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Math Help - how do i solve question 1? and is the answer for qusetion 2 correct?

  1. #1
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    how do i solve question 1? and is the answer for qusetion 2 correct?

    1)The domain of f(x)=
    how do i solve question 1? and is the answer for qusetion 2 correct?-e5.jpg

    2)
    how do i solve question 1? and is the answer for qusetion 2 correct?-e6.jpg

    a)x=2
    b)x=4
    c)x=-2
    d)3

    my answer is a for both qusetions is that right?
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  2. #2
    Senior Member yeKciM's Avatar
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    1. D_p : {(-\infty , -3)}\cup{(-3,4)}\cup{(4, +\infty)}

    2: x=2
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  3. #3
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    Quote Originally Posted by yeKciM View Post
    1. D_p : {(-\infty , -3)}\cup{(-3,4)}\cup{(4, +\infty)}

    2: x=2
    thankx for ur reply ....for question 1 what is the answer yekcim??and what is Dp??
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  4. #4
    Senior Member yeKciM's Avatar
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    that's region where ur function is defined.... u probably don't mark it as D_p sorry
    meaning that ur function is not defined in -3 and 4 ... and everywhere else it is...
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    Quote Originally Posted by yeKciM View Post
    that's region where ur function is defined.... u probably don't mark it as D_p sorry
    meaning that ur function is not defined in -3 and 4 ... and everywhere else it is...
    yeah but this is a multiple choice question where i have to choose only 1 answer... so whats the right answer??
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  6. #6
    Senior Member yeKciM's Avatar
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    sorry ... didn't look at this right... ur function is f_{(x)}=\frac{x-4}{(x+3)(x-4)}=\frac{1}{x+3} so it's just -3 sorry sorry



    it should be a) R{-3}
    Last edited by yeKciM; July 24th 2010 at 01:11 AM. Reason: my mistake :D
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  7. #7
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    No! yeKciM, your first answer was correct.

    \frac{x- 4}{x^2- x- 12}= \frac{x-4}{(x-4)(x+3)}
    is not defined for x= 4 or x= -3.

    \frac{x- 4}{(x- 4)(x+ 3)}= \frac{1}{x+3} only for x\ne 4.

    If x= 4, \frac{x- 4}{(x- 4)(x+ 3)}= \frac{0}{0} which is indeterminant while \frac{1}{x+ 3}= \frac{1}{7}.
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  8. #8
    Senior Member yeKciM's Avatar
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    yes i know but when u plot it ... it's define in x=4....
    just for -3 it's not
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    Quote Originally Posted by yeKciM View Post
    yes i know but when u plot it ... it's define in x=4....
    just for -3 it's not
    ok so i'm confused now?? !! what is the answer which choice is it a,b,c or d??
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  10. #10
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    for the rational function \displaystyle f(x) = \frac{x-4}{(x+3)(x-4)} , there is a vertical asymptote at x = -3 and a point discontinuity at x = 4.
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  11. #11
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by soso4 View Post
    ok so i'm confused now?? !! what is the answer which choice is it a,b,c or d??
    well i think its just -3

    reason why i think that is, if we look at function like

    f_{(x)}=\frac{x-4}{(x-4)(x+3)}

    it would mean that it's not defined in x=4 , x=-3 and that it has stationary point in x=4 but that isn't true ... it hasn't any stationary point's and it's defined in x=4 as u see from plot that i posted .... witch is actually function

    f_{(x)}=\frac {1}{x+3}

    lol... didn't have this much of polemics for such a simple function a long time


    P.S. u shouldn't be just interested in "which choice is it a,b,c or d??"... it's more important to u to understand how and why does i say this, or HelsofIvy that, or skeeter... rather then be just interested in right answer... and soon have more problems (with same type) which u can't solve
    Last edited by yeKciM; July 24th 2010 at 07:45 AM.
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  12. #12
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    Quote Originally Posted by yeKciM View Post
    well i think its just -3
    and you are incorrect.

    note the following graphs ...

    the first graph is a large view of the function showing the vertical asymptote.

    the second graph is the function "blown up" near x = 4 ... note the "hole", a point discontinuity.
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  13. #13
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by skeeter View Post
    and you are incorrect.

    note the following graphs ...

    the first graph is a large view of the function showing the vertical asymptote.

    the second graph is the function "blown up" near x = 4 ... note the "hole", a point discontinuity.
    in which app did u plot it ?
    i zoom my a loot and it doesn't show that ..
    lol I'm not saying that ur wrong just need to know which app do u use
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  14. #14
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    TI-84 emulator
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  15. #15
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    Quote Originally Posted by yeKciM View Post
    yes i know but when u plot it ... it's define in x=4....
    just for -3 it's not
    No, it isn't. The graph of y= \frac{x-4}{(x-4)(x+3)} looks like a hyperbola but with a hole at (4, 1/7). The graph of y= 1/(x+3) is the same hyperbola without the hole. If you plotted it with a calculator or a computer graphing program, your grid was probably too course, and it "jumped over" x= 4.

    It can be crucially important in Calculus to distinguish things like \frac{x-4}{(x-4)(x+3)} from \frac{1}{x+3}.
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