Greatest Integer Functions and odd, even functions

  • Thread starter Hydroxide
  • Start date
  • #1
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Homework Statement


Q1
[|x+2|]=2[|x|]-3

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

Homework Equations





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.
 

Answers and Replies

  • #2
682
1
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:
  • #3
16
0
Q1, case work first to take care of the absolute value (n>=0 and n<0).
notice that if n is in integer,
[x+n]=[x]+n

Q2. fog(-x)=f(g(-x)) by definition.
But that leaves me with [|x+2|]=[|2x-3|] correct?
So can i say [|x-5|]=0 ? hmmm, seems to simple
 
  • #4
Or you could draw the graphs and find the solution graphically if you know some curve sketching.
 
  • #5
fog(-x)=f(g(-x))=f(-g(x))=f(gx)
Hence fog is even.
 
  • #6
16
0
fog(-x)=f(g(-x))=f(-g(x))=f(gx)
Hence fog is even.
Yeah I figured that one out, but thanks anyway
 
  • #7
682
1
But that leaves me with [|x+2|]=[|2x-3|] correct?
So can i say [|x-5|]=0 ? hmmm, seems to simple
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:

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