1. ## discontunity rule

discontunity rule says if a function breaks at certain point , then that point is discontinuity point.

Example : y=1/x if, x=0 ...y breaks , so x=0 is a point of discontinuity.

Now with this idea, lets us try ,

f(x)=(x^2-64/x+8)
= x-8

this can never break for any real value of x ...right ? so this function cant be having discontinious point ....am i right ?

2. ## Re: discontunity rule

Originally Posted by defeated_soldier
discontunity rule says if a function breaks at certain point , then that point is discontinuity point.

Example : y=1/x if, x=0 ...y breaks , so x=0 is a point of discontinuity.

Now with this idea, lets us try ,

f(x)=(x^2-64/x+8)
= x-8

this can never break for any real value of x ...right ? so this function cant be having discontinious point ....am i right ?
1) "discontinuity rule" - What is that? Your definitions is very, very loose.

2) Your notation needs some work. This (x^2-64/x+8) is NOT the same as this (x^2-64)/(x+8). You tell me why.

3) Your conclusion is incorrect.f(x) = x-8 ONLY away from x = -8. If x = -8, f(x) does not exist. Notice, as you observed:

$\frac{x^{2}-64}{x+8}\;=\;\frac{(x+8)(x-8)}{x+8}\;=\;(x-8)*\frac{x+8}{x+8}$

The problem was your next conclusion. This expression is equivalent to x-8 ONLY if $\frac{x+8}{x+8}\;=\;1$. That expression does NOT equal one (1) for x = -8.

You must start such a problem by defining the Domain. x is NOT equal to -8. From there, you may continue with the simplification, but you may not remove the ORIGINAL restriction to the Domain.

3. ## Re: discontunity rule

Hello, defeated_soldier!

Discontunity rule says if a function breaks at certain point,
then that point is discontinuity point.

Example: $\:y\:=\:\frac{1}{x}$
If $x\,=\,0,\:y$ breaks.
So $x\,=\,0$ is a point of discontinuity. .Right!

Consider: $\:f(x)\:=\:\frac{x^2\,-\,4}{x\,-\,2}$
This function requires special treatment.

It is true that it factors: $\:f(x)\;=\;\frac{(x\,-\,2)(x\,+\,2)}{x\,-\,2}$

But we can't cancel if $x\,-\,2\:=\:0$
. . That is, if $x\,=\,2$, we'd be "cancelling zeros" . . . illegal!

So we explain it like this:
. . If $x\,=\,2$, the function does not exist.
. . If $x\,\neq\,2$, the function is: $\,f(x)\:=\:x\,+\,2$ ... a straight line.

The graph looks like this:
Code:
              |
|       *
|     *
|   o
| * :
*   :
* |   :
- - * - + - + - - - -
*     |   2
|

The graph is the straight line: $y\:=\:x\,+\,2\,$ . . . except when $x\,=\,2$.
. . There is a "hole" at $(2,4).$

The function is discontinuous at $x\,=\,2.$

4. amazing ...explanation.

well, i found as we dont know the value of x so, it would be unfair by cancealing x-2 ....yea ...thats good ....right.

so , if we make the denominator zero...then we can find the point of discontinuity ...thats the easiest method probabily.

Thank you

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