Trouble finding the Linear Functions that satisfy the given Identity

AI Thread Summary
To solve the identity f[f(x)] = 2x + 1, assume f(x) is a linear function of the form f(x) = ax + b. By substituting and simplifying, it leads to the equations involving a and b, which are found to be irrational numbers. The solutions for the first identity are f(x) = √2x + (-1 + √2) and f(x) = √2x + (-1 - √2). The second identity, f[f[f(x)]] = 2x + 1, requires a similar approach but has not yet been attempted.
SolXero
Messages
3
Reaction score
0

Homework Statement



Given f[f(x)] = 2x + 1 find all linear functions that satisfy this identity.
Given f[f[f(x)]] = 2x + 1 find all linear functions that satisfy this identity.


2. The attempt at a solution

I have not started to attempt a solution at this because I have no idea how to do these types of problems.
 
Physics news on Phys.org
SolXero said:

Homework Statement



Given f[f(x)] = 2x + 1 find all linear functions that satisfy this identity.
Given f[f[f(x)]] = 2x + 1 find all linear functions that satisfy this identity.


2. The attempt at a solution

I have not started to attempt a solution at this because I have no idea how to do these types of problems.
You're give that f(x) is a linear function, so let f(x) = ax + b.
For the first problem, you know that f(f(x)) = 2x + 1, so f(ax + b) = 2x + 1, or a(ax + b) + b = 2x + 1. Solve for a and b. Hint: a and b are not rational numbers.

Solve the second problem in a similar manner.
 
Thanks a lot!

For the first problem the answer would be:

f(x) = √2x + (-1+√2) and
f(x) = √2x + (-1-√2)

right?

I have yet to do the second one.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Replies
19
Views
2K
Replies
2
Views
2K
Replies
15
Views
2K
Replies
23
Views
3K
Replies
10
Views
2K
Replies
69
Views
8K
Replies
3
Views
2K
Back
Top