# Math Help - Problem with an exercise related to functions

1. ## Problem with an exercise related to functions

I've been wondering about the following for a while:

Function $f: [1, \infty[ \rightarrow R$ is non-negative and implements the equation $x + f(x) = e^{f(x)}$. Find $f(e-1)$ and $f(1)$.

I just cannot get anything out of it.
Do you have any ideas?

2. I for one cannot read your post. The symbols are ‘scrambled’.
Why do you use special fonts? Why can’t you use standard notation?
If you expect reasonable help learn LaTeX.
http://www.mathhelpforum.com/math-he...-tutorial.html

3. Thanks for the tip and sorry! I didn't realize that others couldn't see it as for me it shows it perfectly well. I'm fixing it soon.

4. Originally Posted by Delaop
I've been wondering about the following for a while:

Function $f: [1, \infty[ \rightarrow R$ is non-negative and implements the equation $x + f(x) = e^{f(x)}$. Find $f(e-1)$ and $f(1)$.

I just cannot get anything out of it.
Do you have any ideas?
Let y= f(1). Then $1+ f(1)= e^{f(1)}$ becomes $1+ y= e^y$. There is a "trivial" solution to that- do NOT try to "solve" the equation in any algebraic way.

Let y= f(e-1). Then $1+ f(e-1)= e^{f(e-1)}$ becomes $e- 1+ y= e^y$. Again, there is "one" very simple solution. Solve it "by inspection".

5. Thanks. I found out that obviously $f(1) = 0$ and $f(e-1) = 1$. But as I have now got the values by guessing, I should be able to state that these are the only solutions. If f was strictly monotonic, that would be the case, but can I show that f is strictly increasing or strictly decreasing? Am I completely wrong?

Edit: If $f(1) = 0$ was also something else than 0, f wouldn't be a function. Same with the value 1. Right?