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Math Help - Range and inverse of bijective functions

  1. #1
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    Range and inverse of bijective functions

    Need help with functions, please.

    For each of the following bijective functions find the range S and the inverse:

    a.) f : x |→ x - 1 (x ∈ R, x ≥ 0)

    b.) f : x |→ (x + 1) ((x ∈ R, x ≤ -2)
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  2. #2
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    Quote Originally Posted by Hellbent View Post
    Need help with functions, please.

    For each of the following bijective functions find the range S and the inverse:

    a.) f : x |→ x - 1 (x ∈ R, x ≥ 0)

    b.) f : x |→ (x + 1) ((x ∈ R, x ≤ -2)
    so, u wanna find the range from the function....
    The simplest way is to write down/construct the graph from the function, the range of the is the interval of y-axis that fit the graph.

    example:

    - the range (S) for the graph is (0,inf)

    and to find the inverse from the function,

    1)
    let, y=x^2-1
    the objectives to arrange the function until you get the equation for x=...


    ........hope it help..........
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    Quote Originally Posted by pencil09 View Post
    so, u wanna find the range from the function....
    The simplest way is to write down/construct the graph from the function, the range of the is the interval of y-axis that fit the graph.

    example:

    - the range (S) for the graph is (0,inf)

    and to find the inverse from the function,

    1)
    let, y=x^2-1
    the objectives to arrange the function until you get the equation for x=...


    ........hope it help..........
    Helped a little, double-checked with someone else who said the range isn't (0,inf).
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  4. #4
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    Quote Originally Posted by Hellbent View Post
    Helped a little, double-checked with someone else who said the range isn't (0,inf).
    The range of y = x^2 - 1 is clearly [-1, oo). Drawing the graph is straightforward and from the graph the range is obvious. The previous poster was (I think) giving you an example of what to do, not answering the exact question you posted.
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    Hello
    \forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1
    so the range must be [-1,\infty ).
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    Quote Originally Posted by mr fantastic View Post
    The range of y = x^2 - 1 is clearly [-1, oo). Drawing the graph is straightforward and from the graph the range is obvious. The previous poster was (I think) giving you an example of what to do, not answering the exact question you posted.
    Help with part b.) please.
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  7. #7
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    Quote Originally Posted by Hellbent View Post
    Help with part b.) please.
    Have you drawn the graph? Where are you stuck?
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    Quote Originally Posted by Raoh View Post
    Hello
    \forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1
    so the range must be [-1,\infty ).
    How would the inverse be found?

    Quote Originally Posted by mr fantastic View Post
    Have you drawn the graph? Where are you stuck?
    Not sure how to draw the graph of this one.

    Quote Originally Posted by Raoh View Post
    Hello
    \forall x\in \mathbb{R},x^{2}\geq 0\Leftrightarrow x^2-1\geq -1
    so the range must be [-1,\infty ).
    Inverse: y = x - 1
    x = √(y + 1)
    Last edited by mr fantastic; November 16th 2009 at 06:49 PM. Reason: Merged posts
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  9. #9
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    Hello
    the geometric figure of (x+1)^2 is a parabola ,and since -1 is a root...u know what to do
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    Quote Originally Posted by mr fantastic View Post
    Have you drawn the graph? Where are you stuck?
    Would it be an upside down parabola, with x ≤ -2?
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  11. #11
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    Quote Originally Posted by Hellbent View Post
    Inverse: y = x - 1
    x = √(y + 1)

    f(x)=x^2-1\Rightarrow y=x^2-1\Rightarrow x^2=y+1\Leftrightarrow \left [x=\sqrt{y+1},x=-\sqrt{y+1}  \right ]
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  12. #12
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    Quote Originally Posted by Raoh View Post
    Hello
    the geometric figure of (x+1)^2 is a parabola ,and since -1 is a root...u know what to do
    Hello
    I hope, replace the x with -1.
    Got 0 as my answer?!
    (-1 + 1)(-1 + 1)
    1 - 1 - 1 + 1
     = 0
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  13. #13
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    Quote Originally Posted by Hellbent View Post
    Hello
    I hope, replace the x with -1.
    Got 0 as my answer?!
    (-1 + 1)(-1 + 1)
    1 - 1 - 1 + 1
     = 0
    Yes,which means -1 is a ROOT.
    and hence the graph should be like this,
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  14. #14
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    Do I follow the same steps with this one f : x |→ (x + 1) (x ∈ R, x ≤ -2)?
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  15. #15
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    Yes the same steps
    f(x)=(x+1)^2\Rightarrow y=(x+1)^2\Leftrightarrow \sqrt{y}=x+1\Leftrightarrow x=\sqrt{y}-1.
    therefore f^{-1}(y)=\sqrt{y}-1.
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