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One of the requirements of the MVT is that the function has to be continuous at each point on the closed interval $a\leq x\leq b$. If we call the function $f(x)$, I am confused about how the function is continuous at points $(a, f(a))$ and $(b, f(b))$. The definition of a function being continuous at a point (let's call that point $(c, f(c)))$ is that $ \lim_{x\to c} f(x) = f(c).$ If we take the point $(a, f(a))$, in order for $f(x)$ to be continuous at $(a, f(a))$, $ \lim_{x\to a} f(x) = f(a)$ must hold true. However, it can't because the two sided limit doesn't exist. $x$ can approach $a$ from the right but not from the left, and $x$ can approach $b$ from the left but not from the right. So shouldn't "has to be continuous at each point" actually be "has to be defined at each point"?

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    $\begingroup$ Look up one-sided continuity. $\endgroup$ – Randall Jun 24 at 2:23
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    $\begingroup$ Generally I think the mean value theorem is phrased as "continuous on the closed interval $[a, b]$" rather than "continuous at each point on the closed interval $[a, b]$" precisely for this reason: continuity on a closed interval is defined to mean two-sided continuity at the interior points and one-sided continuity at the endpoints. $\endgroup$ – Micah Jun 24 at 2:56
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The function $f:[a,b]\to R$ is continuous at $x=a$ if

$$\lim_{x\to a^+} f(x)=f(a)$$

Similarly $f:[a,b]\to R$ is continuous at $x=b$ if

$$\lim_{x\to b^-} f(x)=f(b)$$

The continuity at the interior points is two sided while the continuity at endpoints is one sided.

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The other answers do a good job addressing the question from an analytic perspective. Topologically, there is an analogous explanation (which exists because metric spaces like $\mathbb{R}$ are also topological spaces, a more general class of objects):

Definition: A function $f:X \to Y$ between topological spaces is continuous if, for every open set $U \subset Y$, its preimage $f^{-1}(U)$ is open in $X$.

Note: When $X$ and $Y$ are metric spaces, this definition of continuity agrees with the epsilon-delta definition from analysis. The open sets in a metric space are unions of open "balls"; see here.

So user532874's concern can be expressed as follows:

Consider a purportedly continuous function $f:[a,b] \to \mathbb{R}$ and the open set $\Big(f(a) \! - \! \varepsilon, \ f(a) \! + \! \varepsilon\Big)$ for a small $\varepsilon >0$. If $f$ is continuous, then the preimage of this set should be an open set as well. But the preimage instead includes a half-open interval $[a,c)$ or even, if $f$ is constant, the entire closed interval $[a,b]$. Ostensibly, this preimage is not open: in particular, there doesn't seem to be any "wiggle room" around $x=a$. What gives?

And the answer is that, if $U$ is a subset of a topological space $X$ and we have a continuous function $g:U \to Y$, then the preimage of open sets in $Y$ will be open not with respect to $X$, but to $U$ which has been endowed with the subspace topology. Under this topology, the open sets of $U$ are $\{V \cap U \ | \ \text{V is an open set w.r.t. X}\}$.

In our scenario, we have $X = \mathbb{R}$ and $U = [a,b]$. Given the subspace topology, note that a half-open interval $[a,c)$ actually is "fully" open in $[a,b]$ because $[a,c) = [a,b] \cap (a \! - \delta, \ c)$ for any $\delta > 0$, and this interval $(a \! - \delta, \ c)$ is open in $\mathbb{R}$. Likewise, $[a,b]$ is open with respect to itself since $[a,b] = [a,b] \cap (a \! - \! \delta, \ b \! + \! \delta)$.

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  • $\begingroup$ I know this subspace topology appears arbitrary, but it's meaningfully chosen to be the coarsest topology so that the canonical injection $U \hookrightarrow X$ is a homeomorphism, and this has "desirable" implications; might add a section on this when I'm awake again. $\endgroup$ – Kaj Hansen Jun 24 at 10:18
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Using the definition of limit,"approaching" $a$ means that for any real number $\delta >0$ (no matter how "tiny"), we consider the elements $x\in [a,b]$ such that $|a-x|\le\delta$, and the similarly for $b$. This implies that this elements as close to $a$ (or $b$) as they might be, must be in the domain of $f$ in the first place, and as a result the limit exists if it exists from the side we can approach $a$ (or $b$). (Note that $f$ is indeed defined at $a$ and $b$ because they are in the domain of $f$ by hypothesis)

Hope this helps!

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We usually say that a function is continuous at an endpoint of its domain if it's continuous as one approaches the endpoint from all possible directions (which is just from within the interval here).

An example is the function $\sqrt x,$ defined on $[0,+\infty),$ which is continuous at $x=0.$ However, the function defined on the same domain by $0$ when $x=0$ and $1/x$ elsewise is discontinuous there.

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For instance, consider $f:[0,1]\to\Bbb R$ given by $f(x)=\begin{cases}x\,,x\in[0,1)\\2\,,x=1\end{cases}$.

Then $f$ wouldn't be continuous at $x=1$, and MVT wouldn't apply.

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Let $S \subset \mathbb R$, and let $f:S \to \mathbb R$. To say that $f$ is continuous at a point $a \in S$ means that if $\epsilon > 0$, then there exists $\delta > 0$ such that if $x \in S$ and $|x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$.

In the above definition, $S$ can be any subset of $\mathbb R$ (not necessarily an interval), and $a$ can be any point in $S$ (including points that are on the boundary of $S$). So this definition still makes perfect sense in the case where $S$ is a closed interval and $a$ is one of the endpoints of $S$.

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