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Thread: discontunity rule

  1. #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. #2
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    Re: discontunity rule

    Quote 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:

    [tex]\frac{x^{2}-64}{x+8}\;=\;\frac{(x+8)(x-8)}{x+8}\;=\;(x-8)*\frac{x+8}{x+8}[/tex]

    The problem was your next conclusion. This expression is equivalent to x-8 ONLY if [tex]\frac{x+8}{x+8}\;=\;1[/tex]. 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.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
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    Re: discontunity rule

    Hello, defeated_soldier!

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

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

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

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

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

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

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

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

    The function is discontinuous at [tex]x\,=\,2.[/tex]

    I'm the other of the two guys who "do" homework.

  4. #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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