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Math Help - Points not included in Domain

  1. #1
    Junior Member cupid's Avatar
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    Points not included in Domain

    image >>>

    For 21

    book gives answer: x \in (2,4)
    but the function at x = 2 becomes \sqrt{\frac{0}{-8}} \implies \sqrt{0}
    so shouldn't point 2 be included in the domain?

    For 23

    book gives answer: x \in (2,infinity)

    but here also at x=2 function is defined

    i think in both questions point 2 should be included ... just need confirmation or reason why it is not included?
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  2. #2
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    \displaystyle x^2-2x-8 = (x-4)(x+2) \implies x \neq -2,4
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  3. #3
    Junior Member cupid's Avatar
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    Its 2 not -2
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  4. #4
    Master Of Puppets
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    Maybe the book's answer is incorrect.
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  5. #5
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    Hello, cupid!

    I don't understand their answers.


    21.\;f(x) \;=\;\sqrt{\dfrac{\log_{0.3}(x-1)}{x^2-2x-8}}

    First, note that: . x\,>\,1


    The radicand must be positive.

    . . [1] The numerator and denominator are both positive, or
    . . [2] The numerator and denominator are both negative.


    Case [1]

    \log_{0.3}(x-1) \:>\:0 \quad\Rightarrow\quad x-1 \:>\:0.3^0 \quad\Rightarrow\quad x \:>\:2

    (x+2)(x-4) \:>\:0 \quad\Rightarrow\quad x \,<\,\text{-}2\,\text{ or }\,x \,>\,4

    . . Hence: . x\,>\,4


    Case [2]

    \log_{0.3}(x-1)\,\le\,0 \quad\Rightarrow\quad x-1\,\le\,0.3^0 \quad\Rightarrow\quad 1\,<\,x \,\le\,2

    (x+2)(x-4)\:<\:0 \quad\Rightarrow\quad -2 \,<\,x\,<\,4

    . . Hence: . 1\,<\,x\,\le\,2


    \text{The domain is: }\;\left(1,2\,\right] \cup (4,\infty)

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  6. #6
    Junior Member cupid's Avatar
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    your domain include point 6

    so log_{0.3}5 = -1.337
    and denominator is 16

    so root will be of negative number
    which is not defined (we're not talking about complex numbers)

    so this domain is wrong

    right now i have to go to my class ... will post my solution later.
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  7. #7
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    Quote Originally Posted by cupid View Post
    your domain include point 6
    so log_{0.3}5 = -1.337
    and denominator is 16 so root will be of negative number
    which is not defined (we're not talking about complex numbers)
    so this domain is wrong.
    The domain is indeed [2,4).
    I suspect whoever supplied the book's answer was just careless.
    If you graph the function, it is easy to see this.
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  8. #8
    Junior Member cupid's Avatar
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    can you provide me with the the graph ??

    And what about the 23rd ?
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  9. #9
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    Points not included in Domain-untitled.gif
    Quote Originally Posted by cupid View Post
    can you provide me with the the graph ?
    .
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  10. #10
    Junior Member cupid's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, cupid!

    I don't understand their answers.



    First, note that: . x\,>\,1


    The radicand must be positive.

    . . [1] The numerator and denominator are both positive, or
    . . [2] The numerator and denominator are both negative.


    Case [1]

    \log_{0.3}(x-1) \:>\:0 \quad\Rightarrow\quad x-1 \:>\:0.3^0 \quad\Rightarrow\quad x \:>\:2

    (x+2)(x-4) \:>\:0 \quad\Rightarrow\quad x \,<\,\text{-}2\,\text{ or }\,x \,>\,4

    . . Hence: . x\,>\,4


    Case [2]

    \log_{0.3}(x-1)\,\le\,0 \quad\Rightarrow\quad x-1\,\le\,0.3^0 \quad\Rightarrow\quad 1\,<\,x \,\le\,2

    (x+2)(x-4)\:<\:0 \quad\Rightarrow\quad -2 \,<\,x\,<\,4

    . . Hence: . 1\,<\,x\,\le\,2


    \text{The domain is: }\;\left(1,2\,\right] \cup (4,\infty)


    Here's my solution >>> PIC
    [img][/img]
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