# How can I prove that no derivative exist withing this function?

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.

• Could words like neighborhood be acceptable? – JB King Oct 24 '14 at 3:50

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.
• How did you deduce discontinuity of $f$ from that of $f'$? – Jyrki Lahtonen Oct 24 '14 at 3:54
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 $0$.
• @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. – Ross Millikan Oct 24 '14 at 4:48