Greatest Integer Functions and odd, even functions

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Homework Help Overview

The discussion revolves around two questions involving greatest integer functions and properties of even and odd functions. The first question involves solving an equation with absolute values and greatest integer functions, while the second question explores the composition of an even function and an odd function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss case work for handling absolute values in the first question and mention properties of greatest integer functions. There is an exploration of the implications of function composition in the second question, with some participants questioning the correctness of their reasoning.

Discussion Status

Several participants have provided insights into the properties of greatest integer functions and the behavior of even and odd functions. There is an ongoing exploration of different approaches, including graphical methods and case analysis. Some participants express uncertainty about specific steps or properties, indicating a productive exchange of ideas.

Contextual Notes

Participants are navigating the complexities of greatest integer functions and absolute values, noting the need for careful consideration of cases and properties. There is an emphasis on understanding the definitions and behaviors of the functions involved.

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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