The number of real roots of continuous $f(x)$

Question

Is there a way to know the number of real roots, or to have some bound at least to the number of real roots of a continuous function $f(x)$.

I think, intuitively, that the number of real solutions to $f(x)=0$ should be bounded by the number of roots of $f'(x)$ plus one.

Answer 1

Assuming you mean roots over $\mathbb{R}$, the number of roots of $f(X)$ is bounded above by the number of roots $f^\prime(x)$ plus one. This can be seen by using Rolle's theorem. (Assuming $f$ is differentiable)

That is suppose, that $f(x)$ has $k$ roots $x_1,\cdots, x_k$. By Rolles theorem, we know that $f^\prime(x)$ has at least one root between $x_i$ and $x_{i+1}$ for each $i=1,\cdots, k-1$. Hence $f^\prime(x)$ has at least $k-1$ roots and so the statement follows.

Answer 2

If $f$ is continuously differentiable you are correct. This is a consequence of Rolle's theorem If $f(x)$ takes the same value at two different values of $x$, there is a stationary point between them, so between any two roots there is a root of $f'$. You need to make sure your function satisfies the hypotheses of the theorem. There is no lower bound, though. $f(x)=2+\sin x$ has no roots, but its derivative has infinitely many.