Calculus I › Applications of derivatives › full formula sheet
Rolle’s theorem
Start and end at the same height? Somewhere in between, you turned around.
Requirements: f continuous on [a, b], differentiable on (a, b), and f(a) = f(b). Named for Michel Rolle (1691).
Before this lesson: Definition of the derivative
Where it comes from
The problem: throw a ball straight up; it leaves and returns to your hand at the same height. The naive claim — “f(a) = f(b) means the function is constant” — dies instantly: sin x on [0, 2π] starts and ends at 0 but waves up and down in between. The true claim: somewhere the motion turned around — at the top of the throw, velocity is zero. Between any two equal heights sits a flat tangent.
But again the hypotheses are load-bearing. The corner strikes back — f(x) = |x| on [−1, 1]:
The intuition for why it should be true: if f isn’t constant, it must rise above or dip below the endpoint level somewhere inside — and at that highest or lowest interior point, the tangent has nowhere to tilt. Flat.
Before reading on: a non-constant continuous f on [a, b] with f(a) = f(b) attains a max or min somewhere. Where must that extreme point lie — and what is the slope of the tangent there?
Derivation
Two cases, then one classic theorem each: the Extreme Value Theorem (a continuous function on [a, b] attains a max and a min) and Fermat’s theorem (an interior max/min of a differentiable function has derivative zero).
Why must the extreme be interior? If both M and m occurred only at the endpoints, then every value of f would lie between f(a) and f(b) = f(a) — forcing f constant, contradicting Case 2. So non-constancy pushes an extreme inside.
Before reading on: f(x) = x³ on [−1, 1] has f(−1) ≠ f(1), yet f′(0) = 0. Does that contradict Rolle? Why not?
How to use it
The procedure, every time:
- Check all three conditions: continuous on [a, b], differentiable on (a, b), and f(a) = f(b). All three — the |x| example shows why.
- Conclude: some c in (a, b) has f′(c) = 0. That existence claim is often the whole point (e.g. “prove f′ has a zero”).
- To locate c: solve f′(x) = 0 and keep solutions strictly inside (a, b). Expect possibly more than one.
When to reach for it
Whenever equal values appear: between any two zeros of f lies a zero of f′ — the workhorse corollary. If a polynomial has roots at 1, 2, and 3, its derivative has a root in (1, 2) and another in (2, 3). It is also the untilted special case that proves the Mean Value Theorem.
At least one, maybe more
Rolle promises ≥ 1 interior flat point, never exactly 1. A wavy function with f(a) = f(b) can turn around many times.
Worked examples
Four problems, easiest first. In each one, read every step — the why of each move is the lesson.
Example 1 — the basic move: f(x) = x² − 4x on [0, 4]
- Conditions. Polynomial: continuous on [0, 4], differentiable on (0, 4). f(0) = 0, f(4) = 16 − 16 = 0. Equal ✓
- Conclude. Some c in (0, 4) has f′(c) = 0.
- Locate. f′(x) = 2x − 4 = 0 → x = 2. Check: 0 < 2 < 4 ✓ — the parabola’s vertex.
Your turn: f(x) = x² − 6x on [0, 6] — find c.
Answer: c = 3.
f(0) = 0 = f(6) ✓; f′(x) = 2x − 6 = 0 gives x = 3, and 0 < 3 < 6 ✓.
Example 2 — trigonometry: f(x) = sin x on [0, π]
- Conditions. sin x is continuous and differentiable everywhere. f(0) = 0 = f(π). ✓
- Conclude. Some c in (0, π) has f′(c) = 0.
- Locate. f′(x) = cos x = 0 → x = π/2 (in (0, π)). ✓ — the top of the arch.
Your turn: f(x) = cos x on [−π/2, π/2] — find c.
Answer: c = 0.
f(−π/2) = 0 = f(π/2) ✓; f′(x) = −sin x = 0 gives x = 0 ✓.
Example 3 — two turnarounds: f(x) = x³ − 3x on [−√3, √3]
- Conditions. Polynomial ✓. f(−√3) = −3√3 + 3√3 = 0; f(√3) = 3√3 − 3√3 = 0. Equal ✓
- Locate. f′(x) = 3x² − 3 = 0 → x² = 1 → x = ±1.
- Check. Both −1 and 1 lie strictly inside (−√3, √3) ✓ — two valid c’s (a local max and a local min).
Your turn: f(x) = x³ − 4x on [−2, 2] — find every c.
Answer: c = ±2/√3 ≈ ±1.155.
f(−2) = 0 = f(2) ✓; f′(x) = 3x² − 4 = 0 gives x = ±2/√3, both strictly inside (−2, 2) ✓.
Example 4 — when it breaks: f(x) = |x| on [−1, 1]
- Conditions? Continuous on [−1, 1] ✓; f(−1) = 1 = f(1) ✓ — but not differentiable at 0 ∈ (−1, 1). ✗ Stop.
- What goes wrong. f′(x) = −1 for x < 0, +1 for x > 0 — never 0. The V bottoms out at a corner, turning around without ever going flat.
- The lesson. Equal endpoints alone don’t force a flat tangent — differentiability is what rules out the sharp turn.
Your turn: Does Rolle apply to f(x) = √x on [0, 4]?
Answer: No.
f(0) = 0 ≠ 2 = f(4) — the equal-endpoints hypothesis fails. (f is also not differentiable at 0.)
Memorization tips
- Say it: “same height at both ends, flat tangent somewhere inside.”
- Three conditions, chant them: continuous on closed, differentiable on open, equal endpoint values.
- The |x| alarm: equal endpoints + a corner = no guarantee. Check differentiability before concluding.
- Between zeros of f, a zero of f′: the corollary you’ll actually use — roots of f force roots of f′ between them.
- At least one: never “exactly one” — x³ − 3x gave two. Solve f′ = 0 fully.
- Rolle is MVT at slope zero: f(a) = f(b) makes the average slope 0. Learn Rolle and you’re halfway to the Mean Value Theorem.
Final challenge
Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and Rolle’s theorem is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What does Rolle’s theorem say in plain English?
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then at some interior point c the tangent is flat: f′(c) = 0. Start and end at the same height, and somewhere you turned around.
Does f(a) = f(b) mean f is constant?
No. sin x on [0, 2π] has f(0) = f(2π) = 0 but waves up and down — with flat tangents at π/2 and 3π/2. Equal endpoints only force a turnaround somewhere.
Why does Rolle’s theorem need the hypotheses?
Without differentiability the conclusion can fail: f(x) = |x| on [−1, 1] has f(−1) = f(1) = 1 but no c with f′(c) = 0, because of the corner at 0.
How is Rolle’s theorem related to the Mean Value Theorem?
Rolle’s is the special case of MVT with average slope 0: f(a) = f(b) makes [f(b)−f(a)]/(b−a) = 0. And MVT itself is proved by tilting f down to Rolle’s setup.
Can Rolle’s theorem give more than one c?
Yes — it guarantees at least one. f(x) = x³ − 3x on [−√3, √3] has f′ = 0 at both x = 1 and x = −1.
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