---
title: "Rolle's theorem with discontinuity at end points"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:rolle-end-point-discontinuity"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-end-point-discontinuity
site: BCT Engineering
---

# Rolle's theorem with discontinuity at end points

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2074 Chaitra, 1+2+2 marks
- Video: [RollesTheorem Differential Calculus Mean Value Theorem Continuity and Differentiability of functions](https://www.youtube.com/watch?v=Oq9wZT5wdnU) — dy by dx
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-end-point-discontinuity
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

State Rolle's theorem. Is the theorem true when the function is not continuous at the end points? Justify your answer. Verify Rolle's theorem for f(x) = x² − 5x + 6 on [2,3].

## Short answer

Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b) and f(a)=f(b), then ∃c∈(a,b) with f'(c)=0. The theorem fails if continuity at endpoints is missing; e.g., f(x)=x on [0,1) with f(1)=0 gives f'(x)=1≠0. For f(x)=x²−5x+6 on [2,3], f(2)=f(3)=0, f'(x)=2x−5, so c=5/2=2.5 ∈ (2,3) verifies the theorem.

## Answer

#### Rolle's theorem
If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f′(c) = 0 for at least one c in (a, b).

#### Discontinuity at the end points
The theorem is **not true** in general if f fails to be continuous at an end point, because continuity on the closed interval is a necessary hypothesis. Counterexample: f(x) = x for 0 ≤ x < 1 and f(1) = 0 on [0, 1]. Here f(0) = f(1) = 0 and f is differentiable on (0, 1), but it is discontinuous at x = 1. Still f′(x) = 1 ≠ 0 everywhere in (0, 1), so no c exists.

#### Verification for f(x) = x² − 5x + 6 on [2, 3]
1. f is a polynomial, so continuous on [2, 3] and differentiable on (2, 3).
2. f(2) = 4 − 10 + 6 = 0 and f(3) = 9 − 15 + 6 = 0, so f(2) = f(3).
3. f′(x) = 2x − 5. Setting f′(c) = 0 gives c = 5/2 = 2.5, which lies in (2, 3).

**Rolle's theorem is verified with c = 5/2.**

## Check yourself

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

1. Is Rolle's theorem true if f is discontinuous at an end point?
   A. Always
   B. Not in general
   C. Only if f' exists
   D. Only for polynomials
   Answer: B. Continuity on the closed interval is a hypothesis. (Key point: Necessity)

2. In the counterexample f(x) = x on [0, 1) with f(1) = 0, f'(x) is:
   A. undefined
   B. 0 everywhere
   C. varies
   D. 1 everywhere on (0,1)
   Answer: D. So f' never vanishes although f(0) = f(1). (Key point: Counterexample)

3. For x^2 - 5x + 6 on [2, 3], f'(c) = 0 gives:
   A. c = 5
   B. c = 5/2
   C. c = 2
   D. c = 3
   Answer: B. 2c - 5 = 0. (Key point: Verification)

4. f(2) and f(3) are:
   A. 0 and 6
   B. both 0
   C. both 6
   D. 2 and 3
   Answer: B. 4 - 10 + 6 = 0 and 9 - 15 + 6 = 0. (Key point: End values)

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