Results 1 to 3 of 3

Math Help - Square root's inverse right but restrictions?

  1. #1
    s3a
    s3a is offline
    Super Member
    Joined
    Nov 2008
    Posts
    597

    Square root's inverse right but restrictions?

    I get the answer which is f-1(x) = (-1/9(x+2)^2) + 1 but the solution guide says:

    " f-1(x) = (-1/9(x+2)^2) + 1 x >/= 2 OR f-1(x) = (-1/9(x+2)^2) + 1 [-2, infinty["

    How do I get such restrictions? This seems to be false when checked with my graphing software but I could be wrong.

    Any help would be greatly appreciated!
    Thanks in advance!
    Attached Thumbnails Attached Thumbnails Square root's inverse right but restrictions?-sqrt.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    let f(x)=y and the equation with respect to x

    y=3\sqrt{-(x-1)}-2

    y+2=3\sqrt{-(x-1)}

    (y+2)^2=9(-(x-1))

    \frac{(y+2)^2}{9}=-(x-1)

    -\left(\frac{(y+2)^2}{9}\right)=x-1

    1-\left(\frac{(y+2)^2}{9}\right)=x

    change the variables

    f(x)=1-\left(\frac{(x+2)^2}{9}\right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    f(x) = 3\sqrt{-(x - 1)} - 2

    The inverse, as Amer worked out, is
    f^{-1}(x) = -\frac{(x + 2)^2}{9} + 1
    (I like to put the variables first. )

    But it wouldn't be right to leave it as it is without stating the restrictions. If we consider this function without knowing beforehand that it was the inverse of some other function, the domain would be all reals and the range would be (-∞, 1], but that's wrong in our case. This function f^{-1}(x) is the inverse of f(x), and since f(x) is one-to-one, the domains and ranges switch.

    The domain of f(x) is (-∞, 1] and the range is [-2, ∞). Therefore the domain of f^{-1}(x) is [-2, ∞) and the range is (-∞, 1]. The graph of f^{-1}(x) would be the right half of a parabola, opening downward.


    01
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square root inside square root equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 10th 2011, 04:17 PM
  2. Replies: 12
    Last Post: November 22nd 2008, 12:41 PM
  3. Replies: 2
    Last Post: June 14th 2008, 11:50 PM
  4. Simplifying Square Root w/in square root
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 7th 2006, 08:01 AM
  5. Replies: 2
    Last Post: April 29th 2006, 01:13 AM

Search Tags


/mathhelpforum @mathhelpforum