Timeline for Good exercises that force you to apply the definition of the derivative, without explicitly telling you to do so?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 30, 2022 at 12:52 | comment | added | Snaw | @Raciquel The motivation is that they won't be able to use the theorem that says that if $\lim_{x\to x_0} f'(x)=L$ and $f(x)$ is continuous around $x_0$ then $f'(x_0)=L$, because then they aren't using the definition of the derivative. | |
| Jan 30, 2022 at 0:50 | comment | added | Snaw | @Raciquel I think these first three examples with $f(x)=(x-c)^2g(x)$ give us a function which is not twice differentiable but still continuously differentiable. If we take $g(x)=1$ at rationals, $0$ at irrationals we get a function that is not differentiable around $0$ (this is a variant of the function I linked to in #1 in the list above). Maybe there's a way to take these functions and tweak them in order that we get functions that are differentiable but not continuously so? | |
| Jan 29, 2022 at 23:31 | comment | added | user1815 | A favorite variant is to use ${1\over2}x + x^2 \sin(1/x)$ & show that $f'(0)>0$ but $f$ is not increasing over any interval containing $0$. | |
| Jan 29, 2022 at 16:29 | comment | added | Snaw | Oh, that's a very good point. I forgot about these and should've listed a 4th case for when when $\lim_{x\to x_0} f'(x)$ doesn't exist. I usually show this in class as an example of a function which is differentiable yet not continuously differentiable, and I think that modifications would result in a pretty similar exercise. Do you think it's possible to easily generate functions with this property but that are not very similar to this one? I haven't yet given this enough thought. | |
| Jan 28, 2022 at 12:40 | history | answered | Jochen Glueck | CC BY-SA 4.0 |