Results 1 to 3 of 3

Math Help - domain of (fofof)(x)

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    112

    domain of (fofof)(x)

    how to find the domain of (fofof)(x)
    where f(x)=sqrt(x-1)
    Book answer is [2,infinity).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    226
    The domain of f(g(x)) is the intersection of the domain of f with the domain of g. Extending the concept to three functions, the domain of f(g(h(x)) is the intersection of the domain of f with the domain of g with the domain of h.

    f(x) = \sqrt{x - 1}

    The domain of f(x) is all x such that x - 1 \geq 0. This implies that x \geq 1.

    f(f(x)) = \sqrt{\sqrt{x - 1} - 1}

    The domain of f(f(x)) is all x such that \sqrt{x - 1} - 1 \geq 0 and x \geq 1 (from the domain of f(x)). Let's solve the inequality in terms of x.

    \sqrt{x - 1} - 1 \geq 0

    \implies \sqrt{x - 1} \geq 1

    \implies x -1 \geq 1

    \implies x \geq 2

    The intersection of x \geq 1 and x \geq 2 is x \geq 2, so the domain of f(f(x)) is x \geq 2.

    f(f(f(x))) = \sqrt{\sqrt{\sqrt{x - 1} - 1} - 1}.

    The domain of f(f(f(x))) is all x such that \sqrt{\sqrt{x - 1} - 1} - 1 \geq 0 and x \geq 2 (from the domain of f(f(x))). Let's solve the inequality in terms of x.

    \sqrt{\sqrt{x - 1} - 1} - 1 \geq 0

    \implies \sqrt{\sqrt{x - 1} - 1} \geq 1

    \implies \sqrt{x - 1} - 1 \geq 1

    \implies \sqrt{x - 1} \geq 2

    \implies x - 1 \geq 4

    \implies x \geq 5

    The intersection of x \geq 2 and x \geq 5 is x \geq 5, so the domain of f(f(f(x))) is x \geq 5.

    I believe the answer in your book is incorrect assuming I did not err.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,779
    Thanks
    1519
    Misinterpreted the question.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Converting from the S domain to the Z domain
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: February 24th 2011, 10:15 PM
  2. Domain
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 29th 2010, 12:20 AM
  3. domain
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: January 8th 2010, 11:49 AM
  4. Replies: 6
    Last Post: September 16th 2009, 06:25 AM
  5. help with this domain
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: June 20th 2009, 02:38 PM

Search Tags


/mathhelpforum @mathhelpforum