Why does Rolle's theorem require continuity on a closed interval?

[omri3012]

Hallo.
If we consider Rolle's Theorem:
"If f is continuous on [a, b], differentiable in
(a,b), and f (a) = f (b), then there exists a point c in (a, b) where f'(c) = 0."

Why do we need to state continuity of f in interval and differentiability of f in open segment? Why can't we say f differentiable on [a,b]?

Thanks,

Omri

 

[LeonhardEuler]

If you said f is differentiable on [a,b], the result would still be true, just slightly less useful. As stated the theorem shows that the result is true even if f is not differentiable at a or b. If it was restated to require f to be differentiable on [a,b], it could not be used in these cases.

 

[g_edgar]

for example... $f(x) = \sqrt{(1-x^2)}$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ but not differentiable on $[-1,1]$.

 

[HallsofIvy]

Similarly, f(x)= |1- x^2| is continuous on [-1, 1], differentiable on (-1, 1) but not differentiable on [-1, 1].

 

[g_edgar]

Similarly, f(x)= |1- x^2| is continuous on [-1, 1], differentiable on (-1, 1) but not differentiable on [-1, 1].

A possible definition of "differentiable on [-1,1]" requires only one-sided derivatives at the two endpoints.