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Rolle's theorem with discontinuity at end points

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.

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].

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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