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Homework Help: Greatest Integer Functions and odd, even functions

  1. Mar 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Q1
    [|x+2|]=2[|x|]-3

    Q2
    If f is even and g is odd, is fog even, odd or neither

    2. Relevant equations



    3. The attempt at a solution
    Q1
    Not sure. Can someone please give me a start on this?
    I think if I knew some properties of greatest integer functions I could work it out

    Q2
    Let f(-x)=f(x) and g(-x)=-g(x)

    All I need to know is what fog(-x) equals. The rest I can do myself.

    Thanks a lot.
     
  2. jcsd
  3. Mar 13, 2007 #2
    Q1, case work first to take care of the absolute value (x>=0 and x<0).
    notice that if n is in integer,
    [x+n]=[x]+n

    Q2. fog(-x)=f(g(-x)) by definition.

    edit: sry typo, it should be x>=0 and x<0
     
    Last edited: Mar 13, 2007
  4. Mar 13, 2007 #3
    But that leaves me with [|x+2|]=[|2x-3|] correct?
    So can i say [|x-5|]=0 ? hmmm, seems to simple
     
  5. Mar 13, 2007 #4
    Or you could draw the graphs and find the solution graphically if you know some curve sketching.
     
  6. Mar 13, 2007 #5
    fog(-x)=f(g(-x))=f(-g(x))=f(gx)
    Hence fog is even.
     
  7. Mar 13, 2007 #6
    Yeah I figured that one out, but thanks anyway
     
  8. Mar 13, 2007 #7
    No!
    [tex]2[x]\neq [2x][/tex]
    furthermore,
    [tex]|x|-3 \neq |x-3| [/tex]
    watch what you are dealing with, greatest integer function and absolute values have different properties.
    [tex][x]+[y]\neq [x+y][/tex]
    watch what identities you are using! I suggest you get a deep understand of what greatest integer function means. [x] is the greatest integer that is less than or equal to x.
    for instance, [1.01]=1 and [1.9999999999999999999]=1.

    suppose you have [1.5]=1, but 2[1.5]=2 while [2*1.5]=3.
    for absolute value, suppose you have -3, |-3|=3, but |-3|+2=5, while |-3+2|=1. they are not equal!

    get difference cases, for x>0, |x|=x, that is one case
    for x<=0, |x|=-x, that is another case, similarly for |x+2|.

    get rid of the absolute values first, break it up into two equations.

    then get rid of complications in [], since it is not easy to work with [x]. so make [x+2]=[x]+2. the rest is simple algebra.
     
    Last edited: Mar 13, 2007
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