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Math Help - Function restriction.

  1. #1
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    Function restriction.

    f : R-> R

    f(x)= x^{2}+2x+3

    a) Define a restriction of f that admits inverse.


    Can u help me ?? I dont have any idea, how to do that.
    Last edited by mr fantastic; November 26th 2011 at 03:57 PM. Reason: Title.
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  2. #2
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    Re: Function restriction help plz

    Quote Originally Posted by Fabio010 View Post
    f : R-> R

    f(x)= x^{2}+2x+3

    a) Define a restriction of f that admits inverse.


    Can u help me ?? I dont have any idea, how to do that.
    f(x) = x^2 + 2x + 3 = x^2 + 2x + 1 + 2 = (x+1)^2 + 2

    note that the graph of f(x) is a parabola with vertex at (-1,2)

    for f(x) to have an inverse, the domain must be restricted so that f(x) is 1-1 ... that would be values of x such that x > -1
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  3. #3
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    Re: Function restriction help plz

    Quote Originally Posted by Fabio010 View Post
    f : R-> R

    f(x)= x^{2}+2x+3

    a) Define a restriction of f that admits inverse.


    Can u help me ?? I dont have any idea, how to do that.

    f(x)=y=x^2+2x+3

    Start by solving x^2+2x+(3-y)=0
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  4. #4
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    Re: Function restriction help plz

    I already know the inverse function.

    x = +/-\sqrt{y-2}-1

    The inverse graph is a parabole with vertex(2,-1)

    I dont understood the part f(x) is 1-1

    btw: Sorry for posting in wrong section(this problem is from a university calculus test)
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  5. #5
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    Re: Function restriction help plz

    A function is 1-1 (or injective) if it maps different arguments to different values. For a function to have an inverse, it must be injective: if y = f(x_1) = f(x_2) where x_1\ne x_2, then f^{-1}(y) is not defined because it cannot be both x_1 and x_2. Therefore, you need to find a restriction where the function is 1-1, as skeeter said.
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  6. #6
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    Re: Function restriction help plz

    Hum....so to be injective:

    we know that vertex of f is (-1,2) so,

    the restriction is [-1, +oo[ ??? as skeeter said...
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  7. #7
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    Re: Function restriction help plz

    Quote Originally Posted by Fabio010 View Post
    so to be injective:

    we know that vertex of f is (-1,2) so,

    the restriction is [-1, +oo[ ???
    Yes, or ]-oo, -1], or any subset of those two.
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  8. #8
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    Re: Function restriction help plz

    Ok, so to f admit inverse we must restrict the domain in order to make f injective. (Sorry my english)
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  9. #9
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    Re: Function restriction help plz

    Yes, that's right.
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