Rolle's Theorem
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.4: “The Mean Value Theorem” (Rolle’s Theorem is Theorem 4.4) |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-4-the-mean-value-theorem |
| Textbook used in class | Stewart, Calculus, Section 3.2: “The Mean Value Theorem” |
Opening Scenario
Suppose a runner completes a lap and returns to exactly the starting line. No matter how the runner accelerated or changed pace during the lap, there must have been at least one moment when the forward velocity along the track was zero -- the runner was at a turning point. That is the physical content of Rolle’s Theorem: a quantity that starts and ends at the same value must have a zero rate of change somewhere in between.
Before reading the formal statement, sketch a smooth curve that starts and ends at the same height. Notice that the curve must either reach a peak and come back down, or dip to a valley and come back up, or stay completely flat. In every case there is at least one point where the tangent line is horizontal.
Quick Reference
Rolle’s Theorem. Let $f$ satisfy all three conditions:
- $f$ is continuous on the closed interval $[a, b]$,
- $f$ is differentiable on the open interval $(a, b)$,
- $f(a) = f(b)$.
Then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
The theorem guarantees existence; it does not say $c$ is unique, and it does not compute $c$.
Key Concepts
1. Why All Three Hypotheses Matter
Removing any single hypothesis can make the conclusion fail.
Continuity required. Consider $f$ defined on $[0, 1]$ by $f(x) = 1$ for $x \in (0, 1)$ and $f(0) = f(1) = 0$. Then $f(0) = f(1)$ and $f$ is differentiable on $(0,1)$ with $f'(x) = 0$. But if instead we use a function that jumps discontinuously from $0$ at both endpoints to $1$ on the interior, the derivative on the interior ($0$) does not connect back to the endpoint values in a smooth way, and we can construct examples where no zero of $f'$ lands in the expected place.
A cleaner example: $f(x) = \begin{cases} 0 & x = 0 \\ 1 - x & 0 < x \leq 1 \end{cases}$ has $f(0) = 0 = f(1)$ and $f'(x) = -1 \neq 0$ on $(0, 1)$, yet continuity fails at $x = 0$. The conclusion fails.
Differentiability required. Let $f(x) = |x|$ on $[-1, 1]$. Then $f(-1) = f(1) = 1$ and $f$ is continuous on $[-1,1]$. But $f'(x) = -1$ for $x < 0$, $f'(x) = 1$ for $x > 0$, and $f'(0)$ does not exist. There is no $c$ in $(-1, 1)$ with $f'(c) = 0$.
Equal endpoint values required. If $f(a) \neq f(b)$, then a strictly monotone function such as $f(x) = x$ is continuous and differentiable everywhere, yet $f'(x) = 1 \neq 0$ on all of $(0, 1)$.
“Two of the three conditions are enough.” All three are needed. Each one blocks a specific type of counterexample.
2. Why the Conclusion Follows: A Proof Sketch
Here is the reasoning that makes Rolle’s Theorem true, not just a rule to memorize.
Since $f$ is continuous on $[a, b]$, the Extreme Value Theorem guarantees $f$ attains its absolute maximum and its absolute minimum on $[a, b]$.
Case 1. Both the absolute max and the absolute min occur at the endpoints. Since $f(a) = f(b)$, the function’s largest and smallest values are equal. The only way a function’s max equals its min is if $f$ is constant on $[a, b]$. Then $f'(x) = 0$ for every $x \in (a, b)$, and any $c$ works.
Case 2. At least one of the absolute extrema occurs at an interior point $c \in (a, b)$. At that interior point, $f$ has a local extremum. Since $f$ is differentiable at $c$, Fermat’s Theorem gives $f'(c) = 0$.
Those two cases exhaust all possibilities. So in every situation, there is a $c \in (a, b)$ with $f'(c) = 0$.
3. Applying Rolle’s Theorem
Example 1. Verify that Rolle’s Theorem applies to $f(x) = x^2 - 4x + 3$ on $[1, 3]$, then find all $c$.
Verification:
- $f$ is a polynomial, so continuous on $[1, 3]$ and differentiable on $(1, 3)$.
- $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$. Equal endpoint values.
All conditions hold.
Finding $c$: $f'(x) = 2x - 4 = 0 \Rightarrow x = 2$. Since $2 \in (1, 3)$:
Boxed answer: $c = 2$.
Recap. The parabola opens upward, touches zero at $x = 1$ and $x = 3$, and dips to its minimum at $x = 2$. Rolle’s Theorem predicted a horizontal tangent inside $(1, 3)$, and the vertex at $x = 2$ provides it.
4. Using Rolle’s Theorem to Bound the Number of Roots
Between any two roots of a differentiable function, there must be at least one critical number of $f$. This gives a way to prove that a function has at most $k$ roots by showing its derivative has at most $k - 1$ zeros.
Example 2. Show that $f(x) = x^3 + x + 1$ has exactly one real root.
Existence. $f(-1) = -1 - 1 + 1 = -1 < 0$ and $f(0) = 0 + 0 + 1 = 1 > 0$. By the Intermediate Value Theorem there is at least one root in $(-1, 0)$.
Uniqueness. $f'(x) = 3x^2 + 1 \geq 1 > 0$ for all real $x$. The derivative is never zero.
If there were two roots $r_1 < r_2$, Rolle’s Theorem on $[r_1, r_2]$ would produce a $c \in (r_1, r_2)$ with $f'(c) = 0$. But $f'(x) > 0$ everywhere, so no such $c$ exists. Contradiction.
Boxed answer: $f(x) = x^3 + x + 1$ has exactly one real root, located in $(-1, 0)$.
Recap. The argument is a proof by contradiction. Assume two roots; apply Rolle; arrive at a contradiction with the derivative information. This pattern appears throughout analysis.
“Rolle’s Theorem tells you where the root of $f$ is.” No. Rolle’s Theorem concerns $f'$, not $f$ itself. The root of $f$ is found by other means (IVT for existence; uniqueness from $f'$ never vanishing).
5. Predict-and-Check
Consider $f(x) = \sin x$ on $[0, \pi]$.
Predict. The sine starts at $0$, rises to a peak, and returns to $0$. The peak should be at the midpoint $x = \pi/2$.
Check. $f'(x) = \cos x = 0$ at $x = \pi/2$. Confirmed: $c = \pi/2 \in (0, \pi)$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting to verify $f(a) = f(b)$ | Applying Rolle to $f(x) = x$ on $[0,1]$ | Check: $f(0) = 0 \neq 1 = f(1)$; Rolle does not apply |
| Assuming $c$ is unique | Claiming “the one $c$” | The theorem says at least one; there can be many |
| Applying Rolle when $f$ is not differentiable on $(a,b)$ | Using $f(x)=\lvert x\rvert$ on $[-1,1]$ | $f'(0)$ does not exist; the differentiability hypothesis fails |
| Confusing the root of $f$ with the zero of $f'$ | Saying Rolle finds a root of $f$ | Rolle finds where $f' = 0$, not where $f = 0$ |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Does Rolle’s Theorem apply to $f(x) = x^2 - 6x + 5$ on $[1, 5]$? If so, find all $c$.
Show answer
$f(1) = 1 - 6 + 5 = 0$ and $f(5) = 25 - 30 + 5 = 0$. Equal endpoint values. $f$ is a polynomial (continuous and differentiable everywhere). All conditions hold.
$f'(x) = 2x - 6 = 0 \Rightarrow x = 3$.
Boxed answer: $c = 3$.
Problem 2. Explain why Rolle’s Theorem does not apply to $f(x) = \tan x$ on $[0, \pi]$.
Show answer
$\tan x$ is not defined at $x = \pi/2 \in [0, \pi]$, so $f$ is not continuous on the closed interval $[0, \pi]$. The continuity hypothesis fails.
Boxed answer: Rolle’s Theorem does not apply because $f(x) = \tan x$ is discontinuous at $x = \pi/2 \in [0, \pi]$.
Level 2 -- Multiple Steps
Problem 3. Show that $x^5 - 5x + 4 = 0$ has at most two real roots in $[0, 2]$.
Show answer
Let $f(x) = x^5 - 5x + 4$. Suppose for contradiction that there are three roots $r_1 < r_2 < r_3$ in $[0, 2]$.
Apply Rolle’s Theorem on $[r_1, r_2]$: there exists $c_1 \in (r_1, r_2)$ with $f'(c_1) = 0$. Apply Rolle’s Theorem on $[r_2, r_3]$: there exists $c_2 \in (r_2, r_3)$ with $f'(c_2) = 0$.
So $f'$ has at least two zeros in $(0, 2)$.
But $f'(x) = 5x^4 - 5 = 5(x^4 - 1) = 5(x^2 - 1)(x^2 + 1)$. On $(0, 2)$, $f'(x) = 0$ only at $x = 1$ (since $x = -1$ is outside). That is at most one zero, not two.
Contradiction. So there are at most two roots in $[0, 2]$.
Boxed answer: At most two real roots in $[0, 2]$.
Level 3 -- Deeper Problems
Problem 4. Suppose $f$ is continuous on $[0, 1]$, differentiable on $(0, 1)$, $f(0) = 0$, and $f(1) = 0$. Must $f'(c) = 0$ for some $c \in (0, 1)$? What if $f$ is only continuous on $[0, 1]$ with no differentiability assumption?
Show answer
With differentiability on $(0, 1)$: yes. All three hypotheses of Rolle’s Theorem hold, so there exists $c \in (0, 1)$ with $f'(c) = 0$.
Without the differentiability assumption: no guarantee. The function $f(x) = |x - 1/2| - 1/2$ is continuous on $[0,1]$, has $f(0) = 0$ and $f(1) = 0$, but $f'(1/2)$ does not exist and $f'(x)$ is $-1$ on $(0,1/2)$ and $+1$ on $(1/2, 1)$, never zero. Rolle’s Theorem does not apply without differentiability, and the conclusion fails.
Boxed answer: Yes with differentiability; not guaranteed without it.
Mastery Checklist
Mental Model
Rolle’s Theorem is the turning-point guarantee. A smooth curve that starts at a certain height, wanders around, and returns to the same height must have at least one peak or valley in the middle. At any peak or valley of a differentiable function, the tangent is horizontal.
The three hypotheses make the scenario well-defined: continuity keeps the curve connected, differentiability ensures a tangent exists at every interior point, and equal endpoint values force the curve to turn around rather than drift to a new level.
Connections
Within Chapter 3
- Extreme Value Theorem (Section 3.1): Rolle’s proof uses it. The EVT guarantees an interior extremum exists; Fermat’s Theorem then says its derivative is zero.
- Mean Value Theorem (Section 3.2): The MVT is Rolle’s Theorem applied to a tilted version of $f$. Rolle is the special case where the tilt is zero.
- Increasing/Decreasing Test (Section 3.3): If $f' > 0$ on an interval, $f$ cannot return to a previous value, so Rolle cannot apply. This is how you rule out extra roots.
Back to Applications of Differentiation | Previous: Absolute and Local Extrema | Next: MVT Applications