The Derivative and the Mean Value Theorems: Examples and Counterexamples

About this page

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 9 results · all verified · 8 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

1 · Prerequisites

2 · Logical flowchart

3 · Items

Worked derivatives from the algebra of derivatives and the power rule: (3x⁴ - 5x + 2)' = 12x³ - 5, and the quotient rule applied to (x²+1)/(x-1) on ℝ ∖ {1}

x ↦ |x| is continuous everywhere and not differentiable at 0: the difference quotient equals 1 on the right and -1 on the left, so the two one-sided limits differ

x ↦ x³ is increasing on ℝ although its derivative vanishes at 0, which is the witness for the false statement that a vanishing derivative forbids strict increase, and which makes its inverse non-differentiable at 0

The chain rule applied to x ↦ (x²+1)⁵ and to x ↦ ((3x-1)²+2)³, with the Carathéodory factor written out in closed form in the first case

For a natural n ≥ 1, the derivative of x ↦ x^1/n on (0,∞) is 1/ι(n)x^1/n - 1, obtained from the inverse rule applied to x ↦ xⁿ; in particular (√x)' = 1/(ι(2)√x)

f(x) = x on [0,1) with f(1) = 0 is differentiable at every point of (0,1) with f' ≡ 1, yet no c satisfies f(1) - f(0) = f'(c), so continuity on the closed interval cannot be dropped from the mean value theorem

The identity on [0,1] attains its maximum at 1 and its minimum at 0 with derivative 1 at both, so Fermat's theorem genuinely needs the extremum to be at an interior point

The mean value theorem gives |√x - √y| ≤ 1/ι(2) |x - y| for x, y ≥ 1, so the square root is Lipschitz with constant 1/2 on [1,∞)

x ↦ √x on (0,1] is differentiable with unbounded derivative and is not Lipschitz there, so the boundedness hypothesis in the Lipschitz corollary cannot be dropped

With f(x) = x³ and g(x) = x² on [-1,1] the quotient form f(b)-f(a)/g(b)-g(a) = f'(c)/g'(c) is meaningless because g(b) = g(a), while the product form of Cauchy's theorem still holds