Greatest Integer Functions and odd, even functions

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The discussion revolves around solving two problems involving greatest integer functions and the properties of even and odd functions. For the first question, participants suggest using case work to handle absolute values and emphasize understanding the properties of greatest integer functions, noting that expressions like [x+n] do not equal [x]+n in certain cases. In the second question, it is established that if f is even and g is odd, then the composition fog is even, as demonstrated through the relationship fog(-x) = f(g(-x)) = f(-g(x)). Participants stress the importance of correctly applying properties of absolute values and greatest integer functions to avoid common mistakes. Overall, the thread highlights the need for careful analysis and understanding of mathematical functions to solve these types of problems effectively.
Hydroxide
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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.
 
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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:
tim_lou said:
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
 
Or you could draw the graphs and find the solution graphically if you know some curve sketching.
 
fog(-x)=f(g(-x))=f(-g(x))=f(gx)
Hence fog is even.
 
chaoseverlasting said:
fog(-x)=f(g(-x))=f(-g(x))=f(gx)
Hence fog is even.

Yeah I figured that one out, but thanks anyway
 
Hydroxide said:
But that leaves me with [|x+2|]=[|2x-3|] correct?
So can i say [|x-5|]=0 ? hmmm, seems to simple

No!
2[x]\neq [2x]
furthermore,
|x|-3 \neq |x-3|
watch what you are dealing with, greatest integer function and absolute values have different properties.
[x]+[y]\neq [x+y]
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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