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 ...
$\endgroup$ 43 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $f:\Bbb R\backslash \{a\}\to\Bbb R$.
$\endgroup$ $\begingroup$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 $\Bbb R$.
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