Prove a Monotonicity Theorem

From Derivative Signs to Monotonicity

The Mean Value Theorem connects a function’s change between two points to its derivative at an intermediate point. This makes it a natural tool for proving that a function is monotone: if the derivative has a consistent sign, then the change between any two points has the corresponding sign. The hypotheses matter. In particular, continuity on the interval, including any endpoints that belong to it, cannot be omitted.

We will use the derivative sign criterion established earlier in the course: if a function is continuous on an interval and differentiable at every interior point, then a nonnegative derivative throughout the interior implies that the function is nondecreasing. Here we use the Mean Value Theorem to prove the strict version, establish the converse restriction that monotonicity places on the derivative, and clarify why strict increase does not require a strictly positive derivative everywhere.

These terms distinguish two different kinds of conclusions. A nondecreasing function may be constant on part of its domain, whereas a strictly increasing function cannot take the same value at two distinct points. We will assume the interval has at least two points when discussing derivative criteria; the order definitions themselves also apply to a one-point interval.

The Strict Derivative-Sign Theorem

The proof follows directly from the Mean Value Theorem. Choose any two points \(x<y\) in the interval and apply the theorem on \([x,y]\). The difference \(f(y)-f(x)\) is the derivative at an intermediate point multiplied by the positive interval length \(y-x\). A strictly positive derivative therefore forces a strictly positive change in the function.

Proof. Take any \(x,y\in I\) with \(x<y\). Since \(I\) is an interval, \([x,y]\subseteq I\). The function is continuous on \([x,y]\) and differentiable on \((x,y)\), so the Mean Value Theorem gives a \(c\in(x,y)\) such that

Here \(f'(c)>0\) by hypothesis, and \(y-x>0\) because \(x<y\). Therefore \(f(y)-f(x)>0\), which means \(f(x)<f(y)\). Since the pair \(x<y\) was arbitrary, \(f\) is strictly increasing on \(I\). \(\square\)

The endpoint cases are included in this argument. If \(x\) or \(y\) is an endpoint belonging to \(I\), continuity there is supplied by continuity of \(f\) on \(I\). Differentiability is needed only at the intermediate point \(c\), which lies strictly between \(x\) and \(y\).

Monotonicity Also Restricts the Derivative

The derivative-sign criterion gives a sufficient condition for nondecrease. The converse holds at every interior point where the derivative exists. The proof uses the definition of the derivative and checks both signs of the difference quotient. Checking only positive increments would not, by itself, justify taking the two-sided derivative limit.

Proof. Because \(a\) is an interior point, there is a \(\delta>0\) such that \(a+h\in I\) whenever \(|h|<\delta\). If \(0<h<\delta\), nondecrease gives \(f(a)\leq f(a+h)\). Dividing by \(h>0\) yields

If \(-\delta<h<0\), then \(a+h<a\), so nondecrease gives \(f(a+h)\leq f(a)\). Now the numerator \(f(a+h)-f(a)\) is nonpositive and the denominator \(h\) is negative. Consequently,

Thus every difference quotient for sufficiently small nonzero \(h\) is nonnegative. Since \(f'(a)\) exists, these quotients converge to \(f'(a)\) as \(h\to0\). The order-preservation property of limits gives \(f'(a)\geq0\), as claimed. \(\square\)

Together with the derivative sign criterion from earlier in the course, this gives a useful characterization: for a function continuous on an interval and differentiable at its interior points, nondecrease is equivalent to having a nonnegative derivative at every interior point. The forward direction is the theorem just proved; the reverse direction is the earlier criterion. The same reasoning applied to \(-f\) shows that a nonincreasing differentiable function has derivative at most zero at each interior point.

Zero Derivative and Flat Intervals

A derivative that is zero at a single point does not prevent strict increase, as the cubic example shows. A derivative that is zero throughout an entire interval has a stronger consequence: the function must be constant there. This follows by applying the Mean Value Theorem to every pair of points in that interval.

Proof. Take any \(x,y\in J\) with \(x<y\). The Mean Value Theorem applied on \([x,y]\) gives a \(c\in(x,y)\) such that

Thus \(f(y)=f(x)\). Since any two distinct points can be ordered in this way, the function has the same value at every point of \(J\); it is constant. \(\square\)

This result also gives a useful refinement of the nonnegative-derivative criterion. If \(f'\geq0\) on the interior of an interval, the earlier derivative sign criterion gives nondecrease. If the function is not constant on any nondegenerate subinterval, then it is in fact strictly increasing. Indeed, if \(x<y\) but \(f(x)=f(y)\), nondecrease would give

for every \(z\in[x,y]\). Both inequalities would have to be equalities, making \(f\) constant throughout \([x,y]\), contrary to the assumption. This is a way to prove strict increase even when the derivative has zeros.

Why Endpoint Continuity Cannot Be Dropped

A derivative condition concerns interior points, but monotonicity compares every pair of points in the interval, including included endpoints. Continuity on the interval is therefore a substantive hypothesis in the Mean Value Theorem argument. Differentiability throughout the interior alone does not control a separately assigned endpoint value.

For example, define \(q:[0,1]\to\mathbb R\) by \(q(0)=1\) and \(q(x)=x\) for \(x>0\). At every interior point \(x\in(0,1)\), the function agrees with the identity function in a neighborhood of \(x\), so \(q'(x)=1>0\). Nevertheless, \(q(0)=1>1/2=q(1/2)\), so \(q\) is not nondecreasing. The function is not continuous at the included endpoint \(0\), and the strict monotonicity theorem does not apply.

When using a derivative to infer monotonicity, first check continuity on the whole interval and differentiability at every interior point. Then apply the Mean Value Theorem to an arbitrary pair of points. To conclude strict increase from a nonnegative derivative, take care: nonnegativity alone gives only nondecrease. One must also rule out a constant subinterval or use another argument, as in the factorization for \(x^3\).