2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ But $\frac{1}{x}$ is defined at $x=3$ and that's the equation that we have in hand. $\endgroup$ – MathEnthusiast May 30 '16 at 14:04
  • $\begingroup$ So we're simplifying the numerator and denominator just to make graphing easier and nothing else? $\endgroup$ – MathEnthusiast May 30 '16 at 14:08
4
$\begingroup$

$f(x)=\frac{x-3}{x^2-3x}=\frac{(x-3)}{x(x-3)}$, 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.


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

$\endgroup$
  • $\begingroup$ This is the actual answer to OP’s question: the simplification $\frac{(x-3)}{x(x-3)} \to \frac{1}{x}$ simply isn’t valid (namely, it fails when $x=3$). $\endgroup$ – Lynn May 30 '16 at 14:14
  • $\begingroup$ So basically, we simplified to make the graphing easier? $\endgroup$ – MathEnthusiast May 30 '16 at 16:32
  • $\begingroup$ @user331377 Indeed $\endgroup$ – Dragonemperor42 May 30 '16 at 16:33
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ But we have simplified the equation to $\frac{1}{x}$, right? Why not just graph that? $\endgroup$ – MathEnthusiast May 30 '16 at 14:03
  • $\begingroup$ Yes but except for $x=3$ $\endgroup$ – David Quinn May 30 '16 at 14:05
  • $\begingroup$ That's kind of my question, why? $\frac{1}{x}$ is defined at $x=3$. $\endgroup$ – MathEnthusiast May 30 '16 at 14:07
  • $\begingroup$ $\frac 1x$ is indeed defined at $x=3$ but the original expression is not $\endgroup$ – David Quinn May 30 '16 at 14:11
1
$\begingroup$

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 $f(0)=1$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.