# Simplification of equation gives different domain and range

I am trying to find the domain and range of:

$$f(x) = \frac {x^2 - 4x + 3 }{x - 1}$$

In the book I am using, it says that the domain (x) is the set of all real numbers except 1 and that the range (y) is the set of all real numbers except -2. However, we can simplify this equation to:

$$f(x) = \frac {(x-3)(x-1)}{x - 1}$$

which is equal to

$$f(x) = x-3$$

Now, the domain and range of this simplified equation is different; they are now the set of all real numbers.

My question is: when determining the domain and range, do we simplify or don't we? Does simplifying the equation really produce different domains and ranges? Am I doing something wrong?

• Welcome to MSE! Its a removable pole. Commented Sep 22, 2018 at 7:56

We should always state our domain when we define our function.

The original function is $$f: \mathbb{R} \setminus \{1 \} \to \mathbb{R}$$

$$f(x) = \frac{(x-3)(x-1)}{x-1}.$$

It is well defiend on $$\mathbb{R} \setminus \{ 1\}$$ but it is not defined when $$x=1$$.

Compared it to function $$g: \mathbb{R} \to \mathbb{R}$$

$$g(x)=x-3$$

$$f$$ and $$g$$ are different functions because the domain is different.

When such question is being asked, the common practice is treat $$f$$ as $$f$$ and not $$g$$.

The second function $$f_2(x)=x-3$$ is equal to the original one $$f(x) = \frac {x^2 - 4x + 3 }{x - 1}$$ unless for a single point $$x=1$$ where $$f(x)$$ is not defined.

If we define $$f(1)=2$$ the two funtions become equivalent, that is they represent the same function.

In that case we talk of a removable discontinuity of $$f(x)$$ at $$x=1$$.