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Let's say the function is defined on $[x,y]$

I just don't know what to think of here. I think that every function is not differentiable at the end points because they are points! How can the limit exist from the other side of the end point?

More importantly, when would a function be differentiable at the end point?

Can anyone provide an example and explain how he came with it?

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$f(x)=\mathbf{1}_{(x,y)}$. – Jp McCarthy Jan 28 '13 at 23:10
@JpMcCarthy: please flesh that out for me. I am not familiar with the symbols you are using. Thanks for understanding. – user43901 Jan 28 '13 at 23:11
The function equal to one for all points between the points but equal to zero at the end points. – Jp McCarthy Jan 28 '13 at 23:12
If it is equal to zero at the end points then the function is not continuous at the end points. I think I need to mention that. – user43901 Jan 28 '13 at 23:15
OK. The answer below is not continuous either. You want something like $f(x)=x\sin(1/x)$. Google x sin sin1\x. – Jp McCarthy Jan 28 '13 at 23:26

For a continuous example take $$f(t) = \sqrt{t-x}+\sqrt{y-t}$$ where the one sided derivatives at the endpoints are infinite.

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thanks a lot! Can you explain the motivation how you formulated it? – user43901 Jan 28 '13 at 23:18
The square root function is one of the simplest examples of a continuous function failing to have a (bounded one-sided) derivative. I just modified it to have two bad points. You should try to build a collection of examples and counterexamples to every mathematical concept you encounter. – mrf Jan 28 '13 at 23:22
You are right, and that is what I am trying to do. But I did not know that square root function was a common culprit. – user43901 Jan 28 '13 at 23:26
What about $t<y$? You want $y-t$. – Jp McCarthy Jan 28 '13 at 23:27
@JpMcCarthy It is. (It was the wrong way for a minute or two while I edited the answer to use the OP's bizarre names for the endpoints of the interval.) – mrf Jan 28 '13 at 23:32

If you can picture the upper half of the unit circle centered at $(0,0)$, try the function $f$ defined by $f(t)=\sqrt{1-t^2}$ for $|t|\leqslant1$ and $f(t)=0$ for $|t|\gt1$.

For the interval $[x,y]$, try $f(t)=\sqrt{(y-t)(t-x)}$ for $x\leqslant t\leqslant y$ and $f(t)=0$ elsewhere, this is the upper half of the circle with diameter $[x,y]\times\{0\}$.

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If you consider the constant function $1$, it is differentiable over the whole closed interval $[x,y]$. True, the derivative at the endpoints is one-sided, but that is usually acceptable. In contrast, the function $$f(z)=\begin {cases} 0 & z=x,y \\ \frac 1{x-z}+\frac 1{y-z} & otherwise \end {cases}$$ is differentiable over $(x,y)$, but not at either endpoint.

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Ah, I never knew that it is usually acceptable that way. Also, would you mind explaining the motivating behind coming up with this function. I am trying to learn how to think about these problems. Thanks – user43901 Jan 28 '13 at 23:13
Not continuous. – Jp McCarthy Jan 28 '13 at 23:30
@user43901: The function is not continuous at the endpoints (it goes off to infinity, but I have defined it to be zero), so there is no derivative. – Ross Millikan Jan 28 '13 at 23:38
@JpMcCarthy: but the requirement of continuity was not there when I answered. – Ross Millikan Jan 28 '13 at 23:57
Ross is right. That was my mistake. – user43901 Jan 29 '13 at 0:25

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