Why aren't these functions the same?

by Whistlekins
Tags: functions
 P: 21 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?
HW Helper
P: 3,557
 Quote by 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?
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.
 P: 21 I understand that. But why can't I write g(x) = (x^2+x-2)/(x-1) = x+2 ?
P: 835
Why aren't these functions the same?

 Quote by 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
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.
Mentor
P: 18,346
 Quote by 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.
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.
P: 835
 Quote by micromass 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.
Mentor
P: 18,346
 Quote by pwsnafu In fact it could also be larger. No body said x couldn't be complex.
Very true!
 P: 21 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?
Mentor
P: 18,346
 Quote by 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?
Yes to both.
 P: 21 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.

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