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Dec
15
awarded  Caucus
Dec
15
revised Condition of the mean value theorem
edited tags
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
15
asked Condition of the mean value theorem
Dec
3
revised Does every differentiable function has an infliction point between a local maximum and minimum?
added 156 characters in body
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
awarded  Commentator
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 $f'''(x) \neq 0$.
Dec
2
revised Does every differentiable function has an infliction point between a local maximum and minimum?
added 34 characters in body
Dec
2
revised Does every differentiable function has an infliction point between a local maximum and minimum?
added 34 characters in body
Dec
2
asked Does every differentiable function has an infliction point between a local maximum and minimum?
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
18
revised Intiutive argument that $\exp' = \exp$
edited tags
Oct
15
revised Intiutive argument that $\exp' = \exp$
edited tags
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 $\exp'(0) = 1$?
Oct
15
asked Intiutive argument that $\exp' = \exp$
Sep
23
revised Proof of continuity of Thomae Function at irrationals.
minor formatting improvement
Sep
23
suggested approved edit on Proof of continuity of Thomae Function at irrationals.