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Math Help - Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5

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    Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5

    Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
    The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
    What is the domain of f-1?

    Would the answer be
    -9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5
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    Quote Originally Posted by azninvasion View Post
    Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
    The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
    What is the domain of f-1?

    Would the answer be
    -9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5
    the domain and range of a function and it's inverse are "swapped".
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    so, does that mean the answer is -4 ≤ x ≤ 5????
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    Quote Originally Posted by azninvasion View Post
    so, does that mean the answer is -4 ≤ x ≤ 5????
    No.

    The domain of f^{-1} equals the range of f
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    Quote Originally Posted by e^(i*pi) View Post
    No.

    The domain of f^{-1} equals the range of f
    -4 ≤ f-1(x) ≤ 5 Its this then?
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    I think its your first answer,
    Last edited by mr fantastic; November 16th 2009 at 04:20 AM.
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    0 ≤ x ≤ 9 maybe its this? haha this is like the hardest math problem on this forum ><
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    Quote Originally Posted by azninvasion View Post
    Suppose f is a one-to-one function whose domain is given by the inequality -4 ≤ x ≤ 5.
    The range of f is given by the inequality 0 ≤ f(x) ≤ 9.
    What is the domain of f-1?

    Would the answer be
    -9 ≤ f-1(x) ≤ 0 or -4 ≤ f-1(x) ≤ 5
    domain are x-values ...

    since the range of f(x) is 0 \le f(x) \le 9 , the domain of f^{-1}(x) is 0 \le x \le 9
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    Quote Originally Posted by skeeter View Post
    domain are x-values ...

    since the range of f(x) is 0 \le f(x) \le 9 , the domain of f^{-1}(x) is 0 \le x \le 9
    oooooo, why didn't you say so in the first place. now I get it.

    +thanked.
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    Quote Originally Posted by skeeter View Post
    the domain and range of a function and it's inverse are "swapped".
    ... that's what I said "in the first place".
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  11. #11
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    Quote Originally Posted by skeeter View Post
    ... that's what I said "in the first place".
    but your post was in the second place

    not that funny now that i read it
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