Corollary $1$. If $f$ is differentiable on the interval $a<x<b$, then the zeros of $f$ are separated by zeros of $f'$.
proof. Let $x_1,x_2,…,x_n$ be the real roots of the function $f$. Without loss of generality, let $x_1<x_2< \cdots <x_n$. Let's first consider the roots $x_1$ and $x_2$. Since $f(x_1)=f(x_2)=0$, it follows that there exists $c_1$ in the interval $(x_1,x_2)$ such that $f'(c_1)=0$. Again, since $f(x_2)=f(x_3)=0$, we can find $c_2$ in the interval $(x_2,x_3)$ such that $f'(c_2)=0$. Repeating the process of applying Rolle's Theorem to the intervals $(x_3,x_4)$ to $(x_{n-1},x_n)$, we find that the zeros of $f$ are separated by zeros of $f'$.
That's the proof i've come up with so far.
This is actually a corollary of Rolle's Theorem. I wanted to know if we can conclude that $f$ is continuous on the interval $[a,b]$ just by the information provided by the corollary. Or can we only say that $f$ is continuous on the open interval $(a,b)$?
Also, how could I further improve the proof?
Best Answer
Just from the fact that $f$ is differentiable on $(a,b)$ you get no information about the continuity of $f$ on $[a,b]$. It may even be undefined on $a$ and on $b$.
And your proof assumes that $f$ only has finitely many zeros. Why? The function $f\colon(0,1)\longrightarrow\mathbb R$ defined by $f(x)=\sin\left(\frac1x\right)$ has infinitely many zeros. But you can still deduce from Rolle's theorem that between any two zeros of $f$ there's (at least) one zero of $f'$.