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    For part b I've got gf = 3/8x-3 but then the questions asks for it's domain and I don't know how to do that..

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    (Original post by creativebuzz)
    For part b I've got gf = 3/8x-3 but then the questions asks for it's domain and I don't know how to do that..

    Well, as x is on the bottom of the fraction, think of the cases where the function would return an undefined number?
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    (Original post by sarcasmrules)
    Well, as x is on the bottom of the fraction, think of the cases where the function would return an undefined number?
    It would be undefined if 8x-3=0 so x=3/8

    But why are we trying to find where the function would be undefined?
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    (Original post by creativebuzz)
    It would be undefined if 8x-3=0 so x=3/8

    But why are we trying to find where the function would be undefined?
    Well domain is all the values of x a function can take
    When x=3/8 we get 0 for the denominator which we can't do.

    When you have a fraction the domain is every x value apart from the one that makes it 0
    So in this case x is an element of the real numbers , x cannot be 3/8
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    Is there a cheat to work out domain and function
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    (Original post by Super199)
    Well domain is all the values of x a function can take
    When x=3/8 we get 0 for the denominator which we can't do.

    When you have a fraction the domain is every x value apart from the one that makes it 0
    So in this case x is an element of the real numbers , x cannot be 3/8
    Ah okay, so where did I go wrong here in part b:



    y = x/x -2

    therefore, horizontal asymptote: x=2
    vertical asymptote: y= 2

    so the range is every value except x = 2
    and the domain is every value except y = 1
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    (Original post by creativebuzz)
    Ah okay, so where did I go wrong here in part b:



    y = x/x -2

    therefore, horizontal asymptote: x=2
    vertical asymptote: y= 2

    so the range is every value except x = 2
    and the domain is every value except y = 1
    Do you know the relationship between the domain of f(x) and the range of f^-1(x)?
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    (Original post by Super199)
    Do you know the relationship between the domain of f(x) and the range of f^-1(x)?
    the domain of f(x) = the range of f^-1(x)

    the range of f(x) = the domain of f^-1(x)

    so if the domain of f(x) is every value except x =2, then the range of f^-1(x) is every value except x =2

    is that correct?
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    (Original post by creativebuzz)
    the domain of f(x) = the range of f^-1(x)

    the range of f(x) = the domain of f^-1(x)

    so if the domain of f(x) is every value except x =2, then the range of f^-1(x) is every value except x =2

    is that correct?
    Y is an element of the real y cannot be 2. Since it's range and not domain so you use y.
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    (Original post by Super199)
    Y is an element of the real y cannot be 2. Since it's range and not domain so you use y.
    I don't understand what you're trying to say..
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    (Original post by creativebuzz)
    I don't understand what you're trying to say..
    You use y instead of x when it's range ( I think). So every y value apart from 2
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    (Original post by Super199)
    You use y instead of x when it's range ( I think). So every y value apart from 2

    Ah yes I see what you mean! I understand it now! Positive rating given for thanks
 
 
 
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