6
$\begingroup$

Say we have the function $f:(0,1)\cup\{2\}\rightarrow \mathbb R$ defined by $f(x)=5$.

Is $f$ differentiable or not differentiable (or neither) at $2$?

Is $f$ a differentiable function?

$\endgroup$
  • $\begingroup$ Differentiabilty at an isolated point is not defined. $\endgroup$ – Kavi Rama Murthy Feb 3 at 7:45
  • $\begingroup$ No. One can only take a limit over such $h\to0$ that $f(a+h)$ is defined. Since for an isolated point all such small enough $h$ must be zero the difference quotients $\frac{f(a+h)-f(a)}{h}$ are not defined for small $h$, and neither is the limit. $\endgroup$ – Conifold Feb 3 at 8:23
3
$\begingroup$

Differentiability is defined in terms of sequences of differences $h_1,h_2,h_3\dots$ that are all nonzero and yet converge to zero. Such a sequence does not exist in the given domain, as there does not exist a number $k$ in the domain where $|2-k|=0.2$ or $0.1$ or indeed any number smaller than $1$. Hence differentiability is not defined at isolated points.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is $f$ a differentiable function? $\endgroup$ – iqntt1s Feb 3 at 9:24
  • $\begingroup$ @iqntt1s Over $(0,1)$, yes. At $\{2\}$, no. $\endgroup$ – Parcly Taxel Feb 3 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.