Show that floor function does't satisfy FTC.

Question

The function is $f(x) = \lfloor 1-x^2 \rfloor$.

$$f(x) = \left \{ \begin{array}{lr} -3 & : x \in [-2,-\sqrt{3})\\ -2 & : x \in [-\sqrt{3},-\sqrt{2})\\ -1 & : x \in [-\sqrt{2},-1)\\ 0 & : x\in [-1,1) \\ 1 & : x \in [1,\sqrt{2} )\\ 2 & : x \in [\sqrt{2}, \sqrt{3})\\ 3 & : x \in [\sqrt{3}, 2)\\ 4 & : x = 2 \end{array} \right.$$

But I don't know how to integrate that and I doesn't find examples.

@AdamSaltz, Of course, but I can't see a formal argument. Plot — David, Apr 27 '13 at 01:52
Your previous question gives you the integral. Its graph has some sharp corners. Does it have a derivative at those corners? — Brian M. Scott, Apr 27 '13 at 02:01
@BrianM.Scott, Its clear that this thing doesn't have derivative in the corners, but how I can construct that corners in a function that I could use to show the left limit and right limit differ? — David, Apr 27 '13 at 02:05
I honestly don’t see the problem. You have a piecewise definition of $F$. Pick one of the corners, say the one at $x=-1$. The derivative on the left is $-1$, and the derivative on the right is $0$. (By the way, I just noticed that here and in the other problem you’ve computed $f(x)$ incorrectly for $x\ge 1$.) — Brian M. Scott, Apr 27 '13 at 02:08
@BrianM.Scott, well. Finally I see your point. Write an answer, please. — David, Apr 27 '13 at 02:14
Here is a related problem. Do not forget to up vote this answer or other answers if youbenefit from it. — Mhenni Benghorbal, Apr 27 '13 at 04:01

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

Start with the piecewise definition of $F$ that you obtained for this problem. (Note, though, that you’ve calculated $f(x)$ incorrectly for $x\ge 1$: $f(1)=0$, and $f(x)$ is negative for all $x>1$.) As you can see, $F$ is piecewise linear, and adjacent pieces have different slopes. Thus, the derivative from the left at any of the ‘corners’ is different from the derivative from the right at that ‘corner’, and consequently $F$ is not differentiable there.

Show that floor function does't satisfy FTC.

1 Answers1