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Say you have a graph for v[t]

Question asks “when is speed constant”

The flat part of v[t] is from $2$ to $3$ seconds.

Is the answer $2 < t < 3$ or is it $2 \le t \le 3$ ?

$$\lim_{t\to 2} v(t)=v(2)$$ so does that mean to include or exclude 2?

I've been told to always exclude the endpoints, but you do come across some textbooks that include them. This seems to be one of those "gray" areas ... you'll also encounter the same issue for intervals where a function is increasing, decreasing, concave upward, or concave downward ... Do I just go with the open intervals?

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I'm not really sure how the limit when $t\to 0$ affects the question... do you mean $t\to 2$? – GPerez Feb 3 '14 at 15:52
Yes, corrected. – JackOfAll Feb 3 '14 at 19:44
up vote 4 down vote accepted

The inclusion, or not, of the endpoints is relevant to the object of study for each situation. For example, with integration, you get the same value for $\int_{(a,b)}f$, $\int_{[a,b]}f$, or any set that differs from $[a,b]$ on finitely many* points. So for integration there's no effective difference in including the endpoints or not.

Contrarily, we could be studying global extrema of a function in an interval, and in this case it makes all the difference. We can guarantee the existence of global a maximum and minimum if we're studying the function in $[a,b]$ (Weierstrass extreme value theorem), but we cannot guarantee global extrema if the interval is open. So here we must be clear about which we are talking about.

There's many other realms where closed-ness or open-ness makes a big difference, and in much more general ways. To give a last example, take the intersection of the sets $(0,\frac1n)$ for all $n$. Clearly this is the empty set, whereas the intersection of $[0,\frac1n]$ for all $n$ is the number $0$. In fact, we can use this as a defining property of the real numbers, namely that any infinite intersection of a sequence of "nested" closed intervals contains a real number. We can not use this property if we replace closed with open.

So we must know whether the endpoints affect whatever problem we're working with. In any case, it's always more correct to write every point for which a property holds, if this is asked. So if your function is "flat" on the endpoints, I would write that.

$\small\text{*More generally, $zero$ $measure$}$

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Thoughtful comment, for sure, but what about my specific question? I am neither integrating or talking extrema. – JackOfAll Feb 3 '14 at 20:09
I thought the last paragraph answered your question, that is, determine whether it's flat on the endpoints, and if so include them. If you're not sure then write the function out and I'll do my best to help. – GPerez Feb 3 '14 at 21:14
It is just a graph of v(t), and yea, it's flat on the ends. Looks like a stepwise function. – JackOfAll Feb 4 '14 at 6:29
In that case we can't tell, because it really lies in how the function is written, as in, whether the "step" (of the entire stepwise function) where $v(x)=c$ is defined as $2\le x \le 3$ or $2<x<3$. – GPerez Feb 4 '14 at 8:45
In any case those are the only two points we don't know about; you know for certain (save microscopic "bumps" of ink) from the graph that it's constant on $2<x<3$, so writing that is safe. – GPerez Feb 4 '14 at 8:47

I think when you mean speed (or velocity) constant, the key thing is v'(t)=0. So at the endpoints in your example, the derivatives don't seem to exist (remember that right hand derivative=left hand derivative). So, yes go for the open interval

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Why does it matter if the derivative exists at the ends? I am not taking the derivative (or discussing acceleration) at those endpoints. I am only discussing where speed is constant. – JackOfAll Feb 3 '14 at 20:10
In your case (definition for constant) v'(t)=0 and hence it should be properly defined. – Abhimanyu Arora Feb 3 '14 at 20:14
But I see your point, at v(2)=v(t)=v(3) for all 2<t<3 so given that it's a plot of v(t), you can also include the endpoints. But not when you have a plot of x(t) – Abhimanyu Arora Feb 3 '14 at 20:30
@AbhimanyuArora The definition for constant is that $v(x)=c$, whereas $v'(x)=0$ is a sufficient but not necessary condition, do you agree? – GPerez Feb 4 '14 at 8:44
@GPerez: Yes :-) – Abhimanyu Arora Feb 4 '14 at 11:11

Since $v = dx/dt$ involves a derivative, it naturally makes sense to talk about it being constant in open intervals.

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To me, it seems to make sense to INCLUDE the endpoints. v(2) = v(2.5) = v(3). They all have the same velocity, so velocity is constant with respect to all of those times.

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