What makes a function not defined?

I have started studying precalculus and would then start up with calculus. While studying about functions I wondered whether this function would be defined at $a$ or not. Take a look at it. $$f(x) = \frac{(x-a)(x-b)(x-c)...(x-n)}{(x-a)}$$

Here if we will simplify it further then the term $\left( x-a\right)$ would cancel out making the function defined at $a$ but if we would leave it as such it would be undefined at that point.

I asked this question because I found in some sources that the graph of such functions have an open dot at that point indicating it discontinuous at that point. But I couldn't explain it. Are the expressions before and after cancelling different or it's something else?

I would be highly obliged for your help and thanks ...

• You seem to have it right. Since you cannot divide by $0,$ the function is not defined when $x = a$. The function is undefined at $a.$ And the graph of the function has an open dot. This sort of discontinuity is called a "removable discontinuity." It is possible to define a value of the function at the case that $x=a$ that would make the function continuous. This would be equivalent to canceling the $(x-a)$ factors. Commented Mar 24, 2017 at 17:58
• It means that after simplifying the expression the function becomes continuous. Right . Thanks a lot. Commented Mar 24, 2017 at 18:01
• You cannot cancel out (x-a)/(x-a) when x=a because at x=a the expression (x-a)/(x-a) is 0/0 which is not a number. Commented Mar 24, 2017 at 21:20
• A function $f$ from $A$ to $B$ is a subset of the Cartesian product $A\times B$ with the property that for each $a\in A$ there is exactly one $(a,b)\in f.$... If $g$ is a function from $A\cup \{\alpha \}$ to $B$ with $\alpha \not \in A,$ then $f$ and $g$ are two different functions, regardless of whether or not $g\cap (A\times B)=f.$ Commented Mar 24, 2017 at 21:25

Consider the following functions: $$f(x) = 5$$

$$g(x) = \dfrac{5(x-2)}{(x-2)}$$

The function $g$ is not defined at $x = 2$, but agrees with $f$ at every other point.

So we would say these functions are not the same, because their domains are different.

• This means that mathematically both the expressions you gave are two different functions, though they don't seem to be intuitively. Thanks a lot as it really cleared one of my fundamentals.. Commented Mar 24, 2017 at 18:03

One cannot simplify $\frac{ca}{cb}$ to $\frac ab$ if $c=0$, because the first expression doesn't make any sense. $\frac 00\neq 1$, and that's what you're saying when you simplify the function by cancelling both $x-a$'s.

However, we don't really encounter this problem if we're being precise; we define

$$f(x)=\frac{(x-a)\cdot p(x)}{x-a}$$

for $f:\Bbb R\backslash \{a\}\to\Bbb R$ anyways (where $p$ is some polynomial, but that's beside the point), so the function

$$f(x)=p(x)$$

is still the same, if we leave $f:\Bbb R\backslash \{a\}\to\Bbb R$.

A function $f$ is a special asymmetric relation between two sets $A$ and $B$, represented by $f:A\to B$. The relation consist in that for every element of $A$ (called the domain of $f$) exists a unique element in $B$ (called the codomain of $f$) defined by the function $f$.

In your example, if the codomain of $f$ is $\Bbb R$, then $f(a)\notin \Bbb R$ because the division by zero is not defined in $\Bbb R$.