Prove that $f$ on $[a,b]$ has only a finite number of zeros. 
Let $f:[a,b]\to\mathbb{R}$ be differentiable. Assume that there exists no $x\in[a,b]$ such  that $f(x)=0=f'(x).$ Prove that the set {$t\in[a,b]:f(t)=0$} of zeros of $f$ is finite.

What I have done is, let the set contain infinite number of points. Then by the Bolzano Weierstrass theorem there exists a sequence $<x_n>$ which has a convergent subsequence $<{x_n}_k>$ convergent to $c$. Then $f(c)=0$. But I got stuck in proving $f'(c)=0$.
 A: Let's suppose that the set $A = \{t\mid t \in [a, b],\, f(t) = 0\}$ is infinite and since it is bounded there is a point $c \in [a, b]$ such that $c$ is a limit point of $A$. Clearly by continuity $f(c) = 0$ and hence $c \in A$.
Now it is obvious that $f'(c) = 0$ otherwise $f$ will be strictly monotone at $c$ and there will be a neighborhood of $c$ where $f$ does not vanish except at $c$. And this will contradict that $c$ is a limit point of $A$. Thus we have $f(c) = f'(c) = 0$. Contradiction!!!
A: Hint
If $\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=L$ and $(x_n)\to c$ then $\lim\left(\frac{f(x_n)-f(c)}{x_n-c}\right)=L$.
A: The condition that we're on a closed, bounded interval hints that we could take advantage of its compactness. Indeed:


*

*Let $N$ be the points on which $f$ is nonzero $$N=\{t \,|\, t\in[a,b], f(t)\neq 0\}$$ As $f$ is continuous, $N$ is open on $[a,b]$.

*Let $z_i$ be all zero points of $f$ and let $Z_i$ be open neighborhoods of them on [a,b] such that $z_i$ is the only zero point of $f$ in $Z_i$: $$\forall x\in Z_i\setminus \{z_i\}:\, f(x)\neq 0$$ Such neighborhoods exists because $f'(z_i)\neq 0$, so sufficiently close to $z_i$ function $f(x)$ behaves like $f'(z_i) (x - z_i)$.

*Obviously $N\cup \bigcup_i Z_i=[a,b]$ so the sets form an open cover of $[a,b]$.

*By the definition of a compact space, there exists a finite subcover. However, since each $z_i$ is contained only in $Z_i$ and nowhere else, every such subcover must include all $Z_i$. Therefore the number of $Z_i$ is finite.

