Why aren't these functions the same?

[Whistlekins]

I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2

Now everyone would agree that f has a domain R\{1} and g has a domain R.

Yet I can write (x^2+x-2)/(x-1) = x+2

So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?

 

[Mentallic]

I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2

Now everyone would agree that f has a domain R\{1} and g has a domain R.

Yet I can write (x^2+x-2)/(x-1) = x+2

So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?

If you evaluate f(1) then you get an undefined value 0/0. Of course we know that

\lim_{x\to 1}f(x) = 3

But just because the limit exists doesn't mean that the function is defined at that point.

 

[Whistlekins]

I understand that. But why can't I write g(x) = (x^2+x-2)/(x-1) = x+2 ?

 

[pwsnafu]

I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2

Now everyone would agree that f has a domain R\{1} and g has a domain R.

Yet I can write (x^2+x-2)/(x-1) = x+2

Only if x is not equal to 1. So you need to write $\forall x \neq 1, \,\frac{x^2+x-2}{x-1} = x+2$

So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?

But you didn't rewrite the expression. If I define
$h : \mathbb{R} \to \mathbb{R}$ with $h(x) = \frac{x^2+x-2}{x-1}$ and $h(1) = 3$ then that is indeed equal to g(x), but not equal to f(x).

There is a difference between you can do something, and you did something.

 

[micromass]

I have f(x) = (x^2+x-2)/(x-1) and g(x) = x+2

Now everyone would agree that f has a domain R\{1} and g has a domain R.

I wouldn't agree with this. The domain is something chosen by the person who defines it. All we can say is that the maximal possible domain of $f$ is $\mathbb{R}\setminus \{1\}$. But the domain can possibly be much smaller if we choose it to be.

Yet I can write (x^2+x-2)/(x-1) = x+2

So why wouldn't g have a domain R\{1} if I rewrite the expression, and vice versa for f? What mathematical principle is behind this?

The equation

\frac{x^2 + x - 2}{x-1} = x+2

is only valid for $x\in \mathbb{R}$ with $x\neq 1$. For $x=1$, it is not true. So we have that $f(x) = x+2$ for all $x\in \mathbb{R}\setminus \{1\}$. The value $f(1)$ still isn't defined.

That $f(1)=3$ somehow, is false. However, this is why limits are invented. So we can say that
\lim_{x\rightarrow 1} f(x) = 3

So although $f(1)$ doesn't make sense, we can take the limit. The limit denotes the value that $f(1)$ would have been if it were defined in $1$ and if $f$ were to be continuous.

 

[pwsnafu]

I wouldn't agree with this. The domain is something chosen by the person who defines it. All we can say is that the maximal possible domain of $f$ is $\mathbb{R}\setminus \{1\}$. But the domain can possibly be much smaller if we choose it to be.

In fact it could also be larger. No body said x couldn't be complex.

 

[micromass]

In fact it could also be larger. No body said x couldn't be complex.

Very true!

 

[Whistlekins]

So if I previously define the domain, I can't change that domain unless I write an entirely new function?

Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h?

 

[micromass]

So if I previously define the domain, I can't change that domain unless I write an entirely new function?

Would it be true that if h(x) = x+2 , for all x in R\{1}, then f = h?

Yes to both.

 

[Whistlekins]

Cool, thanks for helping me clear my confusion. I guess that never really got explained to me by anyone and I never picked up on it.