False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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FALSE: if $f^{'} (c) = 0$ then $f$ is not increasing on any interval containing $c$

Statement

False claim: let $I \subseteq R$ be an interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $f : I \to R$ and let $c \in I$ be a point at which $f$ is differentiable with

$f^{'} (c) = 0$

(The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set). Then $f$ is not increasing on $I$ , in the strict sense of Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences.

Why it is tempting. On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed proves that $f^{'} > 0$ at every interior point gives an increasing function, and one reads the implication backwards: if strict increase comes from a strictly positive derivative, surely a derivative that fails to be strictly positive somewhere must destroy the strict increase there. It does not. Claim 5 of that theorem is the true converse, and it is non-strict: an increasing $f$ has $f^{'} \geq 0$ wherever it is differentiable, and nothing forbids equality at isolated points.

Facts & Assumptions

Given: The interval $I : = R$ , the point $c : = 0$ and the function $f : R \to R$ , $f (x) : = x^{3}$ (Integer powers $a^{m}$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

Power rule (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2): for a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable at every real $c$ with derivative $ι (n) c^{n - 1}$ .

[L2]

Canonical naturals (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing): $ι (n) > 0$ for every natural $n \geq 1$ , so in particular $ι (3) > 0$ .

[L3]

Powers (Integer powers $a^{m}$ ): $a^{0} = 1$ , $a^{2} = a \cdot a$ , and $0 \cdot a = 0$ , so $0^{2} = 0$ .

[L4]

Order arithmetic (Sign rules for products and monotonicity of multiplication, Ordered field): a product of two positive reals is positive and a product of two negative reals is positive; the order is total and transitive, and trichotomy holds.

[L5]

Monotonicity from the derivative (On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed, claim 2): for $J$ order-convex and $h : J \to R$ continuous on $J$ and differentiable at every interior point of $J$ with $h^{'} > 0$ there, $h$ is increasing on $J$ .

[L6]

Continuity (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5): $x \mapsto x^{n}$ is continuous at every point of its domain for every natural $n$ ; and continuity passes to a subset of the domain (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L7]

Restriction of the derivative (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set): if $B \subseteq A$ , if $p \in B$ is a limit point of $B$ and if $h : A \to R$ is differentiable at $p$ , then $h ∣_{B}$ is differentiable at $p$ with the same derivative; every point of an order-convex set with at least two elements is a limit point of it (Limit point, isolated point, adherent point, derived set, and dense subset of $R$ ); and a point $p$ is interior to a set $S$ exactly when $N_{ε} (p) \subseteq S$ for some real $ε > 0$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ , Interior, closure, boundary and exterior of a subset of $R$ ).

[L8]

Increasing on a set $J$ means $h (x) < h (y)$ for all $x, y \in J$ with $x < y$ (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L9]

A positive base has positive natural powers (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claim 1).

Refutation

technique · direct

1.1L1L3

By [L1] with $n : = 3$ , the function $f$ is differentiable at every real $c$ with $f^{'} (c) = ι (3) c^{2}$ . In particular $f^{'} (0) = ι (3) \cdot 0^{2} = ι (3) \cdot 0 = 0$ by [L3].

1.2L2L3L4L9

For every real $c \neq 0$ one has $c^{2} > 0$ : if $c > 0$ this is [L9]; if $c < 0$ then $c^{2} = c \cdot c$ is a product of two negative reals, hence positive by [L3] and [L4]. Therefore $f^{'} (c) = ι (3) c^{2} > 0$ for every $c \neq 0$ , being a product of two positive reals by [L2] and [L4].

1.3L4L7

Put $I_{1} : = (- \infty, 0]$ and $I_{2} : = [0, \infty)$ , both order-convex with at least two elements (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Every real $x < 0$ is interior to $I_{1}$ , since $N_{∣ x ∣} (x) \subseteq (- \infty, 0) \subseteq I_{1}$ ; and $0$ is not interior to $I_{1}$ , since every $N_{ε} (0)$ contains $ε / 2 > 0$ , which is not in $I_{1}$ . As every interior point of $I_{1}$ lies in $I_{1}$ and so satisfies $x \leq 0$ , the interior points of $I_{1}$ are exactly the reals $x < 0$ . The same argument gives that the interior points of $I_{2}$ are exactly the reals $x > 0$ .

2.1step 1.2step 1.3L5L6L7

By [L6] the function $f$ is continuous on $R$ , hence $f ∣_{I_{1}}$ is continuous on $I_{1}$ and $f ∣_{I_{2}}$ is continuous on $I_{2}$ . At every interior point $x$ of $I_{1}$ one has $x < 0$ by step 1.3, so $x$ is a limit point of $I_{1}$ by [L7] and $f ∣_{I_{1}}$ is differentiable at $x$ with derivative $f^{'} (x) = ι (3) x^{2} > 0$ by step 1.2 and [L7]. So [L5] gives that $f ∣_{I_{1}}$ is increasing on $I_{1}$ ; the same argument on $I_{2}$ gives that $f ∣_{I_{2}}$ is increasing on $I_{2}$ .

3.1step 2.1L4L8

Let $a, b \in R$ with $a < b$ . If $b \leq 0$ then $a, b \in I_{1}$ and step 2.1 gives $f (a) < f (b)$ . If $a \geq 0$ then $a, b \in I_{2}$ and step 2.1 gives $f (a) < f (b)$ . Otherwise $b > 0$ and $a < 0$ , so $a, 0 \in I_{1}$ with $a < 0$ gives $f (a) < f (0)$ , while $0, b \in I_{2}$ with $0 < b$ gives $f (0) < f (b)$ , and transitivity gives $f (a) < f (b)$ . The three cases are exhaustive, since failing both $b \leq 0$ and $a \geq 0$ means $b > 0$ and $a < 0$ . So $f$ is increasing on $R$ by [L8].

4.1step 1.1step 3.1∎

The false claim fails on this witness: $R$ is an interval, $f$ is differentiable at $c = 0$ with $f^{'} (0) = 0$ by step 1.1, and yet $f$ is increasing on $R$ by step 3.1. So a vanishing derivative forbids nothing of the kind, and the claim is false.

Remarks

What survives. Claim 5 of On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed is the correct converse and is non-strict: an increasing function differentiable at a point of its interval has $f^{'} \geq 0$ there. This witness saturates that inequality at exactly one point, and no more can be said in general.
How large the vanishing set can be is not settled here. The witness has $f^{'} = 0$ at a single point. Nothing on this page says how big the set ${f^{'} = 0}$ may be for an increasing $f$ , and nothing here should be read as suggesting that it must be small.
The same function is the standard witness for a second false reading, that a vanishing derivative marks a local extremum: $f$ has neither a local maximum nor a local minimum at $0$ , precisely because it is increasing. $x \mapsto x^{3}$ 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 $0$ ↗ on the companion page computes the derivative in full and draws the further consequence, through Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , that the inverse of this function is not differentiable at $0$ .

FALSE: if $f^{'} (c) = 0$ then $f$ is not increasing on any interval containing $c$

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: if f′(c)=0f′(c)=0 then ff is not increasing on any interval containing cc

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: if $f^{'} (c) = 0$ then $f$ is not increasing on any interval containing $c$