Finding f(6) from a composite function

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Homework Help Overview

The discussion revolves around finding the value of f(6) from a composite function f, with participants exploring the properties of the function, including its injectivity and the implications of given equations and inequalities.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the injectivity of the function and its implications, question how to derive f(x), and explore the relationships between function values based on inequalities. Some suggest using specific values and constructing tables to infer results.

Discussion Status

The conversation is active, with various approaches being proposed, including guessing values for f(1) and using function composition properties. While some participants express frustration with their progress, others offer guidance and suggest methods for exploration.

Contextual Notes

There are constraints regarding the definition of f for negative numbers, and participants are working within the framework of the problem's given equations and inequalities. The original poster expresses uncertainty about whether to find a general form for f(x) first.

Akash47
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Homework Statement
If x, a, b are positive integers, f(x) is positive integer too. And if a>b, then
f(a)>f(b).What is f(6)?
Relevant Equations
f(f(x))= x^2+2
It is obvious that the function f is not injective. From the given equation, we get f(f(2))=6.And since,there is an inequality given in the problem, I think we can use that to find f(6).But I have got stuck here and can't move.Do I have to find what is f(x) first?Then how?
 
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Akash47 said:
It is obvious that the function f is not injective. From the given equation, we get f(f(2))=6.And since,there is an inequality given in the problem, I think we can use that to find f(6).But I have got stuck here and can't move.Do I have to find what is f(x) first?Then how?
If you consider that for this exercise, f does not need to be defined for negative numbers, then f can be treated as being injective. However, I don't see that being injective would help.

I don't see how to find f(x) in general.

Use the given information to find some possible values for f.

Start with f(1).
What do you know? f(1) is a positive integer. Also, you know that f(f(1)) = 3 .

You also know that f(1) < f(2) < f(3) < ... : Right?
 
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I know that f(1) < f(2) < f(3) , then f(f(1)) <f(f(2)) <f(f(3)) and that implies, 3<6<11! Who doesn't know this?I had tried this method and that didn't help me that's why I have posted the problem? Can you please help me a little bit more?
 
Akash47 said:
I know that f(1) < f(2) < f(3) , then f(f(1)) <f(f(2)) <f(f(3)) and that implies, 3<6<11! Who doesn't know this?I had tried this method and that didn't help me that's why I have posted the problem? Can you please help me a little bit more?

Use that ##(f\circ f) \circ f == f\circ (f \circ f)## (where ##\circ## is function composition) to conclude that ##f(x^2+2)=f(x)^2+2##. That let's you get more relations. SammyS is suggesting you guess values for ##f(1)## and see if they lead to contradictions. Try it. For example, can ##f(1)=1##?
 
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Try making a table with columns of x, f(x), f(f(x)). Then start with some guesses to start out the f(x) column. You can then infer other table entries resulting from your guess and rule out the ones that violate the given restrictions. Similar to what you did to find 3<6<11.
I'm with SammyS; I don't have a clue how to find a closed form answer to what f(x) is. But I do know what f(6) is.
 
Thanks to all.I have solved the problem.
 
Akash47 said:
Thanks to all. I have solved the problem.
It's nice to know that you solved the problem.

However, the people who help students gain understanding and skills to solve problems here at PF are volunteers .

Our only reward is knowing that you have been helped, so, yes it's good that you indicted that you have a solution. A more satisfying reward for us is to actually see that solution, particularly the details we may have helped you with.
 
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