# A difficult problem on functions

• spaghetti3451
In summary, a difficult problem on functions is a complex mathematical problem that requires a deep understanding of mathematical concepts and techniques. These problems can be challenging because they involve complex functions and require multiple steps to solve. Some common strategies for solving them include breaking them down into smaller parts, using algebraic manipulation, and applying known principles and formulas. To improve your skills in solving these problems, it is important to practice regularly and seek help from tutors or study groups. There are also many online resources and tools available, such as tutorials, practice problems, and mathematical software programs, that can aid in solving difficult problems on functions.
spaghetti3451

## Homework Statement

I've been trying to solve the following problem but can't wrap my head around it.

Let ##x##, ##f(x)##, ##a##, ##b## be positive integers. Furthermore, if ##a > b##, then ##f(a) > f(b)##. Now, if ##f(f(x)) = x^2 + 2##, then what is ##f(3)##?

## The Attempt at a Solution

Well, the table illustrates my current understanding of the problem:

x-------------------f(x)-------------------f(f(x))
1-------------------??-------------------3
2-------------------??-------------------6
3-------------------??-------------------11
4-------------------??-------------------18
5-------------------??-------------------27
6-------------------??-------------------38
7-------------------??-------------------51
8-------------------??-------------------66
9-------------------??-------------------83
10-------------------??-------------------102

##f(x)## is a monotonic function, and is allowed to take integer values, so this means that the values in the third column must be interspersed in the second column in the order in which they appear in the third column.

That's how far I can take my intuition.

What is ##f(1)##? There is only one possibility

In other words, what is f(f(1))?

It's also relevant that the formula ##f(f(x))=x^2+2## makes sense for all x only if the range of f is a subset of the positive integers. So f(1) must be a positive integer. Try a few guesses: f(1)=1, f(1)=2, ... You should find that all but one of the assumptions you try lead to a contradiction. Then you should try to prove that all other values of f(1) lead to a contradiction. Once you have found the only possible value of f(1), it's pretty easy to use it to find f(3).

Hint: Show that $f(1) = 1$ is impossible, so that $1 < f(1)$. Why does it follow that $f(1) < f(f(1))$?

failexam said:
Well, the table illustrates my current understanding of the problem:

x-------------------f(x)-------------------f(f(x))
1-------------------??-------------------3

if ##f(f(1))=3##, then ## f(3) = f(f(f(1))) = (f(1))^2+2 ##

Now we have told him the entire solution to the problem except for how to prove that f(1)=1 is impossible and how to compute ##2^2+2##. Both of these steps are pretty trivial.

Mentallic and micromass
Fredrik said:
Now we have told him the entire solution to the problem except for how to prove that f(1)=1 is impossible and how to compute ##2^2+2##. Both of these steps are pretty trivial.
You're absolutely right!

Hint: 2^2 = 2*2 = 4.

Fredrik
Thanks! I've got it.

failexam said:
Thanks! I've got it.

Don't tell us you have got the answer, tell us the answer you have got!

SammyS
Suppose that f(a) < a for some admissible a, then f(a-1)<f(a) < a, then f(f(a-1)) < f(a), (a-1)^2 + 2 < a so a^2 - 2a + 1 +2 < a, a^2 - 3a +2 < 0, some algebra and end up with (a-3/2)^2 < 1/4, a < 2.15 so a <2, by that fact f(3) > 3, then 3 > f(1), which leaves f(1) = 0 which will make f(0)<0, f(1) = 1 so will yield to 3 = f(1) which is non sense and we re left with f(1) =2, by that fact f(3) = 5 !

Noctisdark said:
Suppose that f(a) < a for some admissible a, then f(a-1)<f(a) < a, then f(f(a-1)) < f(a), (a-1)^2 + 2 < a so a^2 - 2a + 1 +2 < a, a^2 - 3a +2 < 0, some algebra and end up with (a-3/2)^2 < 1/4, a < 2.15 so a <2
The fact that f(x) is positive and strictly increasing already implies f(a) >= a for all a.
, by that fact f(3) > 3,
you haven't eliminated the possibility that f(3) = 3.
then 3 > f(1), which leaves f(1) = 0 which will make f(0)<0, f(1) = 1 so will yield to 3 = f(1) which is non sense
and were left with f(1) =2
I don't really understand this.

and we re left with f(1) =2 by that fact f(3) = 5 !
[/QUOTE]
You better check your arithmetic here.

## 1. What is a difficult problem on functions?

A difficult problem on functions is a mathematical problem that involves manipulating and analyzing mathematical functions to find a solution. These types of problems often require a deep understanding of mathematical concepts and techniques.

## 2. Why are problems on functions considered difficult?

Problems on functions can be difficult because they often involve complex mathematical functions that require multiple steps to solve. These problems also require a strong understanding of mathematical principles and techniques.

## 3. What are some common strategies for solving difficult problems on functions?

Some common strategies for solving difficult problems on functions include breaking down the problem into smaller, more manageable parts, using algebraic manipulation, and applying known mathematical principles and formulas.

## 4. How can I improve my skills in solving difficult problems on functions?

One way to improve your skills in solving difficult problems on functions is to practice regularly with a variety of problems. You can also seek help from a tutor or study group and review mathematical concepts and techniques to strengthen your understanding.

## 5. Are there any online resources or tools that can help with solving difficult problems on functions?

Yes, there are many online resources and tools available to help with solving difficult problems on functions. These include online tutorials, practice problems, and mathematical software programs that can assist with calculations and visualizations.

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