Results 1 to 2 of 2

Math Help - Root

  1. #1
    Super Member
    Joined
    Feb 2008
    Posts
    535

    Root

    Show that for any real constants c and d, the eqation x = c + dcos(x) has at least one root.

    My professor gave the hint: Find an interval [a,b] on which
    f(x) = x - c - dcos(x) changes sign.

    Any advice?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Calculus26's Avatar
    Joined
    Mar 2009
    From
    Florida
    Posts
    1,271
    f(x) = x - c - dcos(x)

    Case 1 dcos(c) > 0

    f(c) = -dcos(c) < 0

    f(c + pi) = pi -dcos(c+pi) = pi + dcos(c) > 0

    We're done

    Case 2 dcos(c) < 0

    f(c) = -dcos(c) > 0

    Let n be such that npi > dcos(c)

    f(c-npi) = -npi -dcos(c-npi) = -npi - dcos(c)cos(npi)

    cos(npi) = + 1

    in either case f(c-npi) < 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 5th 2011, 01:07 PM
  2. square root + cube root of a number
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 5th 2010, 10:35 AM
  3. Replies: 2
    Last Post: April 6th 2009, 03:51 AM
  4. Replies: 12
    Last Post: November 22nd 2008, 12:41 PM
  5. Replies: 1
    Last Post: March 29th 2008, 11:11 AM

Search Tags


/mathhelpforum @mathhelpforum