Results 1 to 5 of 5

Math Help - Math question

  1. #1
    Member
    Joined
    Nov 2006
    From
    chicago
    Posts
    156

    Question Math question

    Math question-amath23.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by Mr_Green View Post
    Click image for larger version. 

Name:	amath23.jpg 
Views:	19 
Size:	11.9 KB 
ID:	1238
    are you the same mr. green that posted this?

    Because you certainly have asked a lot of questions
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    From
    chicago
    Posts
    156

    Talking

    Quote Originally Posted by Quick View Post
    are you the same mr. green that posted this?

    Because you certainly have asked a lot of questions


    yes in deed
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2006
    From
    chicago
    Posts
    156
    a and b will be decimals rite?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,184
    Thanks
    403
    Awards
    1
    Quote Originally Posted by Mr_Green View Post
    Click image for larger version. 

Name:	amath23.jpg 
Views:	19 
Size:	11.9 KB 
ID:	1238
    You want an interval such that \left | f(x) - 6 \right | \leq 2.

    So you want \left | x^2 - 2x - 3 \right | \leq 2

    So
    x^2 - 2x - 3 \leq 2
    and
    x^2 - 2x - 3 \geq -2

    For the first inequality solve:
    x^2 - 2x - 3 = 2

    x^2 - 2x - 5 = 0

    So x = 1 \pm \sqrt{6} by the quadratic formula. We want the "+" solution as the "-" solution is less than 0. So we want x \leq 1 + \sqrt{6} to solve the inequality.

    For the second inequality solve:
    x^2 - 2x - 3 = -2

    x^2 - 2x - 1 = 0

    So x = 1 \pm \sqrt{2} by the quadratic formula. We again want the "+" solution since the "-" solution is less than 0. So we want x \geq 1 + \sqrt{2}.

    Putting the two together gives an interval for x:
    1 + \sqrt{2} \leq x \leq 1 + \sqrt{6}

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Math question
    Posted in the Business Math Forum
    Replies: 6
    Last Post: December 21st 2009, 10:46 AM
  2. 2 math question help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 5th 2008, 09:43 PM
  3. tricky math math question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 13th 2008, 09:18 PM
  4. Cr. Math and a dumb math question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 4th 2007, 02:55 PM
  5. math question help
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 13th 2006, 09:21 PM

Search Tags


/mathhelpforum @mathhelpforum