[Math] Is the derivative (of graph) undefined at the start/end (edge) of the graph

calculus

Given this velocity function over $ 0 \le t \le 4 $, "how many times is the acceleration undefined?" Recall, there is no derivative at a discontinuity, cusp, corner, or vertical tangent. I would say in 2 places the deriv is undefined (t=1 and t=2.3)

But, can you make the case that the deriv is also undef. at t=0 and t=4? The limit definition of derivative uses f(x+h), which does not exist on the graph beyond t=4. Likewise, you can't do f(x-h) and f(x+h) So, does this mean the derivative does not exist? But does that mean the derivative is "undefined"?

In other words, how many times is the acceleration (derivative) undefined? Is it 2 or 4 ?

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Best Answer

No, the derivative is defined on the edges (if, of course, the function is defined on the edges).

Take $f:[a,b]\to\mathbb R$.

Now, remember the definitions:

  1. The derivative of $f$ at a point $x_0\in[a,b]$ is the limit $$\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$$
  2. Given a function $g:A\subseteq\mathbb R\to\mathbb R$, the limit $$\lim_{x\to a} g(x)$$ is equal to $L$ if and only if

    For every $\epsilon > 0$ there exists $\delta>0$ such that for every $x\in A$ such that $0<|x-a|<\delta$ is true, $|g(x)-L|<\epsilon$ is also true.

Now, looking at the second definition, you can see that if $f$ is defined on $[a,b]$, then the limit $$\lim_{x\to a} f(x)$$ is actually equal to the right limit $$\lim_{x\to a^+} f(x)$$

Taking that into consideration, it is clear that on the left of the interval, you simply take the "right derivative" and it is by definition equal to the derivative.

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