Proof Strategy for Darboux's Theorem

Turn an Intermediate Derivative Value into a Zero

Darboux’s Theorem for Derivatives says that a derivative takes every value between any two of its values. The previous tutorial used this property to rule out certain patterns in proposed derivatives. Here we focus on the proof strategy behind the theorem: how can information about derivative values at two inputs force the derivative to equal a specified number somewhere between them?

The key is to subtract a line with the target slope. If we want \(f'(c)=\lambda\), define \(g(x)=f(x)-\lambda x\). Then \(g'(x)=f'(x)-\lambda\), so the desired equality is equivalent to \(g'(c)=0\). The endpoint derivative values now tell us that \(g\) initially moves in one direction near one endpoint and in the opposite direction near the other. This forces an extremum at an interior point, where the derivative must vanish.

The Endpoint-Sign Principle

The crucial step is not simply that a continuous function has a minimum on a closed interval. A minimum could occur at an endpoint, where the derivative need not be zero. We need the endpoint derivative signs to show that neither endpoint can be a minimum. The corresponding statement for a maximum follows by reversing the inequalities.

Proof. Suppose first that \(q'_+(a)<0\). By the definition of the right derivative, for all sufficiently small \(h>0\),

Since \(h>0\), this gives \(q(a+h)<q(a)\). Thus \(a\) is not a point where the minimum on \([a,b]\) can occur. If \(q'_-(b)>0\), then for all sufficiently small \(h>0\),

Multiplying by \(-h<0\) reverses the inequality, so \(q(b-h)-q(b)<0\). Hence \(q(b-h)<q(b)\), and \(b\) cannot be a minimum either. Continuity on the compact interval \([a,b]\) ensures that \(q\) attains a minimum there. As neither endpoint can attain it, at least one minimum occurs at an interior point.

For the maximum statement, apply the same reasoning with all inequalities reversed. The condition \(q'_+(a)>0\) gives \(q(a+h)>q(a)\) for small \(h>0\), and \(q'_-(b)<0\) gives \(q(b-h)>q(b)\) for small \(h>0\). Neither endpoint can be a maximum. Continuity ensures that a maximum is attained on \([a,b]\), so one is attained in the interior. \(\square\)

At an interior minimum or maximum where \(q\) is differentiable, \(q'(c)=0\). Indeed, at a minimum the difference quotients for positive increments are nonnegative and those for negative increments are nonpositive. If their common limit exists, it must be both nonnegative and nonpositive, and therefore must be zero. At a maximum the inequalities are reversed, with the same conclusion. This is the interior-extremum condition used in the proof below.

Applying the Strategy to Darboux’s Theorem

We now give the proof strategy in full. Let \(I\) be an open interval, let \(f:I\to\mathbb{R}\) be differentiable, and take \(x_1,x_2\in I\) with \(x_1<x_2\). Suppose \(\lambda\) lies strictly between \(f'(x_1)\) and \(f'(x_2)\). We want to find \(c\in(x_1,x_2)\) with \(f'(c)=\lambda\). There are two possible orders for the endpoint derivatives; each determines which type of extremum to force.

Proof. Values equal to either endpoint derivative are already attained at \(x_1\) or \(x_2\), so it remains to consider a value \(\lambda\) strictly between them. Set \(a=x_1\), \(b=x_2\), and define

The function \(q\) is continuous on \([a,b]\) and differentiable there in the interior. Its derivative is \(q'(x)=f'(x)-\lambda\).

First suppose \(f'(a)<\lambda<f'(b)\). Then \(q'(a)<0\) and \(q'(b)>0\). The endpoint-sign extremum principle gives an interior minimum \(c\in(a,b)\) of \(q\). The interior-extremum condition yields \(q'(c)=0\). Substituting the formula for \(q'\), we obtain \(f'(c)-\lambda=0\), so \(f'(c)=\lambda\).

Now suppose \(f'(a)>\lambda>f'(b)\). In this case \(q'(a)>0\) and \(q'(b)<0\), so the endpoint-sign extremum principle gives an interior maximum \(c\in(a,b)\). Again \(q'(c)=0\), and hence \(f'(c)=\lambda\). Both possible orders have been covered, proving the result. \(\square\)

The essential move is the same in both cases: subtract the target-slope line, use endpoint signs to exclude endpoint extrema, and apply the zero-derivative condition at an interior extremum. This is a proof technique to remember, not just a calculation for one particular function.

What the Construction Adds—and What It Does Not

A common false start is to search for a maximum or minimum of \(f\) itself. An extremum of \(f\) would give \(f'(c)=0\), which is useful only when the target derivative value is zero. To prove that \(f'\) takes a general value \(\lambda\), the auxiliary function \(f(x)-\lambda x\) shifts the derivative by precisely that amount. Its stationary points correspond exactly to points where \(f'\) equals \(\lambda\).

The endpoint signs also matter. Merely knowing that \(q\) attains a minimum on a closed interval does not show that a zero of \(q'\) occurs in the interior: the minimum might be at an endpoint. The derivative signs rule out that possibility. In the increasing-order case, \(q\) decreases immediately to the right of the left endpoint and has values below its right-endpoint value just to the left of the right endpoint. In the decreasing-order case, the same reasoning applies to a maximum.

Finally, Darboux’s Theorem gives an intermediate value property, not continuity of the derivative. The argument locates a point for each target value between two derivative values; it does not show that nearby derivative values converge to a limit. The distinction is important: the conclusion is about which values must occur between inputs, not about how the derivative behaves in a neighborhood of each input.