Somewhere in there, the tangent is parallel AB & BC

If $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one $c$ in $(a, b)$ with

$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$

The right side is the average rate of change across the whole interval, which is the slope of the secant line joining the endpoints. The left side is an instantaneous rate at one particular place. The theorem says those two numbers meet somewhere inside.

Notice which interval each hypothesis lives on. Continuity is demanded on the closed interval, endpoints included, because the conclusion is built out of $f(a)$ and $f(b)$ and those values have to be reachable. Differentiability is demanded only on the open interval, because $c$ is guaranteed to be strictly inside, so what the derivative does at the endpoints is never consulted.

That asymmetry is not a technicality, and it cuts both ways. A function with a vertical tangent at an endpoint still satisfies the hypotheses. A function with a corner one step inside the interval does not.

Take $f(x) = x^{3} - 3x$ on $[-2, 2]$. It is a polynomial, so both hypotheses hold everywhere. The average rate is

$$\frac{f(2) - f(-2)}{2 - (-2)} = \frac{2 - (-2)}{4} = 1,$$

and setting $f'(c) = 3c^{2} - 3 = 1$ gives $c^{2} = \frac{4}{3}$, so $c = \pm\frac{2}{\sqrt{3}} \approx \pm 1.155$. Both lie in $(-2, 2)$, so this interval has two of them.

The fastest way to believe that is to watch the conclusion fail when a hypothesis does.

Differentiability fails inside. Let $f(x) = |x|$ on $[-1, 1]$. It is continuous everywhere, and $f(-1) = f(1) = 1$, so the average rate is $0$. But every tangent slope is either $+1$ or $-1$; nothing equals $0$. There is no $c$, and the reason is the corner at $x = 0$, one point inside the interval where the derivative does not exist.

Continuity fails inside. Let $f(x) = \frac{1}{x}$ on $[-1, 2]$. The average rate is $\frac{\frac{1}{2} - (-1)}{3} = \frac{1}{2}$. But $f'(x) = -\frac{1}{x^{2}}$ is negative wherever it exists, so it is never $\frac{1}{2}$. The function is not merely discontinuous at $x = 0$; it is not defined there, so the closed interval hypothesis fails before anything else can be checked.

Both hold, and the endpoint scare is nothing. Let $f(x) = x^{1/3}$ on $[0, 8]$. Then $f'(x) = \frac{1}{3x^{2/3}}$, which blows up at $x = 0$. That looks fatal and is not, because $x = 0$ is an endpoint, and differentiability is only required on $(0, 8)$. So the theorem applies. The average rate is $\frac{2 - 0}{8} = \frac{1}{4}$, and

$$\frac{1}{3c^{2/3}} = \frac{1}{4} \;\Rightarrow\; c^{2/3} = \frac{4}{3} \;\Rightarrow\; c = \left(\frac{4}{3}\right)^{3/2} \approx 1.540.$$

Two of those three are rejected on sight by a careless reader and one is accepted on sight, and in each case the reader is wrong. Check where the failure sits: inside the interval it is fatal, at an endpoint it is fatal only for continuity.

The conclusion is an existence claim about $f'$. Four things it does not say:

1. Not exactly one. It promises at least one $c$. On $[-2, 2]$ the cubic above has two, and $f(x) = \sin x$ on $[0, 2\pi]$ has two as well, at $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
2. Not the midpoint. For a quadratic the guaranteed $c$ really is the midpoint every single time, which is exactly why the habit forms. For $f(x) = x^{3}$ on $[1, 3]$ the average rate is $\frac{27 - 1}{2} = 13$, and $3c^{2} = 13$ gives $c = \sqrt{\frac{13}{3}} \approx 2.082$, close enough to the midpoint $2$ to be mistaken for it and not equal to it.
3. Nothing about $f(c)$. The theorem produces a slope, not a height. If a question asks for the value of $f'(c)$, the answer is the average rate itself and no solving is required. If it asks for $c$, solve $f'(c) = \text{average rate}$. If it asks for $f(c)$, find $c$ first and then substitute back.
4. Nothing about where. Only that $c$ is strictly between $a$ and $b$. A $c$ that lands on an endpoint is not what the theorem gave you.

A useful discipline: before solving anything, write down the average rate as a number. It is the target that $f'(c)$ has to hit, and half the wrong answers on this topic come from setting $f'(c) = 0$ out of habit when the target was something else.

Rolle's Theorem is the special case where the endpoints happen to sit at the same height. Same two hypotheses, plus $f(a) = f(b)$, and then the average rate is $0$, so the conclusion becomes $f'(c) = 0$ for some interior $c$.

The extra hypothesis is the one that gets skipped. For $f(x) = x^{2} - 4x + 1$ on $[0, 3]$: $f(0) = 1$ and $f(3) = -2$, so Rolle does not apply. The Mean Value Theorem does, and it gives $2c - 4 = -1$, so $c = 1.5$. There is also a point where $f'$ is zero, namely $x = 2$, but Rolle is not what found it, and on a different function that coincidence would not be there.

Read in plain language, the theorem is a statement about averages catching up with instants. A car covers 120 miles in 2 hours, so its average speed was 60 miles per hour. Position is continuous and differentiable, so at some instant the speedometer read exactly 60. Not on average 60, which was already given, and not at least 60, which is weaker. Exactly 60, at some instant, at least once.

Three consequences the rest of Unit 5 leans on, all of them proved from this one theorem: if $f' = 0$ on an interval then $f$ is constant there; if $f' > 0$ on an interval then $f$ is increasing there; and two functions with the same derivative differ by a constant. Every one of those inherits the hypotheses, which is why 5.3 will still be asking whether $f'$ exists before reading anything off its sign.