Results 1 to 8 of 8

Math Help - Help Finding the Inverse!!!

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    21

    Help Finding the Inverse!!!

    Hi i have a question to find the inverse, and i am kinda stuck. Can anyone please help me.

    Y = Srt of X + 2X

    Find the inverse.

    Thanks in advance =]
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Question: Find the inverse of y = \sqrt{x} + 2x.

    The first step is to switch x and y:

    x = \sqrt{y} + 2y

    Then it's time for a little equation rearrangement:

    x - 2y = \sqrt{y}

    (x - 2y)^2 = y

    x^2 - 4xy + 4y^2 = y

    4y^2 - 4xy - y + x^2 = 0

    4y^2 - (4x + 1)y + x^2 = 0

    y = \frac{4x + 1 \pm \sqrt{(4x + 1)^2 - 4(4)(x^2)}}{8}

    y = \frac{4x + 1 \pm \sqrt{(4x + 1)^2 - 16x^2}}{8}

    y = \frac{4x + 1 \pm \sqrt{16x^2 + 8x + 1 - 16x^2}}{8}

    y = \frac{4x + 1 \pm \sqrt{8x + 1}}{8}

    Now I'm not sure how to proceed with this, but perhaps someone else can give a little insight.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    21
    Thanks!

    However, i actually figured it out all the way up to that part. I know the inverse is also a function, but i do not know how to figure out which answer i reject and which answer to accept.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    I think, after thinking about it, that the correct solution is y = \frac{4x + 1 - \sqrt{8x + 1}}{8}. My reasoning is that the original function y = \sqrt{x} + 2x is always above the line y = 2x, so the inverse function must always be below the line y = 0.5x. Note that the inverse is only defined for positive values of x, because the original function, which is also defined for only positive values of x, always produces positive outputs.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by icemanfan View Post
    I think, after thinking about it, that the correct solution is y = \frac{4x + 1 - \sqrt{8x + 1}}{8}. My reasoning is that the original function y = \sqrt{x} + 2x is always above the line y = 2x, so the inverse function must always be below the line y = 0.5x. Note that the inverse is only defined for positive values of x, because the original function, which is also defined for only positive values of x, always produces positive outputs.
    Another way I think you can find out which is the inverse is that the domain and range just switch. f(x) = Domain A and range B and the inverse function of x equals Domain B and Range A. I'm not sure though.
    Last edited by 11rdc11; October 24th 2008 at 02:25 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Haven't you guys considered inyection and surjection? Does make sense to find the inverse?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by 11rdc11 View Post
    Another way I think you can find out which is the inverse is that the domain and range just switch. f(x) = Domain A and range B and the inverse function of x equals Domain B and Range A. I'm not sure though.
    This method of checking confirms that my intuition was correct. For example: (4, 10) is a point on the original function, and (10, 4) is a point on the inverse. (1, 3) is a point on the original function, and (3, 1) is a point on the inverse. Edit: 11rdc11, your proof is comprehensive and thank you.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Also remember that
    f(f^{-1}(x))=x

    So

    y = \frac{4x + 1 - \sqrt{8x + 1}}{8}

    is correct because

    y = \frac{4(\sqrt{x} + 2x) + 1 - \sqrt{8( \sqrt{x} + 2x)+ 1}}{8}

    y = \frac{4\sqrt{x} + 8x + 1 - \sqrt{8\sqrt{x} + 16x+ 1}}{8}

    let

    u = \sqrt{x}

    \frac{8u^2+4u+1 - \sqrt{16u^2+8u+1}}{8}

    \frac{8u^2+4u+1 - \sqrt{(4u+1)^2}}{8}

    \frac{8u^2+4u+1 - (4u+1)}{8}

    \frac{8u^2}{8}

    u^2

    backsub in for u

    y = (\sqrt{x})^2

    y=x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. help finding the inverse
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 16th 2011, 07:39 PM
  2. Finding the inverse
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 6th 2010, 03:15 AM
  3. finding inverse
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 29th 2009, 02:14 PM
  4. inverse trig values and finding inverse
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: April 6th 2009, 12:04 AM
  5. Finding the inverse of f(x)
    Posted in the Pre-Calculus Forum
    Replies: 9
    Last Post: October 6th 2008, 07:06 PM

Search Tags


/mathhelpforum @mathhelpforum