differentiability and closed interval If $f$ is a real-valued differentiable function in $[a,b]$, is it true that $f'$ maps $[a,b]$ into a closed interval? Here $f'$ means first derivative of function $f$.
I tried to use DARBOUX theorem but it is not helpful.
 A: The image of $f'$ need not be a closed interval. As an example, consider
$$g(t) = \begin{cases}e^{-\lvert t\rvert}\sin \frac1t &, t \neq 0\\ 0 &, t = 0\end{cases}$$
and 
$$f(x) := \int_0^x g(t)\,dt.$$
It is immediate that $f'(x) = g(x)$ for $x \neq 0$. It is less evident, but can be verified by a little computation (see below), that $f$ is also differentiable at $x = 0$ with $f'(0) = 0$. Consequently, $f'([a,\,b]) = (-1,\,1)$ for all intervals with $a < 0 < b$.

To compute $f'(0)$, we rewrite the integral to get something easier to handle,
$$f(x) = \underbrace{\int_0^x \sin \frac1t\,dt}_{a(x)} - \underbrace{\int_0^x (1 - e^{-\lvert t\rvert})\sin \frac1t\, dt}_{b(x)}.$$
By symmetry ($g$ is an odd function, hence $f$ is even), we need only consider $x > 0$. $b(x)$ is very simple to treat:
$$\lvert b(x)\rvert \leqslant \int_0^x \left\lvert (1-e^{-t})\sin \frac1t\right\rvert \, dt \leqslant \int_0^x \lvert 1-e^{-t}\rvert\,dt \leqslant \int_0^x t\,dt = \frac{x^2}{2},$$
thus $\frac{b(x)-b(0)}{x} \leqslant \frac{x}{2} \to 0$.
For $a(x)$, we substitute $u = 1/t$ and obtain
$$\begin{align}
a(x) &= \int_0^x \sin \frac1t\,dt\\
&= \int_{1/x}^\infty \frac{\sin u}{u^2}\, du\\
&= \left[\frac{-\cos u}{u^2}\right]_{1/x}^\infty - 2\int_{1/x}^\infty \frac{\cos u}{u^3}\,du\\
&= x^2\cos \frac1x - 2\int_{1/x}^\infty \frac{\cos u}{u^3}\,du.
\end{align}$$
The integral can be estimated
$$2\left\lvert \int_{1/x}^\infty \frac{\cos u}{u^3}\,du\right\rvert \leqslant \int_{1/x}^\infty \frac{2}{u^3}\,du = x^2$$
and therefore
$$\left\lvert \frac{a(x)-a(0)}{x}\right\rvert \leqslant 2x,$$
which, together with the estimate for $b$ yields
$$\left\lvert\frac{f(x)-f(0)}{x}\right\rvert \leqslant \frac52 x.$$
A: More easily let us consider a constant function on a closed interval. Its derivative will be $\{0\}$, which is a closed set in $\mathbb{R}$ but not a closed interval. So the answer.
