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 Jan 20 comment strict local extremum of $f'$ that is neither saddle nor inflection value of $f$ With extremum I mean "strict extremum", then the derivative of constant function would be constant without a strict extremum, so it doesn't match the first condition. Dec 21 comment Relation between points of inflection and saddle points Why did you delete the worked out mean value argument. Was there something wrong with it? Dec 19 comment Relation between points of inflection and saddle points Since the function is differentiable, strictly mono. increasing on the half open intervall implies the same property on the closed intervall. Dec 11 comment $f'$ changes strict monotonicity but $f''$ isn't of strictly opposite signs Thanks, but I noticed that I wasn't precise enough. I want that the property with the second derivative is violated on every neighborhood of $x_0$. See my edit. Dec 11 comment $f'$ changes strict monotonicity but $f''$ isn't of strictly opposite signs Thanks, but I noticed that I wasn't precise enough. I want that the property with the second derivative is violated on every neighborhood of $x_0$. See my edit. Dec 9 comment $f'$ changes strict monotonicity but $f''$ isn't of strictly opposite signs @ClementC. Ideally I want a simple example where one can "write down" a term for $f$ and not only saying that there exists some (maybe very complicated) primitive of your $g$. Mar 19 comment Define the Riemann integral via trapezoids instead of rectangles Yes I am still interested. Mar 17 comment Define the Riemann integral via trapezoids instead of rectangles By the "nontrivial direction" I mean the part that integrability in the trapezoidal sense implies Riemann integrability. Mar 17 comment Define the Riemann integral via trapezoids instead of rectangles Thanks. Can you give more details for the nontrivial direction of the equivalence? Mar 16 comment Define the Riemann integral via trapezoids instead of rectangles Thanks for the idea. How to get it for general functions? Feb 10 comment Continuity of third derivative in extremum test @rubik Just the topological interior of $I$. Dec 15 comment Condition of the mean value theorem However if you know only the less general version you could say, that $f$ has the MVT on every closed sub-interval of $]-1,1[$. What about the theorems (see question above). Dec 2 comment Does every differentiable function has an infliction point between a local maximum and minimum? Furthermore $x$ beeing a local extremum doesn't imply that $f''(x) \neq 0$ (only that $f'(x)=0$). Dec 2 comment Does every differentiable function has an infliction point between a local maximum and minimum? Thanks. But in the non strict case, is the theorem true or is there another counterexample? Dec 2 comment Does every differentiable function has an infliction point between a local maximum and minimum? This is not the definition of inflection point. It is just a neccesary condition if $f$ is twice differentiable. It would be a sufficient condition for example if $f$ is three times differentiable and additionally $f'''(x) \neq 0$. Oct 19 comment Asymptote of solution of a differential equation without solving it I don't really understand why it is sufficient to show that $u'(x) > 0$ for all $x$ and why this it the case. If $u'(x_0) = 0$ for some $x_0$, why is this automatically the case for all $x$? Oct 15 comment Intiutive argument that $\exp' = \exp$ @mookid It's for a high school course. The real numbers are known only on a heuristic middle school level (via examples like $\sqrt{2}$ and $\pi$, nested intervals etc. Powers are introduced usually just for rational numbers and real powers via nested intervals (or sometimes just with reference to the calculater - which is really sad, but I don't have the time to rework those foundations, I just want to give a nice argument for the derivative of $\exp$ despite of the spongy foundations :-() Oct 15 comment Intiutive argument that $\exp' = \exp$ But how to show that $\exp'(0) = 1$? Jul 24 comment Asymptote of solution of a differential equation without solving it Would be nice if you could make it more rigerous May 9 comment Proof: Force always perpendicular and motion in a plane implies that the trajectory is a circle @DanielV No I mean that the force which corresponds to $x''$ and $x'$ which corresponds to the velocity are perpendicular.