1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fixing domains with mods

  1. Jun 20, 2012 #1
    mod(x-6) > mod(x^2 - 5x + 9)

    Can anyone tell me about domain fixtures with mods using the above inequation?

    A real beginner, so have mercy, be elementry .:D
     
  2. jcsd
  3. Jun 20, 2012 #2

    Curious3141

    User Avatar
    Homework Helper

    The word "mods" can have different meanings in Maths. I think you're using it in the sense of "modulus", or absolute value here.

    So just to be clear, are you asking how to solve this inequality?

    [tex]|x-6| > |x^2 - 5x + 9|[/tex]

    You should start by making a careful sketch of both curves. Remember to consider the discriminant of the quadratic expression, its minimum point and the fact that [itex]|x-6|[/itex] is actually represented by two lines at right angles to each other.
     
  4. Jun 20, 2012 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I am going to assume you are using "mod" to mean "absolute value" (unfortunately "mod" has many different meanings in different circumstances.)

    To find values of x for which [itex]|x- 6|> |x^2- 5x+ 9|[/itex] consider the possible signs for the quantities inside the absolute values. Solve x- 6> 0 and x^2- 5x+ 9> 0 to determine where those are positive or negative.

    Where x- 6 and [itex]x^2- 5x+ 9[/itex] are both positive, the absolute values can be taken off to give [itex]x- 6> x^2- 5x+ 9[/itex] or [itex]0> x^2- 5x+ 9[/itex]. Solve that inequality and take the values of x that satisfy it as well as [itex]x- 6> 0[/itex] and [itex]x^2- 5x+ 9> 0[/itex].

    Where x- 6 is positive and [itex]x^2- 5x+ 9[/itex] is negative, the absolute values can be taken off to give [itex]x- 6> -(x^2- 6x+ 9)[/itex] which is the same as [itex]x^2- 5x+ 3> 0[/itex].

    Where x- 6 is negative and [itex]x^2- 5x+ 9[/itex] is positive, the absolute values can be taken off to give [itex]-(x- 6)> x^2- 6x+ 9[/itex] which is the same as [itex]0>x^2- 5x+ 3> 0[/itex].

    Where both x- 6 and [itex]x^2- 5x+ 9[/itex] are negative, the absolute values can be taken off to give [itex]-(x- 6)> -(x^2- 6x+ 9)[/itex] which is the same as [itex]x^2- 7x+ 15> 0[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fixing domains with mods
  1. Domain and Range (Replies: 4)

  2. Domain of a function (Replies: 4)

  3. Domain of a function (Replies: 5)

  4. Domain of the function (Replies: 2)

  5. Finding domain (Replies: 6)

Loading...