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

    For 23

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

    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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    $\displaystyle \displaystyle x^2-2x-8 = (x-4)(x+2) \implies x \neq -2,4$
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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.


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

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


    The radicand must be positive.

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


    Case [1]

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

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

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


    Case [2]

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

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

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


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

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

    so $\displaystyle log_{0.3}5 = -1.337$
    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 $\displaystyle log_{0.3}5 = -1.337$
    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 $\displaystyle [2,4)$.
    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: .$\displaystyle x\,>\,1$


    The radicand must be positive.

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


    Case [1]

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

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

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


    Case [2]

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

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

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


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


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