Consequences of the Mean Value Theorem

Consequences Beyond Secant-Slope Estimates

The Mean Value Theorem relates a secant slope to a derivative somewhere between the endpoints. That relation yields several consequences that are not immediately visible from the theorem’s statement. In particular, derivatives have a striking property: between any two values they take, they also take every value in between. This remains true even when the derivative is not continuous.

We will prove this intermediate value property using the Mean Value Theorem, the Extreme Value Theorem, and Fermat’s Theorem. We will also establish a useful converse direction for monotonicity: if a function is nondecreasing and differentiable at an interior point, its derivative there cannot be negative. These results help describe what derivatives can do, including when they are discontinuous.

The Intermediate Value Property of Derivatives

A continuous function takes every intermediate value between any two of its values. Derivatives need not be continuous, but they still obey that same intermediate-value restriction. The theorem below is usually called Darboux’s Theorem.

Proof. Define \(g(x)=f(x)-\lambda x\) for \(x\in I\). By the difference rule and the derivative of the identity function, \(g'(x)=f'(x)-\lambda\). First suppose \(f'(x_1)<\lambda<f'(x_2)\). Then

Because \(g'(x_1)<0\), the definition of the derivative ensures that for all sufficiently small positive \(h\),

Since \(h>0\), this implies \(g(x_1+h)<g(x_1)\). Thus \(x_1\) cannot be a point where \(g\) attains its minimum on \([x_1,x_2]\). At the other endpoint, \(g'(x_2)>0\) ensures that for all sufficiently small negative \(h\),

Multiplying by \(h<0\) gives \(g(x_2+h)-g(x_2)<0\). Thus \(x_2\) cannot be a point where \(g\) attains its minimum on \([x_1,x_2]\), either.

The function \(g\) is continuous on \([x_1,x_2]\), since it is differentiable on \(I\). By the Extreme Value Theorem, it attains a minimum on this closed interval. Neither endpoint can be a point of minimum, so at least one point \(c\) where the minimum is attained belongs to \((x_1,x_2)\). Fermat’s Theorem gives \(g'(c)=0\). Therefore

Now suppose \(f'(x_2)<\lambda<f'(x_1)\). For the same function \(g\), we have \(g'(x_1)>0\) and \(g'(x_2)<0\). The derivative sign at \(x_1\) gives \(g(x_1+h)>g(x_1)\) for sufficiently small positive \(h\). The derivative sign at \(x_2\), using sufficiently small negative \(h\), gives \(g(x_2+h)>g(x_2)\). Neither endpoint can therefore be a point where \(g\) attains its maximum on \([x_1,x_2]\). The Extreme Value Theorem gives a point of maximum, which must lie in \((x_1,x_2)\). Fermat’s Theorem then gives \(g'(c)=0\), and hence \(f'(c)=\lambda\). This proves both cases. \(\square\)

The proof does not assume that \(f'\) is continuous. Instead, it adjusts \(f\) by subtracting the line \(x\mapsto\lambda x\). The assumed inequalities make the adjusted function slope downward at one endpoint and upward at the other, forcing it to have an interior minimum; with the inequalities reversed, it must have an interior maximum.

A Sign Consequence of Monotonicity

The earlier result “Nonnegative Derivative Implies Nondecreasing” gives one direction of a useful relationship between derivative signs and monotonicity. The reverse direction at each differentiability point follows directly from difference quotients. Unlike the first direction, this reverse implication does not require differentiability throughout the interval.

Proof. For every sufficiently small \(h>0\), both \(x\) and \(x+h\) belong to \(I\), and \(x<x+h\). Since \(f\) is nondecreasing, \(f(x+h)\geq f(x)\), so

For every sufficiently small \(h<0\), we have \(x+h<x\), and nondecreasingness gives \(f(x+h)\leq f(x)\). Both \(f(x+h)-f(x)\) and \(h\) are then nonpositive, with \(h<0\), so

Thus every sufficiently nearby difference quotient with \(h\ne0\) is nonnegative. Since \(f\) is differentiable at \(x\), these quotients converge to \(f'(x)\). A limit of nonnegative real numbers is nonnegative, so \(f'(x)\geq0\), as claimed. \(\square\)

Together with the earlier theorem, this gives a sign test on an open interval: a differentiable function is nondecreasing there if its derivative is everywhere nonnegative, and a nondecreasing function has nonnegative derivative wherever the derivative exists. The theorem also explains why a negative derivative at even one interior point rules out nondecreasingness.

Worked Examples

What These Consequences Do—and Do Not—Say

Darboux’s Theorem rules out a particular kind of discontinuity: a derivative cannot omit a whole interval of values between two values it attains. It does not say that every derivative is continuous. The oscillating example shows that a derivative can fail to be continuous dramatically while retaining the intermediate value property.

It is also important to distinguish assumptions in the monotonicity results. If a function is nondecreasing and differentiable at an interior point, the derivative-sign theorem gives \(f'(x)\geq0\) at that point. Conversely, the earlier “Nonnegative Derivative Implies Nondecreasing” theorem assumes differentiability throughout the open interval and a nonnegative derivative everywhere there. Neither statement allows the conclusion to be inferred from a derivative sign known at just one point.

When applying Darboux’s Theorem, check that both points lie in the interval where the function is differentiable and that the proposed value is strictly between the two derivative values. The theorem then guarantees an intermediate point, but it does not generally identify that point. Solving for it, as in the polynomial example, requires additional information about the function.