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Thread: domain of a 1-1 function and its inverse

  1. #1
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    domain of a 1-1 function and its inverse

    Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

    how do i find the domain and inverse? thanks!
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    Quote Originally Posted by break View Post
    Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

    how do i find the domain and inverse? thanks!
    Its actual domain is R.
    To make it a 1-1 function you should let x \leq -4 or x \geq -4. Do you know why?

    To find the inverse function, Let y=f(x) and solve this equation for x.
    Last edited by Miss; Mar 6th 2010 at 02:43 PM. Reason: -4 not 0. TYPO!
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    Quote Originally Posted by break View Post
    Find the domain on which y = x^2 + 8x + 12 is a 1-1 function, and find the inverse function y=f^-1.

    how do i find the domain and inverse? thanks!
    Since the graph of that function is a parabola, it will be one to one on the two intervals: x< x-coordinate of vertex and x> x-coordinate of vertex.

    You can complete the square to find the vertex.

    To find the two inverses, write the equation as x^2+ 8x+ 12-y= 0 an use the quadratic formula to solve for x.
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  4. #4
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    Quote Originally Posted by Miss View Post
    Its actual domain is R.
    To make it a 1-1 function you should let x \leq 0 or x \geq 0. Do you know why?

    To find the inverse function, Let y=f(x) and solve this equation for x.
    The idea of this is correct, but the correct value to "split" the domain at is not x=0.
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  5. #5
    Member Miss's Avatar
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    Quote Originally Posted by drumist View Post
    The idea of this is correct, but the correct value to "split" the domain at is not x=0.

    Oops. Its a typo
    It shoud be at x=-4
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