Mean Value Theorem — proof, Rolle's theorem: Worked examples

This page is a drill ground. The parent note proved why the Mean Value Theorem (MVT) and Rolle's theorem are true. Here we hunt down every kind of situation they can throw at you — nice curves, curves with hidden traps, degenerate cases, word problems, and exam twists — and solve each one from zero.

Two symbols we lean on the whole time, defined once so we never sneak them in:

The scenario matrix

Every problem in this topic lands in one of these cells. The examples below are labelled with the cell they cover.

Cell	What makes it special	Example
A. Plain MVT	all hypotheses hold, find $c$	Ex 1
B. Plain Rolle	$f (a) = f (b)$ , find flat point	Ex 2
C. Multiple $c$	several valid points inside	Ex 3
D. Hypothesis fails — not differentiable	corner/cusp breaks it	Ex 4
E. Hypothesis fails — not continuous	jump/hole breaks it	Ex 5
F. Degenerate / limiting	$b \to a$ , endpoints merge	Ex 6
G. Bounding inequality (both signs)	turn a difference into a derivative	Ex 7
H. Word problem (speed)	average vs instantaneous, units	Ex 8
I. Exam twist — count roots	use Rolle to bound number of solutions	Ex 9

We prerequisite Continuity and Differentiability throughout, and lean on the Extreme Value Theorem and Fermat's Theorem (interior extrema) whenever a hypothesis is questioned.

Example 1 — Plain MVT (Cell A)

Worked example Find every

$c$ for $f (x) = x^{3}$ on $[0, 2]$ Forecast: guess before reading — cube grows steeply near $x = 2$ , so is $c$ closer to $2$ or the middle?

Step 1 — Compute the secant slope (the target). $\frac{f ( 2 ) - f ( 0 )}{2 - 0} = \frac{8 - 0}{2} = 4.$ Why this step? The MVT says $f^{'} (c)$ must equal this exact number; we need the target before we hunt for $c$ .

Step 2 — Write the derivative and set it equal to the target. $f^{'} (x) = 3 x^{2}$ , so we solve $3 c^{2} = 4$ . Why this step? $c$ is defined as the place where instantaneous slope $=$ average slope.

Step 3 — Solve, keep only interior roots. $c^{2} = \frac{4}{3} \Rightarrow c = \pm \frac{2}{3} \approx \pm 1.1547.$ The negative root is outside $(0, 2)$ , so discard it. Keep $c = \frac{2}{3} \approx 1.155$ . Why this step? MVT only promises a $c$ in the open interval $(a, b)$ ; roots outside don't count.

Verify: $c \approx 1.155$ sits between $0$ and $2$ ✔, and $3 (1.155)^{2} \approx 4.00$ ✔ — matches the secant slope. Your forecast: $c \approx 1.15$ is right of the midpoint $1$ , because the cube is flatter on the left.

Figure s01 shows the white curve $y = x^{3}$ on $[0, 2]$ , the pale-yellow dashed secant joining $(0, 0)$ to $(2, 8)$ (slope $4$ ), and the pink tangent drawn at $c = 2/ 3 \approx 1.155$ . Notice the pink tangent is exactly parallel to the yellow secant, and the pink touch-point sits to the right of the midpoint $x = 1$ — the visual proof of the forecast.

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

Example 2 — Plain Rolle (Cell B)

Example 3 — Multiple $c$ (Cell C)

Figure s02 shows the white sine wave over $[0, 2 π]$ , the yellow dashed horizontal secant (slope $0$ , since both endpoints sit on the $x$ -axis), and two short flat tangents: a pink one at the peak $c = \frac{π}{2}$ and a blue one at the trough $c = \frac{3 π}{2}$ . Both are horizontal — two separate points where the promise of MVT is met.

Example 4 — Hypothesis fails: not differentiable (Cell D)

Worked example

$f (x) = ∣ x - 1∣$ on $[0, 2]$ — does Rolle apply? Forecast: endpoints look equal. Will there be a flat tangent?

Step 1 — Check endpoints. $f (0) = ∣0 - 1∣ = 1$ and $f (2) = ∣2 - 1∣ = 1$ , so $f (a) = f (b)$ ✔. Continuous ✔. Why this step? Two of three hypotheses pass — it's tempting to declare victory.

Step 2 — Check differentiability at the corner $x = 1$ . Left of $1$ the slope is $- 1$ ; right of $1$ it's $+ 1$ . The two one-sided slopes disagree, so $f^{'} (1)$ does not exist. Why this step? Rolle demands differentiability on all of $(0, 2)$ ; a single corner voids the guarantee.

Step 3 — Confirm no flat tangent exists. Everywhere except $x = 1$ , $f^{'} (x) = \pm 1 \neq = 0$ . There is no $c$ with $f^{'} (c) = 0$ . Why this step? Shows the theorem's conclusion genuinely fails — the hypotheses aren't decoration.

Verify: slopes are $\pm 1$ , never $0$ ✔. This is the parent note's $∣ x ∣$ warning, shifted. Fix: always test the "spiky" point.

Figure s03 shows the white V-shape of $∣ x - 1∣$ , the yellow dashed line joining the two equal-height endpoints $(0, 1)$ and $(2, 1)$ , and a pink marker at the sharp corner $(1, 0)$ . The annotation flags that the slope jumps abruptly from $- 1$ to $+ 1$ there — there is no smooth in-between value $0$ , so no horizontal tangent exists anywhere.

Example 5 — Hypothesis fails: not continuous (Cell E)

Example 6 — Degenerate / limiting case (Cell F)

Worked example What is the MVT slope as

$b \to a$ ? Forecast: shrink the interval to a point — what does the "average slope" become?

Step 1 — Write the secant slope for $f (x) = x^{2}$ on $[a, a + h]$ with $h > 0$ . We take $h > 0$ so the interval $[a, a + h]$ has its endpoints in the correct order (left $= a$ , right $= a + h$ ); we will let $h \to 0^{+}$ to collapse it from the right. $\frac{f ( a + h ) - f ( a )}{h} = \frac{( a + h ) ^{2} - a ^{2}}{h} = \frac{2 ah + h ^{2}}{h} = 2 a + h .$ Why this step? We want to see what MVT reports as the interval collapses ( $h \to 0^{+}$ ); fixing the sign of $h$ keeps the interval well-defined.

Step 2 — Find the $c$ that MVT hands us. Set $f^{'} (c) = 2 c = 2 a + h \Rightarrow c = a + \frac{h}{2}$ — the midpoint of the tiny interval. Why this step? For $x^{2}$ the guaranteed point is always the midpoint; watch where it goes.

Step 3 — Take the limit $h \to 0^{+}$ . $c = a + \frac{h}{2} \to a$ , and the secant slope $2 a + h \to 2 a = f^{'} (a)$ . Why this step? The MVT statement degenerates into the definition of the derivative — the average slope becomes the instantaneous slope at the single surviving point.

Verify: at $a = 3$ , midpoint slope $2 (3) = 6$ , and $f^{'} (3) = 2 \cdot 3 = 6$ ✔. Degenerate MVT $=$ derivative. This is the seed of Taylor's Theorem.

Example 7 — Bounding inequality, both signs (Cell G)

Worked example Prove

$ln (1 + x) \leq x$ for all $x > - 1$ Forecast: MVT turns a difference into one derivative value — which two points do we compare, and does the sign of $x$ change the argument?

Step 1 — Case $x = 0$ . $ln (1 + 0) = 0 = x$ , so equality holds. Done for this case. Why this step? The boundary case is where equality is achieved; isolate it first.

Step 2 — Case $x > 0$ : apply MVT on $[0, x]$ . Let $f (t) = ln (1 + t)$ . MVT gives $\frac{l n ( 1 + x ) - l n 1}{x - 0} = f^{'} (c) = \frac{1}{1 + c}, c \in (0, x) .$ Since $c > 0$ , $1 + c > 1$ , so $\frac{1}{1 + c} < 1$ . Multiply by the positive $x$ : $ln (1 + x) < x .$ Why this step? For $c > 0$ the derivative is below $1$ ; multiplying by positive $x$ preserves the inequality direction.

Step 3 — Case $- 1 < x < 0$ : apply MVT on $[x, 0]$ . Now $x < 0$ , so the interval is $[x, 0]$ (left $= x$ , right $= 0$ ). MVT gives $\frac{l n 1 - l n ( 1 + x )}{0 - x} = \frac{1}{1 + c}, c \in (x, 0) .$ Since $- 1 < c < 0$ , we have $0 < 1 + c < 1$ , so $\frac{1}{1 + c} > 1$ . The left side is $\frac{- ln ( 1 + x )}{- x}$ ; setting it $> 1$ : $\frac{- l n ( 1 + x )}{- x} > 1.$ Here $- x > 0$ , so multiply by $- x$ : $- ln (1 + x) > - x$ , i.e. $ln (1 + x) < x$ . Why this step? On the negative side the derivative exceeds $1$ , and the run $- x$ is positive — reversing our earlier reasoning cleanly, we land on the same inequality $ln (1 + x) < x$ .

Verify: $x = 1$ : $ln 2 \approx 0.693 \leq 1$ ✔. $x = 0.5$ : $ln 1.5 \approx 0.405 \leq 0.5$ ✔. $x = - 0.5$ : $ln 0.5 \approx - 0.693 \leq - 0.5$ ✔ (negative side works too). Same MVT technique as the parent note's $sin$ /Lipschitz Continuity example, now covering both signs of $x$ .

Example 8 — Word problem: average vs instantaneous (Cell H)

Worked example Speed camera

A car passes camera $A$ , then camera $B$ located $12 km$ down a straight road, taking $6$ minutes. The speed limit is $100 km/h$ . Prove the driver must have broken the limit at some instant. Forecast: what was the average speed — and what does MVT promise about the speedometer?

Step 1 — Set up notation and compute average speed with clean units. Let $t_{A}$ be the clock time the car passes camera $A$ , and $t_{B}$ the time it passes camera $B$ ; then $t_{B} - t_{A} = 6$ min $= 0.1$ h. Let $s (t)$ be the car's position (km) along the road at time $t$ , so $s (t_{B}) - s (t_{A}) = 12$ km. Average speed $\frac{s ( t _{B} ) - s ( t _{A} )}{t _{B} - t _{A}} = \frac{12 km}{0.1 h} = 120 km/h .$ Why this step? Defining $t_{A}, t_{B}, s (t)$ up front means every symbol in the MVT quotient is earned; this average is exactly the secant slope of position-vs-time.

Step 2 — Model with MVT. $s (t)$ is continuous and differentiable (real motion is smooth). MVT gives a time $c \in (t_{A}, t_{B})$ with $s^{'} (c) = \frac{s ( t _{B} ) - s ( t _{A} )}{t _{B} - t _{A}} = 120 km/h .$ Why this step? $s^{'} (c)$ is the instantaneous speedometer reading — MVT forces it to equal the average somewhere.

Step 3 — Conclude. At that instant $c$ , the speed was exactly $120 > 100$ km/h — over the limit. Guilty. Why this step? The theorem converts an average fact (easy to measure) into an instantaneous fact (the offence).

Verify: units: $\frac{km}{h}$ ✔; $120$ km/h $> 100$ km/h ✔. This is the parent note's "100 km/h speedometer" story made rigorous.

Example 9 — Exam twist: counting roots (Cell I)

Worked example Show

$p (x) = x^{3} + 3 x + 1$ has exactly one real root Forecast: a cubic could have 1 or 3 real roots. Rolle can rule out extras — how?

Step 1 — Existence of at least one root. $p (- 1) = - 1 - 3 + 1 = - 3 < 0$ and $p (0) = 1 > 0$ . By continuity, $p$ crosses zero between $- 1$ and $0$ — at least one real root. Why this step? The Intermediate-Value idea (a consequence of Continuity) guarantees one crossing.

Step 2 — Suppose there were two roots, and invoke Rolle. If $p$ had two roots $r_{1} < r_{2}$ , then $p (r_{1}) = p (r_{2}) = 0$ , so Rolle gives a $c \in (r_{1}, r_{2})$ with $p^{'} (c) = 0$ . Why this step? Two equal-height points (both zero) trigger Rolle — a flat tangent must exist between them.

Step 3 — Show $p^{'}$ is never zero, forcing a contradiction. $p^{'} (x) = 3 x^{2} + 3 = 3 (x^{2} + 1) \geq 3 > 0$ for all $x$ . So $p^{'} (c) = 0$ is impossible. Why this step? No flat point can exist ⇒ two roots are impossible ⇒ at most one root. Combined with Step 1: exactly one.

Verify: $p^{'} (x) = 3 x^{2} + 3 > 0$ always ✔; $p (- 1) = - 3$ , $p (0) = 1$ opposite signs ✔. (Rolle used as a "no second root" filter — a classic Increasing and Decreasing Functions argument in disguise.)

Recall Quick self-test

Rolle fails for $∣ x - 1∣$ on $[0, 2]$ because... ::: it is not differentiable at the corner $x = 1$ . The MVT $c$ for $x^{3}$ on $[0, 2]$ is... ::: $c = \frac{2}{3} \approx 1.155$ . As $b \to a$ , the MVT statement becomes... ::: the definition of the derivative $f^{'} (a)$ . How do you prove a cubic has only one real root with Rolle? ::: assume two roots, get $p^{'} (c) = 0$ between them, then show $p^{'} > 0$ everywhere — contradiction. Average speed $120$ km/h over a $100$ km/h zone proves speeding because... ::: MVT forces the instantaneous speed to equal $120$ km/h at some instant.

Connections

Extreme Value Theorem — the max/min that Rolle stands on.
Fermat's Theorem (interior extrema) — why an interior peak has $f^{'} = 0$ .
Continuity · Differentiability — the two hypotheses tested in Cells D and E.
Increasing and Decreasing Functions — the sign of $f^{'}$ used in the root-counting twist.
Cauchy's Mean Value Theorem · Taylor's Theorem — where the degenerate case (Cell F) leads next.
Lipschitz Continuity — the bounding technique of Cell G.