My father asked me the following question:
A real function $f\colon \mathbb{R}\to \mathbb{R}$ is called $k$-regular if for each $y\in \mathbb{R}$, the equation $f(x)=y$ has precisely $k$ distinct solutions.
- Show that there are no continuous $k$-regular functions for $k$ even.
- For each $k$ odd, provide an example of a continuous $k$-regular function.
Solving this question was fairly easy but led me to the following problem:
For each $a\in \mathbb{R}^+$ (including $0$), consider the function $f_a\colon \mathbb{R}\to \mathbb{R}:x\mapsto x+a\sin(x)$. Clearly $f_0(x)=x$ yields a $1$-regular function. Plotting these functions for $a=4$ and $a=5$, it becomes fairly obvious that there should be a value of $4\leq a\leq 5$ such that $f_a$ is $3$-regular. (The value should be 'close' to 4.5 as can be seen on a plot).
How can we prove that such a value exists and is it feasable to accurately find this value? More generally, given $k=2l+1$, for which value of $a$ is $f_a$ a $k$-regular function?
I tried the following wishful thinking: I want that $f_a'(b)=0=1+a\cos(b)$ and that $f(b)=0=b+a\sin(b)$. From this one finds that $b-1=\tan(b)$. According to wolframalpha, -4.53360... is a solution to this equation and thus $a=\frac{-b}{\sin(b)}$ yields a value close to 4.6 which seems like a decent candidate.
Based on the above equations, there are many more candidates but only two solutions two $b-1=\tan(b)$ seem to yield a $3$-regular function $f_a$. I'm not really seeing why these particular values should work and the others fail.