Answer: Nowhere-Zero Derivative Paradox

Key Idea / Intuition

Your first instinct might be: "If $f'$ is never zero, it can't switch sign, because by the Intermediate Value Theorem it would have to pass through zero." And this is exactly right — but the key insight is subtle: the derivative $f'$ need not be continuous, yet it still satisfies the Intermediate Value Property (Darboux's theorem). So even though $f'$ might be discontinuous, it cannot skip over zero. Therefore yes: $f'$ must have constant sign.

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Formal Proof / Solution

Claim: If $f: \mathbb{R} \to \mathbb{R}$ is differentiable everywhere and $f'(x) \neq 0$ for all $x$, then $f'$ has constant sign.

Step 1: State Darboux's Theorem.

Darboux's Theorem: If $f$ is differentiable on $[a, b]$, then $f'$ satisfies the Intermediate Value Property: for any $a < b$ and any value $k$ strictly between $f'(a)$ and $f'(b)$, there exists $c \in (a, b)$ with $f'(c) = k$.

This is remarkable because $f'$ need not be continuous — yet it cannot "jump" over any value. (The proof uses that $g(x) = f(x) - kx$ achieves its extremum at an interior point where $g' = 0$.)

Step 2: Apply Darboux to conclude constant sign.

Suppose for contradiction that $f'$ takes both positive and negative values. Then there exist $a, b \in \mathbb{R}$ with $f'(a) > 0$ and $f'(b) < 0$.

By Darboux's theorem applied on the interval $[\min(a,b),\, \max(a,b)]$, since $0$ lies strictly between $f'(a)$ and $f'(b)$, there exists $c$ between $a$ and $b$ such that: $f'(c) = 0.$

This contradicts the assumption that $f'(x) \neq 0$ for all $x$.

Conclusion: $f'$ cannot change sign. Since $f'$ is never zero, we must have either $f'(x) > 0$ for all $x \in \mathbb{R}$, or $f'(x) < 0$ for all $x \in \mathbb{R}$. $\blacksquare$

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Why this is surprising:

One might think: "Maybe $f'$ oscillates wildly enough to avoid zero while still changing sign — like $\sin(1/x)$." But Darboux blocks this entirely. The IVP for derivatives is a hidden rigidity that survives even in the absence of continuity.

A classic example of a discontinuous derivative: $f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$) has $f'(0) = 0$ but $f'$ oscillates near $0$. This shows derivatives can be wild — but Darboux guarantees they can never skip a value.