If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable, $f(x) \neq f'(x)$, show that $\{x\in [0,1] \text{ and } f(x) = 0\}$ is finite. If $f:\mathbb{R}\rightarrow\mathbb{R}$ is differentiable, $f(x) \neq f'(x)$ for all $x$, show that $\{x\in [0,1]  \text{ and } f(x) = 0\}$ is finite.
I have shown that there cannot be an interval $[a,b]$ contained in $[0,1]$ such that $f(x) = 0$ since this would imply that $f$ is constant on this interval which implies that $f'(x) = 0$ on this interval and so there is contradiction since for $x\in [a,b], f(x) = 0$ and $f'(x) = 0$ since the interval is constant, but $f(x) \neq f'(x)$ for all $x$.
However I can't prove that for some infinite sequence $\{x_n\}$, say the rational numbers between $0$ and $1$ the statement holds true.
 A: Use that an infinite sequence of zeros would have a converging subsequence, since $[0,1]$ is compact. The limit $x$ of this subsequence fulfills $f(x)=0=f'(x)$.
A: If $X = \{x\in [0,1] \mid f(x)=0\}$ is infinite, then $X$ has a limit point $\bar x \in [0,1]$. By continuity, $f(\bar x) = 0$. By differentiability,
$$
f'(\bar x) = \lim_{\substack{
         x \to \bar x\\ x\in X\setminus\{\bar x\}}}
         \frac {f(x)-f(\bar x)} {x - \bar x} = 
         \lim_{\substack{
         x \to \bar x\\ x\in X\setminus\{\bar x\}}}
         \frac 0 {x - \bar x} = 0,
$$
so $f(\bar x) = f(\bar x)$, contrary to the assumption.
A: If I am not mistaken, we can even show that $f$ has at most one root on $\mathbb{R}$.
To the contrary, assume that we have two zeros, $x_0 < x_2$. Since $f'(x_0) \ne 0$, we may assume (w.l.o.g.) that $f'(x_0) > 0$.
Now we set $x_1 = \inf\{z > x_0 : f(z) = 0\}$. We easily get $f(x_1) = 0$ (by continuity) and $x_1 > x_0$ since $f'(x_0) > 0$.
Now we have $f'(x_1) \ne 0$ and by the intermediate value theorem we can conclude $f'(x_1) < 0$. Now, we define the function $g(z) = f(z) - f'(z)$ and have $g(x_0) < 0 < g(x_1)$. This $g$ is the derivative of $\int f \,\mathrm{d}x - f$. By owing to Darboux's theorem, we find some $z \in (x_0,x_1)$ with $g(z) = 0$, i.e., $f(z) = f'(z)$, which is a contradiction to the assumption.
