The Mean Value Theorem (MVT) is the workhorse that converts a fact about a
derivative
— a local, instantaneous quantity — into a fact about the function over a whole interval. It says
the average rate of change is achieved exactly, at some interior instant.
Geometrically: draw the secant joining the endpoints
(a, f(a)) and (b, f(b)). Somewhere strictly
between, the tangent is parallel to that secant.
\exists\, c \in (a, b) : \quad f'(c) = \frac{f(b) - f(a)}{b - a}.
We reach it through a special case first — Rolle's theorem — and then tilt the picture.
Centrepiece, part 1: Rolle's theorem
Rolle. If f is continuous on
[a, b], differentiable on (a, b), and the ends
are level, f(a) = f(b), then f'(c) = 0 for some
c \in (a, b). We derive it line by line.
Step 1 — get an extremum for free. f is continuous on the
closed bounded interval [a, b], so by the
Extreme Value Theorem
it attains a maximum M and a minimum m on
[a, b].
Step 2 — split into two cases. Either M = m, or not.
Step 3 — the flat case. If M = m, the function is
constant, so f'(c) = 0 at every interior point — done.
Step 4 — the non-flat case forces an interior extremum. If
M \ne m, at least one of them differs from the common endpoint value
f(a) = f(b). So a max or a min is attained at some point
c that is not an endpoint — i.e. c \in (a, b).
Step 5 — Fermat's stationary-point lemma kills the derivative. At an interior
extremum a differentiable function has f'(c) = 0. Why: at an interior
maximum the one-sided difference quotients have opposite signs,
\frac{f(x) - f(c)}{x - c} \le 0 \text{ for } x > c, \qquad \frac{f(x) - f(c)}{x - c} \ge 0 \text{ for } x < c,
so the limit (which exists, since f is differentiable) must be
\ge 0 and \le 0 at once — hence
f'(c) = 0. That proves Rolle.
Centrepiece, part 2: the MVT by tilting
The MVT is Rolle applied to the function with the secant subtracted off.
Step 1 — build the tilt. Let L be the secant line through
the endpoints,
L(x) = f(a) + \frac{f(b) - f(a)}{b - a}\,(x - a),
and define the gap between the curve and the secant,
g(x) = f(x) - L(x).
Step 2 — check Rolle's hypotheses for g.
g is continuous on [a, b] and differentiable on
(a, b) (a difference of such functions). And it is level at the ends —
because the secant touches the curve there:
g(a) = f(a) - L(a) = 0, \qquad g(b) = f(b) - L(b) = 0.
Step 3 — apply Rolle to g. Some
c \in (a, b) has g'(c) = 0.
Step 4 — read off g'. Differentiating
g = f - L and using that L has constant slope,
g'(x) = f'(x) - \frac{f(b) - f(a)}{b - a}.
Step 5 — set it to zero. At the c from Step 3,
0 = g'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} \quad\Longrightarrow\quad f'(c) = \frac{f(b) - f(a)}{b - a}.
The tangent at c is parallel to the secant. That is the Mean Value Theorem.
The corollary the whole of integration leans on
If f' = 0 throughout an interval, then
f is constant. Take any two points
x_1 < x_2 in the interval. By the MVT on
[x_1, x_2] there is a c with
f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1) = 0 \cdot (x_2 - x_1) = 0,
so f(x_2) = f(x_1) — the value never changes. This is exactly why two
antiderivatives
of the same function differ only by a constant: their difference has zero derivative.
Let f be continuous on [a, b] and differentiable on (a, b). Then:
-
Rolle: if f(a) = f(b) then
f'(c) = 0 for some c \in (a, b).
-
Mean Value Theorem:
f'(c) = \dfrac{f(b) - f(a)}{b - a} for some
c \in (a, b) — the tangent is parallel to the secant.
-
Corollary: if f' \equiv 0 on an interval, then
f is constant there; hence two antiderivatives of one function differ
by a constant.
Running two functions at once gives the Cauchy Mean Value Theorem: for
f, g continuous on [a, b] and differentiable
on (a, b), some c \in (a, b) satisfies
\big(f(b) - f(a)\big)\,g'(c) = \big(g(b) - g(a)\big)\,f'(c).
Take g(x) = x and it collapses to the ordinary MVT. Its real payoff is
L'Hôpital's rule. Near a point where f(a) = g(a) = 0, the
Cauchy MVT lets you write, for some c squeezed between
a and x,
\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(c)}{g'(c)}.
As x \to a the trapped c \to a too, so the
indeterminate 0/0 ratio equals the limit of
f'/g' — which is L'Hôpital's rule, falling straight out of a tilted Rolle.