Strict Monotonicity From the Derivative

When Weak Derivative Signs Give Strict Increase

A nonnegative derivative ensures that a function is nondecreasing: its values do not go down as the input increases. But nondecrease allows equal values at distinct inputs. To conclude strict increase, we need to rule out the possibility that the function stays flat across an entire interval. A derivative may vanish at individual points without causing that problem.

The previous tutorial established how derivative signs determine weak monotonicity. We will use the Nonnegative Derivative Implies Nondecreasing theorem, along with the Local Sign from a Nonzero Derivative result. The key idea is that if the derivative is nonnegative everywhere and positive somewhere between two inputs, then the function must make a strict increase somewhere between them. Nondecrease carries that strict gain across the whole pair of inputs.

The word “strictly” changes the conclusion at every pair of distinct ordered inputs: equality is no longer permitted. The derivative test for strict increase must therefore provide more than a nonnegative sign. It must also ensure that the function cannot remain flat throughout any interval.

A Criterion That Allows Zeros

Proof. Take any \(x,y\in I\) with \(x<y\). By the second hypothesis, there is a point \(c\in(x,y)\) such that \(f'(c)>0\). The Local Sign from a Nonzero Derivative result implies that, for all sufficiently small positive \(h\), \(f(c+h)>f(c)\). Choose such an \(h\) with \(c+h<y\); this is possible because \(c<y\). The Nonnegative Derivative Implies Nondecreasing theorem applies on \(I\), so

In particular, \(f(x)<f(y)\). Since this holds for every \(x<y\) in \(I\), the function is strictly increasing. \(\square\)

The condition that every subinterval contain a point where \(f'\) is positive can also be stated by saying that \(f'\) is not identically zero on any nonempty open subinterval of \(I\). Under the first hypothesis \(f'\geq0\), these statements are equivalent: if \(f'\) is not zero everywhere on an interval, then at some point it is nonzero, and nonnegativity makes that value positive.

The positive-derivative result established earlier in the course is the special case in which \(f'\) is positive at every point. The criterion here is more flexible: it permits zeros, provided they do not fill an entire subinterval. The zero set may contain particular points without creating a flat stretch.

Strict Increase and Flat Intervals

For a differentiable function with a nonnegative derivative, the obstruction to strict increase is exactly a subinterval on which the derivative vanishes everywhere. The following characterization makes this precise.

Proof. First suppose \(f\) is strictly increasing. Then it is nondecreasing. By the Derivative Sign for a Nondecreasing Function result, \(f'(x)\geq0\) at every \(x\in I\). Now suppose, for contradiction, that \(f'(x)=0\) throughout some nonempty open subinterval \(J\subseteq I\). The Zero Derivative Implies Constancy theorem says that \(f\) is constant on \(J\). That contradicts strict increase, since \(J\) contains distinct points with equal function values. Thus neither condition fails.

Conversely, suppose \(f'\geq0\) throughout \(I\) and is not identically zero on any nonempty open subinterval. Because \(f'\geq0\), every such subinterval contains a point where \(f'>0\). The Strict Increase from a Nonnegative Derivative theorem now implies that \(f\) is strictly increasing. This proves both directions. \(\square\)

The characterization separates two roles. The nonnegative sign rules out downward movement. The absence of an interval of zero derivative rules out a flat stretch. Neither condition can simply be omitted: nonnegative derivative alone permits plateaus, while a derivative that is sometimes positive but also negative does not meet the sign requirement.

What a Flat Portion Looks Like

A derivative that is nonnegative can still permit the function to be constant on an interval. Here is an example in which the derivative vanishes on two entire regions. Define \(g:\mathbb{R}\to\mathbb{R}\) by

The calculation at the joining points matters: a piecewise formula alone does not guarantee differentiability. Here both one-sided difference quotients agree at each join, so the derivative test applies on all of \(\mathbb{R}\). This example also shows why a strict-increase conclusion cannot be based only on \(g'\geq0\).

Strict Decrease and a Common Pitfall

There is a corresponding criterion for strict decrease. If \(f'(x)\leq0\) everywhere and every nonempty open subinterval contains some \(c\) with \(f'(c)<0\), then \(-f\) has nonnegative derivative and satisfies the strict-increase criterion. Thus \(-f\) is strictly increasing, which means that \(f\) is strictly decreasing. Equivalently, a nonpositive derivative gives strict decrease precisely when it does not vanish identically on any nonempty open subinterval.

A common mistake is to infer strict increase from \(f'(c)>0\) at just one point. That information gives a local increase near \(c\), but it says nothing by itself about the function on distant parts of its domain. The criterion requires a positive derivative somewhere inside every interval of inputs being compared, in addition to nonnegativity everywhere. Conversely, a zero derivative at one point does not rule out strict increase; the examples above show why the full pattern of derivative values matters.

In practice, check the sign of the derivative first. If it is nonnegative, locate its zeros and determine whether they fill any interval. Finitely many zeros, as in the polynomial example, cannot fill a nonempty interval. If the derivative vanishes throughout an interval, the Zero Derivative Implies Constancy theorem identifies the resulting flat portion and rules out strict increase on the whole domain.