I was doing some exercises of Rolle's theorem. But, they didn't check the differentiability the way we checked differentiability normally. I am giving some examples.
When I was checking differentiability with limit (before differentiation) I was just putting $x$ values which given in question. Like $x=2$ check it is differentiable or not.
$$Rf'(x)=\lim_{h \to 0} \frac{f(x-h)-f(x)}{h}$$
I had put $2$ instead of $x$. But, something else is happening in Rolle's theorem.
Verify Rolle's theorem for the function $f(x) = x^2$ in the interval $[-1,1]$.
In answer they didn't change $x$.
Verify Rolle's theorem for the function $f(x) = x^{\frac{2}{3}}$ in the interval $(-1,1)$.
They had written $0$ instead of $x$ this time. I am confused how to find the value. In earlier problem it was closed interval. And, it is open interval now. Which law should I apply to find the value?
For the function $f(x)=x$ when $0 \leq x \leq 1$, $f(x) = 2-x$ when $1 < x \leq 2$, $f(0)=0=f(2)$, according to the Rolle's theorem there exist at least one point in the interval $(0,2)$ such that $f'(x)=0$ but $f'(x) \neq 0$ for any $x \in (0,2)$ why not, explain.
This time they had put $1$ instead of $x$.
I was solving some problems by myself. I noticed in question they gave us two value inside interval. I was trying to solve first problem (which I wrote here) with $-1$ instead of $x$ (as I said they used $x$ but, I tried to use $x$) than, I noticed that function is differentiable. To verify Rolle's theorem I needed a $x$ for that function. For Rolle's condition $f'(c)=0$. Without $x$ I can't solve the condition. Now, I think it's ok if I use $x$ always, isn't it?
I believe it is because of some special condition.
In the first case, it is a common smooth function therefore there is no need to check with differentiability for all points in the domain.
In the second case, only the point x=0 needs further specification.
As in the third case, as x = 1 is the connecting point, we need to further examine the continuity and differentiability condition with respect to that point.
As to the closed and open intervals, it is on the problem of continuity and differentiability, we often require continuous on closed interval and differentiable on open interval.