0
$\begingroup$

Our teacher challenge us a question and it goes like this: The derivative of a function is define such as

$$\begin{cases} 1 & \text{if } x>0 \\ 2 & \text{if } x=0 \\ -1 &\text{if } x<0\end{cases} $$

Now he said that "$2$, if $x=0$" is not possible.He asked us to proved that the derivative at $x=0$ does not exist without using the concept of limit or using graphical intuition such as the slope of tangent but using only verbal words without resorting to use limit.

$\endgroup$
  • $\begingroup$ Could words like neighborhood be acceptable? $\endgroup$ – JB King Oct 24 '14 at 3:50
0
$\begingroup$

Note this function doesn't satisfy the intermediate value property on $[-1,1]$. By Darboux's theorem, there is no function $g$ such that $g'(x) = f(x) \forall x \in [-1, 1]$.In other words, $f$ is not the derivative on $[-1, 1]$ of any function.

$\endgroup$
  • $\begingroup$ How did you deduce discontinuity of $f$ from that of $f'$? $\endgroup$ – Jyrki Lahtonen Oct 24 '14 at 3:54
  • $\begingroup$ @JyrkiLahtonen Sorry, I have modified my solution. $\endgroup$ – John Oct 24 '14 at 4:04
0
$\begingroup$

You might think about the fact that aside from $x=0$ this is the derivative of the absolute value function, which does not have a derivative at $x=0$ You excluded using many things, but did not say what is allowed. You could let $f(0)=a$, then use the fundamental theorem of calculus to get $f(x)$ for all $x$ and show how there is no derivative at $0$.

$\endgroup$
  • $\begingroup$ Just a comment of a neophite but... what happen with distribution theory? I means, it is defined a derivative of absolute value and many others discontinuous functions. $\endgroup$ – Masacroso Oct 24 '14 at 4:42
  • 2
    $\begingroup$ @Masacroso: I think we are at a much lower level with this question. It looks to me first year calculus. Even so, what value would you assign for $f'(0)?$ I think distributions do not care about the value at a point. $\endgroup$ – Ross Millikan Oct 24 '14 at 4:48

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.