Results 1 to 12 of 12

Math Help - Who can?

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    66

    Who can?

    Find the beginning commune

    y=((log2(x-2) ))*1/2 (the first 2 is the base of log)



    Find the end (valuation)commune

    y=4/(x*2+2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by blertta View Post
    Find the beginning commune

    y=((log2(x-2) ))*1/2 (the first 2 is the base of log)



    Find the end (valuation)commune

    y=4/(x*2+2)
    This looks incomprehensible to me.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    Quote Originally Posted by CaptainBlack View Post
    This looks incomprehensible to me.

    CB
    Because my engish is 2 bad and I don't know how to write with math language,for ex. I don't know how to express square root so I wrote it like a base with exponent 1/2.
    Hope u understood me!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    What does "commune" mean?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    Quote Originally Posted by Jameson View Post
    What does "commune" mean?
    For example the "beggining commune" of x*1/2 is R+.
    Are the values of x which the function has realization.
    Hope u understood me?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by blertta View Post
    For example the "beggining commune" of x*1/2 is R+.
    Are the values of x which the function has realization.
    Hope u understood me?
    I'm not sure I do.

    If f(x)=\sqrt{x-1}, is the commune of f(x) x>1? If this is what you mean then commune I take to mean domain of x. I fear though this is not exactly what you mean. Try looking around online for different vocabulary to use. I'm sorry I don't quite get you. Maybe someone else will and I'm missing something.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Ah, I think you're referring to the domain.

    1. y=\sqrt{\log_2{(x-2)}}

    Square roots are defined, like you said, for \mathbb{R}^+\cup \{0\}. Therefore:
    \log_2{(x-2)}\geq 0 \implies x-2\geq 1 \implies x\geq 3

    Therefore, domain set is [3, \infty)


    2. y=\frac{4}{2x+2}
    This expression is defined for 2x+2\neq 0 \implies x\neq -1

    Therefore, domain set is \mathbb{R} \backslash \{-1\}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    Quote Originally Posted by Jameson View Post
    I'm not sure I do.

    If f(x)=\sqrt{x-1}, is the commune of f(x) x>1? If this is what you mean then commune I take to mean domain of x. I fear though this is not exactly what you mean. Try looking around online for different vocabulary to use. I'm sorry I don't quite get you. Maybe someone else will and I'm missing something.
    1/2 is an exponent. Anyway,thank you
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    Quote Originally Posted by Chop Suey View Post
    Ah, I think you're referring to the domain.

    1. y=\sqrt{\log_2{(x-2)}}

    Square roots are defined, like you said, for \mathbb{R}^+\cup \{0\}. Therefore:
    \log_2{(x-2)}\geq 0 \implies x-2\geq 1 \implies x\geq 3

    Therefore, domain set is [3, \infty)


    2. y=\frac{4}{2x+2}
    This expression is defined for 2x+2\neq 0 \implies x\neq -1

    Therefore, domain set is \mathbb{R} \backslash \{-1\}
    i think you understood me.
    At the first, as I know x-2 must be bigger or equal 0 too.
    at the second, the exercise asks not for the domain but for the end "commune".Do you understand me?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by blertta View Post
    i think you understood me.
    At the first, as I know x-2 must be bigger or equal 0 too.
    at the second, the exercise asks not for the domain but for the end "commune".Do you understand me?
    Oh, then I think you're referring to the range. Determining the range may be a bit tricky so we refer to a graph.

    I attached the graph of the second expression.

    As you can see, the y values covered are \mathbb{R}\backslash\{0\}

    I hope I understood you correctly. You may want to refer to these terms as domain ("beginning commune") and range ("end commune").
    Attached Thumbnails Attached Thumbnails Who can?-graph.jpg  
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by blertta View Post
    i think you understood me.
    At the first, as I know x-2 must be bigger or equal 0 too. This is wrong
    A = \{x~|~ x\in \mathbb{R}~\&~x-2>0\} is the set for which the logarithm is defined. But not all of the range is defined for the square root. Therefore, A is not the domain set.

    B = \{x~|~ x\in \mathbb{R}~\&~x-2\geq 1\} however, is the set for which the logarithm are greater than or equal to zero. This is the domain set for the expression. Note that B \subset A
    Last edited by Chop Suey; December 8th 2008 at 01:26 AM.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Sep 2007
    Posts
    66
    Quote Originally Posted by Chop Suey View Post
    Oh, then I think you're referring to the range. Determining the range may be a bit tricky so we refer to a graph.

    I attached the graph of the second expression.

    As you can see, the y values covered are \mathbb{R}\backslash\{0\}

    I hope I understood you correctly. You may want to refer to these terms as domain ("beginning commune") and range ("end commune").
    Thank you very much and Happy Ramadan's day if you are muslim.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum