Mean Value Theorem — proof, Rolle's theorem: Visual walkthrough

We are going to derive, from absolute zero: $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} for some c strictly inside (a, b) .$

Before we can even read that line, let us build every piece of it.

Step 1 — What is a secant, and what is that fraction?

WHAT. Look at the two amber dots in the figure: one at $(a, f (a))$ , one at $(b, f (b))$ . The cyan straight line through them is the secant.

WHY. "Average behaviour" always means start-to-end, ignoring the wiggles in between. The secant is exactly the start-to-end straight-line summary of the curve — like the dashed "as-the-crow-flies" path on a map.

PICTURE. The fraction we care about is the steepness of that cyan line: $secant slope \frac{f ( b ) - f ( a )}{b - a} = \frac{f ( b ) - f ( a ) rise: how much higher}{run: how far across b - a} .$

$f (b) - f (a)$ is the rise — the vertical amber segment on the right of the figure.
$b - a$ is the run — the horizontal amber segment along the bottom.
Their ratio is "rise over run," the standard meaning of slope: how many units up per one unit across.

Figure — Mean Value Theorem — proof, Rolle's theorem

Step 2 — What is a tangent, and what is $f^{'} (c)$ ?

WHAT. In the figure, the cyan tangent kisses the curve at one point $c$ and matches its tilt exactly there.

WHY THIS TOOL — why a limit, not just a slope? A secant needs two points to compute rise/run. A tangent lives at one point, so ordinary "rise over run" would be $\frac{0}{0}$ — undefined. The derivative is the tool built precisely to answer "what is the slope using only one point?" It does so by sliding a second point $c + h$ toward $c$ and watching the secant slope settle: $f^{'} (c) = lim_{h \to 0} \frac{f ( c + h ) - f ( c )}{h} .$

$h$ is a tiny step sideways from $c$ (the amber bracket in the figure).
$f (c + h) - f (c)$ is the tiny rise over that step.
$lim_{h \to 0}$ means "let the step shrink to nothing and report the number the ratio approaches."

PICTURE. Watch the pale secants in the figure fan toward the bold cyan tangent as $h \to 0$ .

Step 3 — The trick: subtract the secant to flatten the ends

WHAT. For every $x$ , drop a vertical amber line from the curve to the secant. Its signed length is $g (x)$ : positive where the curve is above the line, negative where below. The figure shows several of these gap sticks.

WHY THIS STEP. Rolle's theorem (the easy, flat-secant case) needs the two endpoints at the same height. Our curve's endpoints are generally not the same height. So we subtract off the tilt: the secant is the tilt, and removing it lands both ends at height $0$ . We are not changing the shape of the wiggles — subtracting a straight line shifts every point vertically but keeps peaks as peaks and valleys as valleys.

Term by term inside $L (x)$ :

$f (a)$ — the starting height (so the line begins exactly on the curve at $x = a$ ).
$\frac{f ( b ) - f ( a )}{b - a}$ — the secant slope from Step 1.
$(x - a)$ — how far we've walked right of the left post; multiplied by the slope, it gives the extra height the line has climbed.

Step 4 — Check the endpoints of the gap are equal (they're both zero)

WHAT. Compute $g$ at the two posts.

WHY. These are the three Rolle hypotheses we must confirm for $g$ before we're allowed to use Rolle. Here we nail hypothesis 3 (equal endpoint heights).

$g (a) = f (a) - L (a) [f (a) + \frac{f ( b ) - f ( a )}{b - a} (a - a)] = f (a) - f (a) = 0.$

$(a - a) = 0$ kills the slope term, so $L (a) = f (a)$ : the line starts on the curve.

$g (b) = f (b) - L (b) [f (a) + \frac{f ( b ) - f ( a )}{b - a} (b - a)] = f (b) - [f (a) + (f (b) - f (a))] = f (b) - f (b) = 0.$

The $(b - a)$ in the numerator and denominator cancel, leaving $f (a) + f (b) - f (a) = f (b)$ : the line ends on the curve too.

PICTURE. In the figure both amber gap-sticks at the ends have collapsed to length zero — the curve $g$ pins to the axis at $a$ and at $b$ .

Step 5 — Rolle guarantees a flat spot: where does it come from?

WHAT (the engine). Why does a peak or valley force zero slope? Two named tools do the work:

Extreme Value Theorem — a continuous function on a closed interval $[a, b]$ attains a maximum height $M$ and a minimum height $m$ somewhere. So the peak/valley genuinely exists (it isn't a mirage we approach but never reach).
Fermat's Theorem (interior extrema) — if that highest/lowest point sits strictly inside $(a, b)$ , the derivative there is $0$ .

WHY the extremum is interior (covering the case that could break it). If both $M$ and $m$ occurred only at the endpoints, then $M = g (a) = 0 = g (b) = m$ , so $M = m$ — meaning $g$ is constant zero (the flat case, handled above). Otherwise at least one of $M, m$ is a value the endpoints do not share, so it must be achieved at an interior point $c \in (a, b)$ .

WHY Fermat pins it to zero — the squeeze. Sit at an interior peak $c$ . Step right by tiny $h > 0$ : the curve can't be higher, so $g (c + h) - g (c) \leq 0$ , and dividing by $h > 0$ , $\frac{g ( c + h ) - g ( c )}{h} \leq 0 \Rightarrow g^{'} (c) = lim_{h \to 0^{+}} (\dots) \leq 0.$ Step left by tiny $h < 0$ : again $g (c + h) - g (c) \leq 0$ , but now dividing by a negative $h$ flips the inequality, $\frac{g ( c + h ) - g ( c )}{h} \geq 0 \Rightarrow g^{'} (c) = lim_{h \to 0^{-}} (\dots) \geq 0.$ Because $g$ is differentiable, both one-sided limits equal the same number $g^{'} (c)$ . A number that is $\leq 0$ and $\geq 0$ can only be $0$ : $g^{'} (c) = 0.$

PICTURE. The figure marks the interior peak of $g$ with the amber dot; the tangent there is a horizontal cyan line — the flat spot.

Step 6 — Untilt: turn " $g$ flat" into the MVT equation

WHAT. Differentiate the definition $g (x) = f (x) - L (x)$ and set the flat point in.

WHY. A flat spot of $g$ is a hidden statement about $f$ . Undo the subtraction to reveal it.

$g^{'} (x) = f^{'} (x) - slope of L, a constant \frac{f ( b ) - f ( a )}{b - a} .$

$f^{'} (x)$ — the slope of the original curve.
The secant slope is a fixed number, so its derivative contributes that same constant everywhere (the slope of a straight line never changes).

Now plug in the special point $c$ from Step 5, where $g^{'} (c) = 0$ : $0 = f^{'} (c) - \frac{f ( b ) - f ( a )}{b - a} ⟹ f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .$

PICTURE. In the figure the horizontal tangent of $g$ (from Step 5) is tilted back up by the secant slope. A flat line tilted by the secant's steepness becomes a line with exactly the secant's steepness — the cyan tangent of $f$ now runs parallel to the amber secant. That parallelism is the MVT.

Step 7 — The degenerate and edge cases (never let the reader hit an unshown scenario)

Worked example Case-by-case, so nothing is left uncovered

Case A — $f (a) = f (b)$ (Rolle itself). Then the secant slope is $\frac{0}{b - a} = 0$ . The MVT conclusion becomes $f^{'} (c) = 0$ : a horizontal tangent. Rolle is literally MVT with a level secant.

Case B — $f$ constant. Every point is a flat spot; $f^{'} (x) = 0$ for all $x$ , and the secant slope is $0$ too. The equation $0 = 0$ holds at every interior $c$ — infinitely many valid choices.

Case C — $f$ a straight line. Then $f^{'}$ is one fixed number everywhere, and it already equals the secant slope. Every interior point works as $c$ .

Case D — several flat spots (a wavy curve). MVT says "at least one" $c$ , never "exactly one." A curve that crosses the secant slope repeatedly has many valid $c$ 's. See the Increasing and Decreasing Functions link — sign changes of $f^{'}$ create multiple crossings.

Case E — hypotheses violated. If $f$ has a corner (not differentiable) like $∣ x ∣$ on $[- 1, 1]$ , no interior flat tangent need exist; the guarantee dissolves. Continuity/differentiability are load-bearing, not decorative.

The one-picture summary

Everything on one frame: the original curve $f$ with its amber secant; the tilted gap curve $g$ pinned to $0$ at both ends; the flat spot of $g$ (open circle) directly above the parallel tangent of $f$ (cyan). Follow the vertical guide line connecting them — that single vertical line is the proof: flat $g$ upstairs ⇔ secant-parallel $f$ downstairs.

Recall Feynman retelling — the whole walkthrough in plain words

You want to prove: somewhere your steepness matched your average steepness. Trouble is, "average" compares two far-apart points, while "steepness right now" lives at a single point — awkward to line up.

So play a trick. Tip the whole landscape so your start and finish sit at the same height (that's subtracting the secant — the amber straight line). Tipping doesn't erase hills; it just re-levels the ground under them. Now you begin and end at the same height, so like a walk that returns to your friend's level, there's a moment you're moving perfectly flat — the top of a hill or bottom of a dip (that's Rolle, powered by "a top exists" from the Extreme Value Theorem and "the top is flat" from Fermat).

Finally, tip the landscape back to how it really was. That flat moment tilts up by exactly the amount you tipped — the average steepness. So at that instant your real steepness equals the average. Done.

Recall One-line memory hooks

Secant slope :::: $\frac{f ( b ) - f ( a )}{b - a}$ , the average steepness (rise over run). Gap function $g$ :::: $f (x) - L (x)$ , the vertical distance from curve down to secant. Why $g (a) = g (b) = 0$ :::: subtracting the secant makes both endpoints sit on the line. Engine of the flat spot :::: Extreme Value Theorem (top exists) + Fermat (top is flat). Untilt step :::: $g^{'} (c) = 0 \Rightarrow f^{'} (c) =$ secant slope.

Connections

Mean Value Theorem — proof, Rolle's theorem — the parent this page visualizes.
Extreme Value Theorem — guarantees the peak/valley of $g$ actually exists (Step 5).
Fermat's Theorem (interior extrema) — interior peak ⇒ $f^{'} = 0$ , the squeeze in Step 5.
Continuity · Differentiability — the inherited hypotheses for $g$ (Step 4).
Increasing and Decreasing Functions — explains multiple $c$ 's (Case D).
Cauchy's Mean Value Theorem — the same "subtract, apply Rolle" trick with two functions.
Taylor's Theorem — MVT is its degree-one seed.
Lipschitz Continuity — MVT turns a derivative bound into a slope bound (parent Example 3).