---
title: "Rolle vs Lagrange; verify Lagrange's MVT"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:rolle-lagrange-difference-verify"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-lagrange-difference-verify
site: BCT Engineering
---

# Rolle vs Lagrange; verify Lagrange's MVT

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2074 Asoj
- Video: [Problems on Lagrange's Mean Value Theorem/LMVT/First Mean Value Theorem](https://www.youtube.com/watch?v=uXblGx6gOoc) — TEAM Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-lagrange-difference-verify
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

State difference between Rolle's Theorem and Lagrange's Mean value theorem. Verify Lagrange's mean value theorem for f(x) = x(x−1)(x−2) when x ∈ [0, 1/2].

## Short answer

Rolle's theorem requires f(a)=f(b) and gives f'(c)=0; Lagrange's MVT does not require equal endpoints and yields f'(c) = (f(b)−f(a))/(b−a). For f(x)=x(x−1)(x−2) on [0,1/2], f(0)=0, f(1/2)=3/8, slope 3/4. f'(x)=3x²−6x+2; solving f'(c)=3/4 gives c = 1 − √21/6 ≈ 0.236 in (0,1/2), verifying the theorem.

## Answer

#### Difference between Rolle's theorem and Lagrange's mean value theorem

| Point | Rolle's theorem | Lagrange's MVT |
|---|---|---|
| Conditions | f continuous on [a,b], differentiable on (a,b), **and f(a) = f(b)** | Same, but **f(a) = f(b) is not needed** |
| Conclusion | f′(c) = 0 for some c in (a,b) | f′(c) = (f(b) − f(a))/(b − a) |
| Geometry | Tangent parallel to the x-axis | Tangent parallel to the chord joining the end points |
| Relation | A special case of Lagrange's MVT | The general form (Rolle is the case f(a) = f(b)) |

#### Verification of Lagrange's MVT
f(x) = x(x − 1)(x − 2) = x³ − 3x² + 2x on [0, 1/2].

1. f is a polynomial, hence continuous on [0, 1/2] and differentiable on (0, 1/2).
2. f(0) = 0 and f(1/2) = (1/2)(−1/2)(−3/2) = 3/8.
3. Slope of chord = (3/8 − 0)/(1/2 − 0) = 3/4.
4. f′(x) = 3x² − 6x + 2. Set f′(c) = 3/4:

3c² − 6c + 5/4 = 0, so 12c² − 24c + 5 = 0, giving c = [24 ± √(576 − 240)]/24 = 1 ± √21/6.

c = 1 + √21/6 ≈ 1.764 (rejected, outside the interval); c = 1 − √21/6 ≈ 0.236, which lies in (0, 1/2).

**Hence Lagrange's MVT is verified with c = 1 − √21/6 ≈ 0.236.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What extra condition does Rolle's theorem need compared with Lagrange's MVT?
   A. Continuity on [a,b]
   B. f(a) > 0
   C. Differentiability on (a,b)
   D. f(a) = f(b)
   Answer: D. Lagrange drops f(a) = f(b). (Key point: Difference)

2. For f(x) = x(x-1)(x-2) on [0, 1/2], the chord slope is:
   A. 3/4
   B. 3/8
   C. 1/2
   D. 3/2
   Answer: A. (3/8 - 0)/(1/2) = 3/4. (Key point: Chord slope)

3. f'(c) = 3/4 with f' = 3x^2 - 6x + 2 gives:
   A. 3c^2 - 6c + 2 = 0
   B. 12c^2 - 24c + 5 = 0
   C. 12c^2 - 24c - 5 = 0
   D. c^2 - 2c = 0
   Answer: B. 3c^2 - 6c + 5/4 = 0, times 4. (Key point: Equation)

4. Which root is accepted?
   A. Both
   B. c = 1 + sqrt(21)/6
   C. Neither
   D. c = 1 - sqrt(21)/6 ~ 0.236
   Answer: D. Only 0.236 is in (0, 1/2). (Key point: Choosing c)

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