Monotonicity From the Derivative

Reading Global Behavior from Local Derivatives

The previous tutorial showed how a derivative that vanishes throughout an interval forces constancy. A broader principle is that a derivative with one weak sign throughout an interval controls the direction of change: nonnegative derivatives give nondecreasing functions, and nonpositive derivatives give nonincreasing functions. The conclusion is weak monotonicity; a derivative that is sometimes zero may allow flat portions, so it does not by itself guarantee strict increase or decrease.

We will use the Nonnegative Derivative Implies Nondecreasing theorem established earlier in this course. We will also use the Mean Value Theorem and Darboux’s Theorem for Derivatives. These results let us move from information about derivative values to comparisons between function values, and they help us handle sign changes and isolated exceptions without assuming that the derivative is continuous.

Definition: Let \(I\) be an interval. A function \(f:I\to\mathbb{R}\) is nondecreasing on \(I\) if \(f(x)\leq f(y)\) whenever \(x,y\in I\) and \(x<y\). It is nonincreasing on \(I\) if \(f(x)\geq f(y)\) whenever \(x<y\). A function with either property is called monotone.

The order of the inequalities matters. Nondecreasing means that values never go down as the input increases; it does not mean that the values must go up at every step. Similarly, nonincreasing allows equal values. The derivative tests below determine these weak forms of monotonicity.

One-Signed Derivatives

The Nonnegative Derivative Implies Nondecreasing theorem says that if \(I\) is an open interval, \(f\) is differentiable on \(I\), and \(f'(x)\geq0\) for all \(x\in I\), then \(f\) is nondecreasing on \(I\). The nonpositive version follows by applying that theorem to \(-f\): if \(f'(x)\leq0\), then \((-f)'(x)=-f'(x)\geq0\), so \(-f\) is nondecreasing and \(f\) is nonincreasing.

A useful point is that the derivative need not be strictly positive everywhere to give nondecrease. Its value may be zero at some points, or even throughout a subinterval. What the theorem rules out is a net downward change between any two ordered points. Strict conclusions require additional hypotheses, which will be considered in the next tutorial.

Worked Example: A Polynomial That Decreases and Then Increases

Let \(f(x)=x^4-4x^2\) on \(\mathbb{R}\). The power rule gives

f'(x)=4x^3-8x=4x(x^2-2)=4x(x-\sqrt{2})(x+\sqrt{2}).

The derivative can change sign only at \(-\sqrt{2}\), \(0\), and \(\sqrt{2}\). Its factors show that \(f'(x)<0\) on \((-\infty,-\sqrt{2})\), \(f'(x)>0\) on \((-\sqrt{2},0)\), \(f'(x)<0\) on \((0,\sqrt{2})\), and \(f'(x)>0\) on \((\sqrt{2},\infty)\). Therefore \(f\) is nonincreasing on the first and third intervals, and nondecreasing on the second and fourth.

The function values at the sign-change points are

f(-\sqrt{2})=4-8=-4,\qquad f(0)=0,\qquad f(\sqrt{2})=4-8=-4.

Thus the sign information indicates a local minimum at each of \(-\sqrt{2}\) and \(\sqrt{2}\), and a local maximum at \(0\). It does not make \(f\) monotone on all of \(\mathbb{R}\): it decreases on one portion of its domain and increases on another. A derivative sign chart describes monotonicity interval by interval, not automatically across the entire domain.

Worked Example: A Derivative That Vanishes at a Turning Point

Define \(g(x)=(x-3)^4+2\). Then

$$ g'(x)=4(x-3)^3. $$

For \(x<3\), the cube \((x-3)^3\) is negative, so \(g'(x)<0\). For \(x>3\), it is positive, so \(g'(x)>0\), and \(g'(3)=0\). It follows that \(g\) is nonincreasing to the left of \(3\) and nondecreasing to the right. Direct substitution gives \(g(3)=2\); for example, \(g(2)=3\) and \(g(4)=3\). The derivative’s sign change supports the conclusion that \(g\) has a local minimum at \(3\).

The zero derivative at \(3\) is not a difficulty: the sign conditions concern the intervals on either side, and the function is continuous at \(3\) because it is a polynomial. The next theorem makes the connection between such one-sided signs and a local extremum precise.

Sign Changes and Local Extrema

Theorem (Derivative Sign Change Gives a Local Extremum): Let \(I\) be an open interval, let \(a\in I\), and let \(f:I\to\mathbb{R}\) be differentiable. If \(f'(x)\leq0\) for every \(x<a\) in \(I\), and \(f'(x)\geq0\) for every \(x>a\) in \(I\), then \(f(a)\leq f(x)\) for all \(x\) sufficiently close to \(a\). Thus \(f\) has a local minimum at \(a\). If instead \(f'(x)\geq0\) to the left of \(a\) and \(f'(x)\leq0\) to the right, then \(f(a)\geq f(x)\) for all \(x\) sufficiently close to \(a\), so \(f\) has a local maximum at \(a\).

Proof. Choose \(r>0\) such that \((a-r,a+r)\subseteq I\). Fix \(x\in(a-r,a)\). The function is continuous on \([x,a]\), because differentiability implies continuity at every point of \(I\), and it is differentiable on \((x,a)\). By the Mean Value Theorem, there is \(c\in(x,a)\) such that

$$ f(a)-f(x)=f'(c)(a-x). $$

Here \(f'(c)\leq0\) by the assumed sign to the left of \(a\), and \(a-x>0\). Hence \(f(a)-f(x)\leq0\), or \(f(a)\leq f(x)\). Now fix \(x\in(a,a+r)\). The Mean Value Theorem on \([a,x]\) gives some \(c\in(a,x)\) such that

$$ f(x)-f(a)=f'(c)(x-a). $$

In this case \(f'(c)\geq0\) and \(x-a>0\), so \(f(x)-f(a)\geq0\), again giving \(f(a)\leq f(x)\). These two comparisons hold for every \(x\in(a-r,a+r)\), including \(x=a\) by equality. This proves the local minimum claim. For the other sign pattern, apply the result just proved to \(-f\): its derivative is nonpositive to the left of \(a\) and nonnegative to the right, so \(-f\) has a local minimum at \(a\). Equivalently, \(f\) has a local maximum there. \(\square\)

This theorem uses weak derivative signs, so the local extremum need not be unique, and the function need not be strictly monotone on either side. For instance, a function may be constant on a portion next to \(a\). A change from nonpositive to nonnegative derivative guarantees a local minimum, but it does not guarantee that the derivative is nonzero at \(a\); nor is a zero derivative at \(a\) alone enough to establish an extremum.

Worked Example: A Local Maximum from the Opposite Sign Pattern

Consider \(h(x)=-(x+1)^4+5\). Differentiating gives

$$ h'(x)=-4(x+1)^3. $$

If \(x<-1\), then \((x+1)^3<0\), so \(h'(x)>0\). If \(x>-1\), then \((x+1)^3>0\), so \(h'(x)<0\). The local-extremum theorem therefore gives a local maximum at \(-1\). In fact, \(h(-1)=5\); for the test points \(-2\) and \(0\), the calculation gives \(h(-2)=-( -1)^4+5=4\) and \(h(0)=-(1)^4+5=4\). Both nearby values are below \(h(-1)\), as the derivative-sign argument predicts.

Finitely Many Exceptional Points

Sometimes a derivative sign has been checked everywhere except at finitely many points. For a derivative, Darboux’s Theorem for Derivatives can bridge this gap: derivatives have the intermediate value property, even when they are not continuous. Consequently, a derivative cannot be negative at one isolated exceptional point while remaining nonnegative at every nearby nonexceptional point.

Theorem (Finite-Exception Monotonicity): Let \(I\) be an open interval, and let \(f:I\to\mathbb{R}\) be differentiable. Suppose there is a finite set \(E\subseteq I\) such that \(f'(x)\geq0\) for every \(x\in I\setminus E\). Then \(f\) is nondecreasing on \(I\).

Proof. We first show that \(f'(c)\geq0\) at every \(c\in E\). Suppose, to the contrary, that \(f'(c)<0\) for some \(c\in E\). Because \(E\) is finite and \(I\) is open, we can choose a neighborhood of \(c\) contained in \(I\) that contains no other point of \(E\). Choose \(y\neq c\) in that neighborhood. Then \(y\notin E\), so \(f'(y)\geq0\).

The number \(f'(c)/2\) lies strictly between \(f'(c)\) and \(f'(y)\): it is greater than the negative number \(f'(c)\), and it is still negative, hence less than or equal to \(f'(y)\). By Darboux’s Theorem for Derivatives, some point \(z\) strictly between \(c\) and \(y\) satisfies \(f'(z)=f'(c)/2<0\). Our choice of neighborhood ensures \(z\notin E\), contradicting the assumption that \(f'\geq0\) outside \(E\). Thus \(f'(c)\geq0\) for every \(c\in E\). We have now shown that \(f'(x)\geq0\) throughout \(I\). The Nonnegative Derivative Implies Nondecreasing theorem applies and shows that \(f\) is nondecreasing on \(I\). \(\square\)

The finiteness condition matters in this proof: it lets us isolate any exceptional point from all the others. The conclusion is useful when a sign calculation establishes nonnegativity away from a finite list of points, such as roots of a polynomial factor. The argument does not claim that a derivative is continuous; it uses the intermediate value property that derivatives possess.

Worked Example: Checking a Sign Away from One Point

Let \(p(x)=x^4\). Its derivative is \(p'(x)=4x^3\), so this particular function does not have a nonnegative derivative away from zero: \(p'(x)<0\) for \(x<0\). This check illustrates why the finite-exception theorem cannot be applied merely because a derivative vanishes at an exceptional point; its sign condition must hold at every nonexceptional point.

Instead take \(q(x)=x^4+4x\). Then \(q'(x)=4x^3+4=4(x^3+1)\), which is negative for \(x<-1\), so it also fails the required condition on all of \(\mathbb{R}\). To use the theorem correctly, let \(r(x)=x^4\) on \([0,\infty)\) would introduce an endpoint and a different domain issue. A clean example on an open interval is \(s(x)=x^2\) on \((0,\infty)\): its derivative is \(s'(x)=2x>0\) everywhere, so no exception is needed.

For a genuine finite exceptional set, consider \(t(x)=x^3+3x\) on \(\mathbb{R}\). Its derivative \(t'(x)=3x^2+3\) is positive everywhere, so again the conclusion follows directly from the one-signed derivative theorem. This highlights the logical role of the finite-exception result: it is a way to complete a sign argument when there are exceptional points, not a substitute for verifying the sign on the rest of the domain.

Using a Derivative Sign Chart Carefully

For a function whose derivative factors, a sign chart is a compact way to organize the argument. Find the points where the derivative is zero or undefined, divide the domain into intervals at those points, and determine the derivative’s sign on each resulting interval. On every interval where the derivative is nonnegative, the function is nondecreasing; where it is nonpositive, the function is nonincreasing. If the signs on the two sides of a point oppose one another in the appropriate order, the local-extremum theorem may then apply.

A sign chart does not by itself say that the function is monotone on the union of intervals with different derivative signs. The polynomial in the first example decreases, then increases, then decreases, then increases. Nor does a derivative sign change automatically imply strict monotonicity: the sign could be zero throughout a subinterval. These are separate questions from whether a weak monotonicity conclusion follows.

The main discipline is to match each conclusion to its hypotheses. A derivative that is nonnegative throughout an open interval gives nondecrease there. A nonpositive derivative gives nonincrease. Opposite signs on either side of a point give a local extremum under the stated differentiability assumptions. If finitely many points were omitted from a nonnegative-sign check, Darboux’s theorem can sometimes fill those gaps, but only after the sign condition has been verified everywhere else.

Check Your Understanding

Use derivative signs and the results in this tutorial to answer the following.

If \(f'(x)\leq0\) throughout an open interval, what monotonicity conclusion follows, and how can it be obtained from the nonnegative-derivative theorem?
Suppose \(f'\) is nonpositive to the left of \(a\) and nonnegative to the right. What local conclusion follows, and which theorem connects the derivative signs to function values?
Why does a derivative sign chart with different signs on separate intervals not automatically prove monotonicity on their union?
In the finite-exception theorem, where is Darboux’s Theorem for Derivatives used?
Does a nonnegative derivative necessarily imply strict increase? Explain briefly.