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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?

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  • $\begingroup$ Welcome to MSE! Its a removable pole. $\endgroup$
    – Wuestenfux
    Commented Sep 22, 2018 at 7:56

2 Answers 2

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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 $g$.

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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 $x=1$.

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