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You may verify yourself that the function $f$ given by $$f(x) = \begin{cases} x^2 \sin\tfrac1x &\text{if } x \neq 0 \\ 0 & \text{if } x=0.\end{cases}$$ is continuous, is differentiable everywhere, but the derivative is not continuous at $0$. If you integrate this function $n-1$ times, you obtain (using the main theorem of integration) a function ...
A derivative can be discontinuous but it always has the intermediate value property. Thus for instance $f(x)=\operatorname{sign}(x)$ is not the derivative of any function which is differentiable on $[-1,1]$; the antiderivative of $f$ is $F(x)=|x|$ which is simply not differentiable at all at $0$. On the other hand, the function $$g(x)=\begin{cases} 0 & ... 2 For the first part, find a \mathbb Q-basis of the real numbers. This requires the axiom of choice or something similar. Then remove one of the basis element, and you get a subgroup G of \mathbb R with \mathbb R/G\cong\mathbb Q. Show that G is dense. The partition elements are then the cosets of \mathbb G as a subgroup of \mathbb R - that is ... 1 Since another question was linked here as duplicate I would like to give an alternative solution to the problem$$ \max\int_0^1f(x)\,dx,\qquad f(0)=0,\ \int_0^1\dot f^2(x)\,dx\le 1.  If we, for example, did not get the intuition for the artistic solution by CS above, we can try to approach the problem in a more standard way via the variational calculus. ...