Greatest Integer Functions and odd, even functions

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SUMMARY

The discussion focuses on solving two mathematical problems involving greatest integer functions and the properties of even and odd functions. For Q1, participants emphasize the importance of understanding the properties of the greatest integer function and absolute values, particularly noting that [x+n] = [x] + n for integers. In Q2, 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 recommend breaking down the absolute value cases for clearer solutions.

PREREQUISITES
  • Understanding of greatest integer functions
  • Knowledge of properties of even and odd functions
  • Familiarity with absolute value equations
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of the greatest integer function in depth
  • Learn how to solve absolute value equations systematically
  • Explore the implications of function composition with even and odd functions
  • Practice case analysis in mathematical problem-solving
USEFUL FOR

Students studying calculus or algebra, mathematics educators, and anyone interested in understanding the properties of functions and their compositions.

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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