Consider f(x) = |x| on [-1, 1]. We have f(-1) = f(1) = 1 and f is continuous on [-1, 1]. Can Rolle's Theorem be applied to guarantee a point where f'(c) = 0?
AYes — all three hypotheses are satisfied, so Rolle's Theorem applies
BNo — Rolle's Theorem requires f(a) = f(b) = 0, but here f(-1) = f(1) = 1
CNo — f is not differentiable at x = 0, so the differentiability hypothesis fails
DYes — but since |x| has no local max or min in (-1,1), the theorem's conclusion fails, disproving the theorem
Rolle's Theorem requires differentiability on the open interval (a, b). The function |x| has a corner at x = 0 where the derivative does not exist, so the theorem cannot be applied. The fact that f(-1) = f(1) and the function is continuous is not enough — all three hypotheses must hold. Option A is the classic error: students who only remember the equal-endpoints and continuity conditions miss the differentiability check.
Question 2 Multiple Choice
Rolle's Theorem is applied to f(x) = x³ − x on [−1, 1]. The theorem guarantees which of the following?
AExactly one c in (−1, 1) where f′(c) = 0
BAt least one c in (−1, 1) where f′(c) = 0
CAt least one c in [−1, 1] where f′(c) = 0, possibly at an endpoint
DA unique c in (−1, 1) where f(c) = 0
Rolle's Theorem guarantees at least one such c — the conclusion is existential, not uniqueness. For this function, f′(x) = 3x² − 1 = 0 gives x = ±1/√3, so there are actually two such points in (−1, 1). The theorem promised at least one and delivered two. Option C is wrong because the theorem places c strictly inside the open interval (a, b), not at the endpoints.
Question 3 True / False
If f satisfies most three hypotheses of Rolle's Theorem on [a, b], the theorem guarantees that there is exactly one interior point where f′ = 0.
TTrue
FFalse
Answer: False
Rolle's Theorem guarantees the existence of at least one such point, not uniqueness. A function can satisfy all three hypotheses and have multiple horizontal tangents — for example, f(x) = x³ − x on [−1, 1] gives two points where f′(c) = 0. The theorem says 'there exists'; it says nothing about how many such points exist.
Question 4 True / False
A function with a corner (non-differentiable point) on the interior of [a, b] cannot satisfy the hypotheses of Rolle's Theorem, even if it is continuous and f(a) = f(b).
TTrue
FFalse
Answer: True
Correct. Rolle's Theorem requires differentiability on the open interval (a, b). A corner is precisely a point where the derivative does not exist, so the second hypothesis fails. The example f(x) = |x| on [−1, 1] shows this clearly: it is continuous, satisfies f(−1) = f(1) = 1, but has a corner at x = 0, so Rolle's Theorem does not apply — and indeed f′(x) = ±1 wherever it exists, never zero.
Question 5 Short Answer
Why are all three hypotheses in Rolle's Theorem necessary? Briefly describe what goes wrong — and give a specific example — when each hypothesis is dropped.
Think about your answer, then reveal below.
Model answer: Continuity on [a,b]: without it, a function can jump discontinuously from f(a) back to f(a) without any turning point. Example: f(x) = 0 on [0,1] except f(0.5) = 1; f(0) = f(1) = 0 but f′ ≠ 0 wherever defined. Differentiability on (a,b): without it, a function can have a corner that allows it to change direction without a zero derivative. Example: f(x) = |x| on [−1,1] has f(−1) = f(1) but f′(0) is undefined and f′ ≠ 0 elsewhere. Equal endpoint values f(a) = f(b): without it, the function has a net slope and need not turn around. Example: f(x) = x on [0,1] is smooth with f′(x) = 1 everywhere — no horizontal tangent.
Each hypothesis blocks a specific loophole. Continuity blocks teleportation (jumps). Differentiability blocks corners that allow direction changes without a zero slope. Equal endpoints forces the function to 'come back,' creating a turning point. Remove any one and you can construct a counterexample where the conclusion fails.