calculus - Prove that no function exists such that...

The exercise goes like this:

• Find a continous function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall c \in \mathbb{R}$ the equation $f(x)=c$ has exactly 3 solutions;


• Prove that no continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ exists such that the equation $f(x)=c$ has exactly two solutions $\forall c \in \mathbb{R}$;


• For what $n \in \mathbb{N}$ it's true that a continous function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall c \in \mathbb{R}$ the equation $f(x) = c$ has exactly $n$ solutions exists?

For the first point I built kind of a zigzag function which is easily generalizable to every odd $n$. It seems true to me that for even natural numbers such a function doesn't exist, because in some ways it would have to "jump", but I failed to formalize the argument.

Answer

We assume that $ f$ is continuous and that the equation $ f (x)=c $ always has at least two solutions, and find a $ c $ for which the equation has at least three solutions.

Suppose $ f (x_1)=f (x_2)=0$ for $ x_1 0$, and let $ m\in [x_1, x_2] $ satisfy $ f (m)=M $.

By the intermediate value theorem, there must be $ m_1$ and $ m_2$ such that $ f (m_1)=f (m_2) =M/2$ and satisfying
$$ x_1

Now consider an $ n $ such that $ f (n)=2M$. We know that $ n\notin [x_1, x_2 ] $ so without loss of generality assume that $ x_2

If $ M=0$, you can apply the same argument to $-f $.