Darboux's Theorem: The Intermediate Value Property of Derivatives

Darboux's theorem is a small but sharp tool with reach beyond a single textbook exercise. In real analysis itself, it is the standard method for proving that certain explicitly given functions — typically ones with jump discontinuities — cannot be derivatives of anything, which is otherwise difficult to establish directly from the definition of the derivative. It also refines the classical hierarchy continuous $\subsetneq$ differentiable-derivative $\subsetneq$ Darboux, clarifying that "having the intermediate value property" is a genuinely weaker requirement than continuity, useful when classifying pathological functions in courses following The Mean Value Theorem. In the theory of ordinary differential equations, Darboux's theorem is invoked when one studies whether a given right-hand side function can be realized as $f'$ for some antiderivative $f$, relevant to existence questions for primitives and to constructing counterexamples showing that not every function is a derivative even when Riemann integrable functions are. In numerical analysis and root-finding, the theorem justifies bisection-type or sign-change arguments applied directly to a derivative $f'$ (for instance in locating critical points of $f$ by hunting for sign changes of $f'$), even in situations where $f'$ is known to be discontinuous, since continuity of $f'$ is not actually needed for the sign-change argument to guarantee a root. In the broader landscape of mathematical logic and descriptive set theory, the class of "Darboux functions" (those with the intermediate value property, not necessarily derivatives) is studied as an interesting class in its own right; Darboux's theorem shows that the class of derivatives sits inside this larger, well-studied class, linking basic calculus to more advanced classification results about real functions.

:::quiz[Which of the following functions can be ruled out from being a derivative of any function on $[-1,1]$, using Darboux's theorem?]

• $f(x) = x^2$
• $f(x) = \sin(x)$
• A step function taking only the values $-1$ and $1$, jumping at $x=0$ ✓
• $f(x) = x^2\sin(1/x)$ with $f(0)=0$, restricted to its derivative

Ask the tutor: Can you show me why the function $g(x) = f(x) - y_0 x$ must be continuous even though $f'$ itself might not be, and walk through why that continuity is exactly what's needed for the Extreme Value Theorem step?

Continuous functions have the intermediate value property because their graphs cannot "jump" over a horizontal line: the classical proof uses a supremum/completeness argument on the set of points where the function is still below the target value. A derivative $f'$ need not be continuous, so that argument does not directly apply. Yet there is a completely different mechanism forcing the same conclusion for derivatives: a derivative is a limit of difference quotients — of slopes of secant lines — and slopes of secant lines vary continuously as the two points defining them move continuously. Roughly, $f'$ inherits intermediate-value behavior not from its own continuity but from being squeezed by a genuinely continuous family (the difference quotients) as the two evaluation points collide. The proof exploits this by a clever trick: instead of trying to show $f'$ takes a value $y_0$ directly, we tilt the whole function, replacing $f$ by $g(x) = f(x) - y_0 x$. This shifts every value of $f'$ down by $y_0$, so hitting the value $y_0$ for $f'$ becomes equivalent to hitting the value $0$ for $g'$ — that is, finding a critical point of $g$. Since $g$ is differentiable (hence continuous) on a closed bounded interval, it attains a maximum or minimum somewhere; if we can force that extremum to occur at an interior point, Interior Extrema and the Vanishing Derivative tells us $g'$ vanishes there, which is exactly what we want. The heart of the proof is verifying that the sign of $g'$ at the two endpoints forces the extremum to be interior, not at the endpoints themselves.

Example 1: Certifying that a specific piecewise function is not a derivative. Let

We claim there is no function $F$ differentiable on $[-1,1]$ with $F' = h$. Suppose for contradiction such $F$ existed. Then $F' = h$ satisfies $F'(-1/2) = -1$ and $F'(1/2) = 1$. Take $y_0 = 0$, which lies strictly between $-1$ and $1$. By Darboux's theorem applied on $[-1/2, 1/2] \subset [-1,1]$, there must exist $c \in (-1/2,1/2)$ with $F'(c) = 0$. But $F' = h$ takes only the two values $-1$ and $1$ on $[-1,1]$ — it never equals $0$ anywhere, by direct inspection of the formula for $h$. This contradicts the existence of such a $c$. Hence $h$ is not the derivative of any function differentiable on $[-1,1]$, even though $h$ is a perfectly reasonable bounded, Riemann integrable, piecewise-continuous function. This shows the class of derivatives is a strict subclass of, say, the Riemann integrable functions, and that a simple jump discontinuity is an obstruction. Note this matches the Corollary above: $h$ has one-sided limits $-1$ and $1$ at $x_0=0$ that disagree, which is exactly the forbidden situation for a derivative.

Example 2: An explicit differentiable function with discontinuous derivative, and locating a Darboux point directly. Define $f : [-1,1] \to \mathbb{R}$ by

Differentiability at $x \neq 0$: By the product and chain rules (valid since $x^2$ and $\sin(1/x)$ are each differentiable at any $x \neq 0$),

Differentiability at $x = 0$: Compute the difference quotient directly:

Since $|\sin(1/h)| \le 1$, we get $|h \sin(1/h)| \le |h| \to 0$ as $h \to 0$, so by the squeeze theorem

Discontinuity of $f'$ at $0$: Consider $x_n = \dfrac{1}{2\pi n} \to 0$. Then $\sin(1/x_n) = \sin(2\pi n) = 0$ and $\cos(1/x_n) = \cos(2\pi n) = 1$, so

Thus $f'(x_n) \to -1 \neq 0 = f'(0)$ even though $x_n \to 0$, so $f'$ is not continuous at $0$. (In fact $f'$ oscillates between values near $-1$ and near $1$ infinitely often as $x \to 0$, consistent with the Corollary: it has no one-sided limit at $0$ at all, only oscillatory divergence — precisely the only type of discontinuity Darboux's theorem permits.) Locating a genuine Darboux point. Let us verify the theorem concretely on $[a,b] = [x_2, x_1]$ where $x_1 = \frac{1}{2\pi} \approx 0.15915$ and $x_2 = \frac{1}{4\pi} \approx 0.07958$. Compute:

These happen to coincide, so let's instead use $x_3 = \frac{1}{(2n+1)\pi}$ type points to get a sign change. Take $u = \frac{1}{2\pi} \approx 0.15915$ (so $f'(u) = -1$ as above) and $v = \frac{2}{\pi} \approx 0.63662$, at which $\sin(1/v) = \sin(\pi/2) = 1$ and $\cos(1/v)=\cos(\pi/2)=0$, giving

So on $[u,v] \approx [0.15915,\,0.63662]$, we have $f'(u) = -1$ and $f'(v) \approx 1.2732$. Take $y_0 = 0$, strictly between them. Darboux's theorem guarantees some $c \in (u,v)$ with $f'(c)=0$, i.e.

Indeed, at $c = \frac{1}{\pi} \approx 0.31831$ we have $\sin(1/c) = \sin(\pi) = 0$ and $\cos(1/c) = \cos(\pi) = -1$, giving $f'(c) = 0 - (-1) = 1 \neq 0$ — so $c=1/\pi$ itself is not quite the root, but it confirms $f'$ does cross zero somewhere in $(u, 1/\pi)$ or nearby since $f'(u) = -1 <0$ while $f'(1/\pi) = 1 > 0$; a genuine root exists in $(u,\,1/\pi) \approx (0.15915, 0.31831)$ by Darboux's theorem applied to that sub-subinterval, illustrating that the theorem's existence claim is easy to confirm numerically even without solving the transcendental equation $2c\sin(1/c)=\cos(1/c)$ in closed form.