused by most probably is the fact that a vector seems to be multiple elements, but really it's not, it's just a single element which (often) has multiple parts.
You can even have so called set-valued functions. An easy one to think of is the function which takes a positive integer to the set of its prime divisors. Let's call this function $\pi$.
The important thing in both these cases is that a given input only has one output. So in the case of a vector-valued function the output is the full vector (all of its parts). In the case of the set valued function for prime divisors you have to get the set of all the prime divisors. So for example $\pi(15)=\{3,5\}$ you can't have $\pi(15)=\{3\}$ or $\pi(15)=\{5\}$. There is only the one set containing all the divisors. It's the same situation with vector-valued functions. There is only one vector for any input. The function can't return just part of a vector. Thus we still have only one output for every input--it's just a bigger output.
As to why we don't define inverses as vector valued, that has a very easy reason. They wouldn't be inverses anymore. The definition of an inverse to a function $f$ is a function $g$ for which $g(f(x))=x$ if the value of $g$ is a vector containing $x$ that doesn't do what we want. It doesn't tell us the value of the original input to function $f$.
Best Answer
One such function is $f: \Bbb R \to \Bbb R^2\\ x \to (3x+2,4x-1)$
It is important to note that when one says multiple outputs, it is not that there are several alternative outputs, there is a set or ordered sequence of outputs. You can view the output as a single sequence or as several numbers. When you talk of multiple outputs you are viewing it as several numbers. When you talk about a vector valued function, you view the output as one sequence. That is the spirit of the comments you have received.