Question

Say you have a function f(x) = x - 1. It's just a straight line without any restrictions on its domain and range.
Now what if you had g(x) = (x^2 - 1)/(x+1); here, the domain restriction is that x =/= -1 as that would be a 0 denominator.
However, the g(x) can be simplified to f(x) quite easily, but f(x) doesn't have any domain restrictions.
My question is this: does g(x) have a domain restriction, or is it non-existent because it's not in its most simplified form? If it does, why?

Answer 1

yes, \(g\) has the same restrictions
good question though

Answer 2

If the original function (before simplification) has restrictions, they still apply, even after that which caused the restriction has been simplified away.

Answer 3

if you "simplify" aka reduce to lowest terms, your reduction is only valid if the denominator is not zero

Answer 4

That's really an excellent, excellent question if you came up with it on your own!

Answer 5

literally
\[\frac{x^2-1}{x+1}=\frac{(x-1)(x+1)}{x+1}=x-1\iff x\neq -1\]

Answer 6

Yeah I did, thanks. I'm tutoring high-school maths but I just came across it and it made me wonder what the deal is; I put the g(x) equation into Wolfram Alpha and it also told me that x =/= -1, but on the graph it provided, there wasn't a break in continuity at x = -1 so I became even more confused.

Answer 7

graph wont show it, hole is too small

Answer 8

I'd like to provide you both with best response, but I don't think I can.
So if I could elaborate - if you were to draw the graph, would you draw an open circle at x= -1?

Answer 9

of course the real question is what kind of idiot defines a function to be
\[f(x)=\frac{x^2-1}{x+1}\] when they might as well just say \(f(x)=x+1\)
turns out that the form \(\frac{x^2-1}{x+1}\) comes in a very natural setting of computing the slope of a tangent line

Answer 10

oops i meant \(f(x)=x-1\) but you get the idea
and yes, an open circle where \((-1,-2)\) should be

Answer 11

Alright, that all makes sense then, thank you :)

Answer 12

yw