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 $\displaystyle f(x) = \frac{2x+2}{x+1}$ to make $\displaystyle f(x) = 2$ by the above process.

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

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

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

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

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

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

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?