1. ## Question about domains

$f(x) = \frac{2x+2}{x+1}$

$f(x) = \frac{2(x+1)}{(x+1)}$

$f(x) = g(x) = 2$

Now, f(x) in the first case has the domain $\lbrace x \in \mathbb{R} | x \neq -1 \rbrace$, but the second, g(x), has $x \in \mathbb{R}$. Just a bit confused as to why this is, isn't it the same function?

I know that you can define your domain for whatever you wish, but if I algebraically manipulate a set function with a given domain, shouldn't that hold?

2. Originally Posted by rowe
$f(x) = \frac{2x+2}{x+1}$

$f(x) = \frac{2(x+1)}{(x+1)}$

$f(x) = g(x) = 2$

Now, f(x) in the first case has the domain $\lbrace x \in \mathbb{R} | x \neq -1 \rbrace$, but the second, g(x), has $x \in \mathbb{R}$. Just a bit confused as to why this is, isn't it the same function?

I know that you can define your domain for whatever you wish, but if I algebraically manipulate a set function with a given domain, shouldn't that hold?
Hi rowe,

I assume the g(x) function is the f(x) factored, right?

] $f(x) = \frac{2x+2}{x+1}$

$g(x) = \frac{2(x+1)}{(x+1)}$

$f(x) = g(x) = 2$

Remember the domain of $f(x)=\frac{2x+2}{x+1}$ is all real numbers except x = -1 (which would make the denominator zero)

When a value of x sets both the denominator and the numerator of a rational function equal to 0, there is a hole in the graph; that is, a single point at which the function has no value. $g(x)=\frac{2(x+1)}{x+1}$ has a hole at x = -1.

3. Hi, thanks.

Let's just ignore g(x) for now.

I'm aware that f(-1) is undefined, but what I'm saying is that I can algebraically manipulate $f(x) = \frac{2x+2}{x+1}$ to make $f(x) = 2$ by the above process.

The problem is that in doing so, it seems the domain of the function "changes". There is no $x \in \mathbb{R}$ such that $f(x)=2$ is undefined.

Or, do we have to keep the domain of $f$ fixed, so even though I manipulate it, the domain stays the same so even though $f(x)=2$, we have said that we can't allow x to take the value -1?

By that logic, I can algebraically manipulate $f(x)=2$ by:

$f(x)=2\cdot\frac{x+1}{x+1}$

$f(x)=\frac{2(x+1)}{x+1}$

$f(x)=\frac{2x+2}{x+1}$

But in this case, I could have orignally defined the domain of f for all real numbers! In both cases, the domain changes. What's the story?

4. Originally Posted by rowe
Hi, thanks.

Let's just ignore g(x) for now.

I'm aware that f(-1) is undefined, but what I'm saying is that I can algebraically manipulate $f(x) = \frac{2x+2}{x+1}$ to make $f(x) = 2$ by the above process.

The problem is that in doing so, it seems the domain of the function "changes". There is no $x \in \mathbb{R}$ such that $f(x)=2$ is undefined.

Or, do we have to keep the domain of $f$ fixed, so even though I manipulate it, the domain stays the same so even though $f(x)=2$, we have said that we can't allow x to take the value -1?

By that logic, I can algebraically manipulate $f(x)=2$ by:

$f(x)=2\cdot\frac{x+1}{x+1}$

$f(x)=\frac{2(x+1)}{x+1}$

$f(x)=\frac{2x+2}{x+1}$

But in this case, I could have orignally defined the domain of f for all real numbers! In both cases, the domain changes. What's the story?
I'm not sure what your confusion is. Let me state it clearly.

At x = -1, f(x) is undefined. That means there's a hole in the graph at (-1, 2).

F(x) = 2 for all real numbers in the domain except -1.

The graph will be a horizontal line through y = 2 with a point of discontinuity at (-1, 2).

5. I'm not too concerned with where the discontinuity is, I'm asking if I have a function $f:R \to R$, with $f(x)=2$, and I manipulate it so that $f(x)=\frac{2x+2}{x+1}$, is this an "illegal" move so to speak, because the domain doesn't support the discontinuity?

And the same vice versa- the domain inherently denoted for all real x, bar -1 in $f(x)=\frac{2x+2}{x+1}$, that won't change the domain to all R when we change f(x)=2?

Don't worry too much about it, from what you're saying and what I think I've pretty much convinced myself you have to stick to the domain given by the function definition.