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Velocity Graph

Hello, given the velocity graph in the link above, we are being asked to find out when the particle's acceleration is negative. I understand this will be when the velocity graph is decreasing. My question pertains to whether it makes sense to include the endpoints t=0 and t=6 in our intervals. i.e. which intervals below are correct?

[0,1)u(4,6]

or

(0,1)u(4,6)

The answer key says option 2 is correct. If that is accurate, then we are assuming acceleration does not exist at t=0 and t=6. However, I believe one could argue that v(t) is differentiable at t=0 and t=6 if we use the definition of differentiability that says...

"a function is differentiable over a closed interval [a,b] if its differentiable on (a,b) and the one sided limits exits at a and b"

This concept is popping up over a myriad of questions I'm working on. Sometimes the answer key is assuming that a function is not differentiable at the endpoints of a closed interval, and other times it is.

james
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1 Answers1

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This really depends on your notation. Technically, the derivative of a function at $t$ is $$ \lim_{x\to t}\frac{f(x)-f(t)}{x-t} $$ if it exists. Now if you say that your function is defined on the space $[a,b]$, then the above definition is perfectly well-defined, even at the endpoints (there is only one way to approach $a$ and that is from the right). If your function however is defined on all of $\mathbb{R}$ but you only know what it does on $[a,b]$, then you can no longer make any conclusions of what $\lim_{x\to a}f(x)$ is since you can approach $a$ now also from the left.

Adam Martens
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