Well, any function that's discontinuous at an interior point of the interval is going to be non-differentiable as well. So a stronger counter-example is a function that is continuous but not everwhere differentiable which does not satisfy Rolle's Theorem; the classic example is f(x) = |x| on [-1, 1]. Then f is continuous on [-1, 1] and differentiable everywhere except at 0, and f(-1) = f(1), but f'(x) is either 1 or -1 everywhere it is defined.
However, if we want to satisfy the differentiability condition but not the continuity one, we must have the function being discontinuous at one of the end points of the interval. This makes it very easy; you can modify just about any function (that doesn't have a zero derivative on whatever interval you choose) into a counterexample.
For instance, take f(x) = x on [0,1] and modify it by changing one endpoint: let f(x) =
{x, x ≠ 1;
{0, x = 1
Then f is differentiable on (0, 1) but not continuous on [0, 1], and f(0) = f(1) = 0, but f'(x) = 1 on all of (0, 1).
The continuity condition basically ensures that the end point values actually have something to do with the values in the rest of the interval. Otherwise, the end point values are meaningless.
2007-04-09 21:42:56
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answer #1
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answered by Scarlet Manuka 7
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