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Math Help - [SOLVED] Find the domain of the function

  1. #1
    Junior Member moonman's Avatar
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    [SOLVED] Find the domain of the function

    f(x) = log(x^2 - 2x - 3)
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    Quote Originally Posted by moonman View Post
    f(x) = log(x^2 - 2x - 3)
    Find the solution set for x^2 -2x -3 >0.
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  3. #3
    Junior Member moonman's Avatar
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    Quote Originally Posted by Plato View Post
    Find the solution set for x^2 -2x -3 >0.
    x^2 -2x -3 > 0

    (x+1)(x-3)
    x=-1 x=3 is this the answer?
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    Quote Originally Posted by moonman View Post
    x^2 -2x -3 > 0

    (x+1)(x-3)
    x=-1 x=3 is this the answer?
    You have solved x^2 - 2x - 3 = 0. Is that what Plato instructed you to solve ....?
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  5. #5
    Junior Member moonman's Avatar
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    Quote Originally Posted by mr fantastic View Post
    You have solved x^2 - 2x - 3 = 0. Is that what Plato instructed you to solve ....?
    Yes?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moonman View Post
    Yes?
    i recall seeing a ">" sign in Plato's post, as opposed to an "=" sign
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    Junior Member moonman's Avatar
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    Quote Originally Posted by Jhevon View Post
    i recall seeing a ">" sign in Plato's post, as opposed to an "=" sign
    ahhh~!!!
    How do I solve when those > < things are there?
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    Quote Originally Posted by moonman View Post
    ahhh~!!!
    How do I solve when those > < things are there?
    Have you been taught anything about solving inequations? Have you seen equations like this before?
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  9. #9
    Junior Member moonman's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Have you been taught anything about solving inequations? Have you seen equations like this before?
    I've been away from math for so long
    I know that > means greater than and this < means less than.
    The question has to do with domain, so is the domain -1 to 3?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moonman View Post
    ahhh~!!!
    How do I solve when those > < things are there?
    ok, what you did is a start, we found where it IS zero, we want to find where it's bigger. so we can check the intervals that our solution breaks up the real line into

    we have x = -1 and x = 3

    so draw a number line and mark this on it. now, test numbers on both ends of the number line, and inbetween the numbers. if you get a positive answer, then that region is where we are greater than zero, if you get a negative number, that's where it's less than zero


    incidentally, domain is just the set of x-values for which a function is defined. we have to solve for where this expression is positive, because the logarithm is only defined where its argument is positive. so once you answer the above question, that's your domain
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  11. #11
    Junior Member moonman's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok, what you did is a start, we found where it IS zero, we want to find where it's bigger. so we can check the intervals that our solution breaks up the real line into

    we have x = -1 and x = 3

    so draw a number line and mark this on it. now, test numbers on both ends of the number line, and inbetween the numbers. if you get a positive answer, then that region is where we are greater than zero, if you get a negative number, that's where it's less than zero


    incidentally, domain is just the set of x-values for which a function is defined. we have to solve for where this expression is positive, because the logarithm is only defined where its argument is positive. so once you answer the above question, that's your domain

    How do I test the numbers?
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moonman View Post
    How do I test the numbers?
    plug them into the original quadratic, (x + 1)(x - 3), evaluate and see if your answer is positive or negative
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  13. #13
    Junior Member moonman's Avatar
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    Quote Originally Posted by Jhevon View Post
    plug them into the original quadratic, (x + 1)(x - 3), evaluate and see if your answer is positive or negative

    Okay, so I plugged in a -3:
    (-3 + 1) (-3 - 3)
    so that gave me positive 12


    Then I plugged in a -1
    (-1 + 1) (-1 - 3)
    so that gave me neutral 0

    Then I plugged in a 1
    (1 + 1) (1 - 3)
    so that gave me negative 4

    Then I plugged in a 3
    (3 + 1)(3 - 3)
    so that gave me neutral 0

    Then I plugged in a 5
    (5 + 1)(5 - 3)
    so that gave me positive 12

    Did i do this right? What's next?
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    Quote Originally Posted by moonman View Post
    Okay, so I plugged in a -3:
    (-3 + 1) (-3 - 3)
    so that gave me positive 12


    Then I plugged in a -1
    (-1 + 1) (-1 - 3)
    so that gave me neutral 0

    Then I plugged in a 1
    (1 + 1) (1 - 3)
    so that gave me negative 4

    Then I plugged in a 3
    (3 + 1)(3 - 3)
    so that gave me neutral 0

    Then I plugged in a 5
    (5 + 1)(5 - 3)
    so that gave me positive 12

    Did i do this right? What's next?
    There are big gaps in your understanding. Start filling them in by reading this: Solving Quadratic Inequalities: Concepts
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  15. #15
    Junior Member moonman's Avatar
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    Domain is all real numbers? ?
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