Graphing $\frac{(x-3)}{(x^2-3x)}$ and $1/x$

Question

For graphing the first equation, in the solved example given in the textbook, we proceeded as follows:

$\frac{(x-3)}{x^2-3x} \implies \frac{(x-3)}{x(x-3)} \implies \frac{1}{x}$ for $x \neq 3$.

Now the graph of this equation is exactly the same as that of $\frac{1}{x}$ except that there's a hole in $f(3)$ because the equation is not defined for $x=3$. What I don't get is, if we have simplified the equation to $\frac{1}{x}$, why do we still need to consider for the values of $x$ where the value of the original equation was undefined? Both the equations are exactly the same and yet, have different graphs, why so?

Answer 1

$f(x)=\frac{x-3}{x^2-3x}=\frac{(x-3)}{x(x-3)}$, only when $x \neq 3$, as division by $0$ is not possible.

• If $x \neq 3$, then we can divide both sides by $(x-3)$ to get $f(x)=\frac{1}{x}$. For example, for $x=2$, we have $f(2)=\frac{(2-3)}{2(2-3)}=\frac{1}{2}$(obtained by cancelling $(2-3) \neq 0$ )

• If $x =3$, then $f(3)=\frac{(3-3)}{3(3-3)} \neq \frac{1}{3}$, as we can't cancel $(3-3)=0$ on account of division by zero.

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Now, as to why division by $0$ is not possible. Read this

Answer 2

If $x=3$ the original expression is not the same as $\frac 1x$ but is undefined.

Answer 3

Consider 2 functions:

$f: D_1 \to R, f(x) = x+1$, for $x \in D_1=R$

$g: D_2 \to R, g(x) = x+1$, for $x \in D_2 =R^+$

Although they have the same values for common regions in their domain, these functions are not same. You don't use the same graph for them.

You have a similar case in the original question.

Answer 4

I suppose you are talking about so-called "removable singularity".

As mentioned in previous answers, $f(x)=(x-3)/(x^2-3x)$ is in principle undefined at $x= 0,3$, because in that case the denominator is zero. Yet, if $x\ne 3$, $f(x)$ can be simplified into $1/x$, and this expression itself is defined even at $x=3$. So, if we re-define the function so that$$g(x)=\left\{\begin{array}{cl}\frac{x-3}{x(x-3)}; & x\ne3,\\\frac{1}{3}; &x=3,\end{array}\right.$$we obtain a new function $g(x)$ which is equivalent to $f(x)$ for $x\ne 3$ but is also continuous for $x=3$. In this sense, the singular point $x=3$ is removable. Note, however, that $f(x)$ and $g(x)$ is equivalent only for $x\ne 3$, and thus the above argument does not allow us to conclude that $f(x)$ is continuous at $x=3$.

Similar situation happens when we consider $f(x)=\frac{\sin x}{x}$, where $f(0)$ is undefined in the original form but we may "get rid of" the singularity by redefining $f(0)=1$.