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I'm currently taking calculus bc. Our definition of differentiability at a point is that

  1. it is continuous at that point.
  2. the limit of the derivative function exists at that point.

I don't see how a point can be continuous at an endpoint because the one sided limits don't agree (no right sided limit or left sided limit depending on the function). If we are given this piece wise function:

piece wise function

I'm not familliar with latex so sorry for the image.

When we are evaluating f'(x) and writing that as a piece wise function. Should I make the domain 0 < x < 3 or 0 <= x < 3? When graphing it on desmos it does the latter but I don't see how that doesn't break our definition of continuity.

Omar Aboutaleb
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  • You say "our definition of continuity"---what definition is this? (I ask in part because it's a bit curious to require continuity in the definition of differentiability since differentiability implies continuity.) – Jakob Streipel Sep 18 '22 at 22:23
  • I don't know how to reply with an image in the comment section but this is the definition of differentiability in our cpm textbook verbatim: A function, f, is differentiable at x = c if: f is continuous at x = c and lim f'(x) exists as x->c – Omar Aboutaleb Sep 18 '22 at 22:39
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    That is a very strange definition---how then is the $f'(x)$ in that definition defined? – Jakob Streipel Sep 18 '22 at 22:41
  • As for endpoint issues, do these definitions say anything about $c$ being an interior point of the domain of $f$? – Jakob Streipel Sep 18 '22 at 22:42
  • If you're asking how our book defines a derivative, it's just the limit as h -> 0 of (f(x+h)-f(x))/h which I believe is the standard definition. Neither the definition of continuity nor the definition of differentiablility say anything about interior points or even anything about end points. – Omar Aboutaleb Sep 18 '22 at 22:49

2 Answers2

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Continuity at an endpoint refers to the continuity of a function, say $f(x)$, at a left endpoint $a$ or continuity of a right endpoint $b$ of its domain if

$$\lim_{x \to a^+} f(x) = f(a) \text{ or } \lim_{x \to b^-} f(x) = f(b).$$

Your question about excluding $0$ and $5$ depends on the definition of differentiability you have in your textbook, or how your teacher defines it. Some real analysis textbooks define it for only interior points and not endpoints. It seems like Wolfram Alpha is using that definition too. I wouldn't trust Desmos because it's saying the derivative exists at $x=3$, which we've proven false. (Plus, Desmos misinterprets improper integrals sometimes.)

Please let me know via downvoting or the comments what you think and hopefully it's shed some light.

Accelerator
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  • I don't know what the correct answer is but I know that f isn't differentiable at 3. Wolfram alpha exludes 0,3 and 5 from the domain but desmos doesn't exclude 0 and 5. Excluding 0 and 5 seems intuitive to me because of our definition but I don't know if I am missing something. – Omar Aboutaleb Sep 18 '22 at 23:19
  • Hold on, I think I made a mistake, so I edited my answer. I'll get back to this problem later. @EffieJohnson – Accelerator Sep 18 '22 at 23:23
  • Ok. Thank you for the help. – Omar Aboutaleb Sep 18 '22 at 23:24
  • You are correct in that it's not differentiable at $x=3$. My bad, I accidentally said it is. I'll check out what Desmos gives me later and see where you're confused. @EffieJohnson – Accelerator Sep 18 '22 at 23:28
  • @EffieJohnson I have edited my entire answer. – Accelerator Sep 19 '22 at 02:36
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    Thank you for your help. It felt like I found contradictory answers online so it makes sense that it is definition dependent. – Omar Aboutaleb Sep 19 '22 at 02:46
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The domain of $f(x)$ is $0\le x\le 5$.

Also $f(x)$ is continuous in its domain even at $x=3$, because for every $x$ is in its domain, it has a value of $f(x)$ equal at left and right sides. For example at $x=3$, we have $(x+1)^{1/2}=2=5-x$.

For endpoints, only one side matters which is OK as well.

But $f(x)$ is not differentiable at $x=3$, because left derivative is not equal with the right one. Because we have:

$\frac{d(x+1)^{1/2}}{dx} = \frac{1/2}{(x+1)^{1/2}}=1/4<>-1=d(5-x)/dx$