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This question came up in class, and I'm having trouble parsing out precisely what it means as far as the Mean Value Theorem is concerned. If the two intervals are disjoint, does that not imply a discontinuity in the domain?

In the Mean Value Theorem we have: if $f: [a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$, and differentiable on $(a,b)$, then there exists a point $c \in (a,b)$ where $f'(c) = \dfrac {f(b)-f(a)}{b-a}$. Where does this disjoint union $[a,b]\cup[c,d]$ even fit in there? I would appreciate a little guidance here, please.

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    "Discontinuity in domain" is a bit ambiguous and misleading. As a domain is by definition all input it hypothetically can't be discontinuous. But if it "jumps" between closed intervals the function is still continous. Meanwhile the mean value theorem obvious need not apply as the function and its derivativecan "jump" to a completely value. – 2017-02-02

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Let's say you have a function on $[a,d]$ with an increasing derivative such that $$f'(x)=\frac{f(d)-f(a)}{d-a}$$ for some $b

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    The interval between $b$ and $c$ is arbitrary, I suppose that's what was throwing me off. Thanks. – 2017-02-02
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Suppose you define $f: [0,1] \cup [2,3] \to \mathbb{R}$ as:

$$ f(x)= \begin{cases} 0 \quad x \in [0,1] \\[5pt] 1 \quad x \in [2,3] \end{cases} $$

Then $f$ is continuous on $[0,1] \cup [2,3]$ and differentiable on $(0,1) \cup (2,3)$ but $f'(x) = 0$ on its domain. Therefore the MVT-like equality will fail whenever $a \in [0,1]$ and $b \in [2,3\,]$.