What does "can define a function" mean? I'm struggling with the following question : "Prove that $f(x) = \sqrt{(x^3-1)(x-1)}$ can define a $f : R → R$ function."
I don't even really understand what I'm being asked. I thought showing that $(x^3-1)(x-1)$ is always greater than or equal to 0 would prove that $R$ is indeed the correct domain, but I don't know if that's what I'm supposed to do. Or perhaps should I show that there is a single $y$ such that $f(x) = y$? Or maybe both?
Can somebody explain to me how to correctly answer the question? Many thanks in advance.
 A: To prove something is a function, you need to show that given every element in the domain,  there exists EXACTLY one element in the codomain that the function outputs.
In this case,  there is no ambiguity that two possible outputs could be given at any step,  since everything involved in $f(x)$ is itself a function, everything that is a valid input has a unique output.  So the only thing left to prove is that everything in $\mathbb{R}$ is indeed a valid input.  Since the only domain issue involved is the $\sqrt()$ function,  you indeed need to show that you are never putting a negative number into that,  ie $(x^3-1)(x-1)\geq 0$
As a side note,  issues where you have to show that you get a UNIQUE value for $y$ aka $f(x)$  come up when you define a function implicitly, either off an equation or for a non-unique representation.   Examples include, can you make a function for $y=f(x)$ out of $x^2+y^2=1$?  The answer is no,  because, for example, at x=0 there are two valid outputs,  $1$ and $-1$.
The other way it can fail if you try to define a function on something that is a non-unique representation.
For example, if you tried to define a function on rational numbers such that
$$f(p/q)=p$$
Ie a function that takes in fractions and outputs the numerator, this would FAIL to be a function because $1/2=2/4$   ,  but $f(1/2)=1$ and $f(2/4)=2$, so you don't get unique outputs.
