Why Strict Monotonicity Rules Out a Local Turn
A local maximum or minimum is a point where a function’s values, in a small neighborhood, stay on one side of the value at that point. Strict monotonicity rules out this behavior at an interior point: if a function is strictly increasing, values to the right are larger and values to the left are smaller. If it is strictly decreasing, the inequalities are reversed. In either case, nearby values occur both above and below the function value.
This conclusion is useful when classifying extrema. Once strict monotonicity is established on an interval, there is no need to test each interior point of that interval for a local extremum. The conclusion concerns interior points only; endpoints and points where the direction of change switches require separate consideration.
Definitions and Scope
Let \(I\) be an interval, and let \(f:I\to\mathbb{R}\). The function is strictly increasing on \(I\) if, whenever \(x_1,x_2\in I\) and \(x_1<x_2\), we have \(f(x_1)<f(x_2)\). It is strictly decreasing on \(I\) if, whenever \(x_1,x_2\in I\) and \(x_1<x_2\), we have \(f(x_1)>f(x_2)\). A function that is either strictly increasing or strictly decreasing on an interval is called strictly monotone there.
Local extrema are defined by comparisons with nearby points in the domain. Absolute extrema, by contrast, compare \(f(c)\) with the values at every point in the domain. The result below concerns local extrema and requires an interior point so that there are domain points on both sides.
The No-Interior-Extrema Theorem
The essential feature is the availability of points on both sides of an interior point. Strict increase forces the function value at a point to be greater than all nearby values on its left and less than all nearby values on its right. It therefore cannot be at least as large as all nearby values, or at most as large as all nearby values. Strict decrease gives the opposite ordering, with the same consequence.
Let \(c\) be any interior point of \(I\). We show that it cannot be either kind of local extremum.
Every proposed neighborhood of \(c\) contains a point \(y>c\). Strict increase gives \(f(y)>f(c)\), contradicting the inequality required for a local maximum.
Every proposed neighborhood contains a point \(x<c\). Strict increase gives \(f(x)<f(c)\), contradicting the inequality required for a local minimum.
If \(f\) is strictly decreasing, a nearby point to the right has a smaller value than \(f(c)\), and a nearby point to the left has a larger value. This contradicts the local-minimum and local-maximum conditions, respectively.
For completeness, the neighborhood points used in the proof can be chosen within any proposed local-extremum neighborhood. Since \(c\) is interior, there is an \(\eta>0\) such that \((c-\eta,c+\eta)\subset I\). Given any \(\delta>0\), choose \(0<\varepsilon<\min(\eta,\delta)\). Then \(c-\varepsilon\) and \(c+\varepsilon\) both lie in \(I\), are within \(\delta\) of \(c\), and lie on opposite sides of \(c\). Strict monotonicity compares their function values with \(f(c)\), so the required local-extremum inequality cannot hold.
Notice the scope of the proof: it uses only the order of inputs and outputs. Neither continuity nor a derivative is part of the theorem’s assumptions. A differentiable function may have a derivative equal to zero at an interior point and still be strictly monotone; the theorem still applies.
Worked Examples
Worked Example: A Strictly Increasing Cubic with a Zero Derivative
Show that \(f(x)=x^3\) has no interior local extrema on \(\mathbb{R}\), even though its derivative is zero at \(x=0\).
Take any \(u,v\in\mathbb{R}\) with \(u<v\). The difference in outputs factors as
Here \(v-u>0\). The other factor is also positive for \(u<v\), because
The sum of the two squared terms could be zero only if \(u=0\) and \(v=0\), which would contradict \(u<v\). Thus \(f(v)-f(u)>0\), so \(f\) is strictly increasing on \(\mathbb{R}\). By the theorem, it has no interior local maximum or minimum. As a check at the point in question, \(f'(x)=3x^2\), so \(f'(0)=0\); nevertheless, the strict-increase argument applies at \(0\) just as it does at every other real number.
Worked Example: A Strictly Decreasing Reciprocal Function
Show that \(f(x)=1/x\) has no interior local extrema on \((0,\infty)\), and check the direction of change using two inputs.
Take \(u,v\in(0,\infty)\) with \(u<v\). Since \(uv>0\),
Therefore \(f(u)>f(v)\), so \(f\) is strictly decreasing on \((0,\infty)\). Every point of this interval is interior, and the theorem shows that there are no interior local maxima or minima.
For a numerical check, \(1/4<1/2\), while
The smaller input gives the larger output, as required for a decreasing function. These values are checks of the general comparison, not a replacement for the argument for arbitrary \(u<v\).
Worked Example: A Turning Point Separates Two Monotone Intervals
Analyze \(f(x)=x^2\) on \(\mathbb{R}\), distinguishing what happens on each side of \(0\) from what happens at \(0\).
For negative inputs \(u<v<0\), the difference is
because \(v-u>0\) and \(v+u<0\). Thus \(f\) is strictly decreasing on \((-\infty,0)\), and has no local extrema at points in that interval. For positive inputs \(0<u<v\), the same factorization gives
because both factors are positive. Thus \(f\) is strictly increasing on \((0,\infty)\), and has no local extrema at points there.
At \(0\), however, \(f(0)=0\), and for every real \(x\),
So \(0\) is a local minimum (in fact, an absolute minimum). There is no contradiction: \(f\) is not strictly increasing or strictly decreasing on an interval containing both negative and positive inputs. For example, \(f(-1)=1\) and \(f(0)=0\) show a decrease, while \(f(0)=0\) and \(f(1)=1\) show an increase. The change in direction occurs at the point where the minimum is located.
Endpoints and the Meaning of “Interior”
The interior-point condition is essential. On a closed interval, a strictly monotone function may have extrema at its endpoints. For example, \(f(x)=x\) is strictly increasing on \([0,1]\). Its smallest value is \(f(0)=0\), and its largest value is \(f(1)=1\). The theorem excludes local extrema at interior points \(0<c<1\); it does not exclude endpoint extrema.
At an endpoint, only one side of the interval is available for comparison. The proof of the theorem depends on having nearby inputs both less than and greater than \(c\). This is why endpoint behavior must be considered separately when an optimization problem asks for extrema on a closed interval.
The same care is needed when a function is strictly monotone only on separate subintervals. The theorem can be applied on each such interval, but it says nothing by itself about a point where the direction of change switches. The square-function example illustrates this: each side of \(0\) is strictly monotone, while \(0\) is a local minimum.
A Common Mistake: Treating a Zero Derivative as an Extremum
A common error is to reason that if \(f'(c)=0\), then \(c\) must be a local maximum or minimum. A zero derivative alone does not establish either conclusion. For \(f(x)=x^3\), the calculation \(f'(0)=0\) is correct, but the claim that \(0\) is a local extremum is false.
Indeed, if \(h<0\), then \(h^3<0=f(0)\), while if \(h>0\), then \(h^3>0=f(0)\). Values near \(0\) therefore occur both below and above \(f(0)\). The function is strictly increasing on all of \(\mathbb{R}\), so the theorem rules out a local extremum at \(0\). A derivative condition may help identify points to investigate, but it must not be mistaken for a classification.
Using the Theorem Efficiently
When applying the result, make the interval and the direction of monotonicity explicit. A complete argument has three parts:
- Specify an interval on which the function is defined.
- Establish that it is strictly increasing or strictly decreasing on that interval.
- Apply the theorem only to interior points of that interval.
Strict increase and strict decrease each exclude both types of local extremum, not just one. The direction does, however, tell you how nearby values compare: a strictly increasing function has lower values to the left and higher values to the right; a strictly decreasing function has higher values to the left and lower values to the right. That ordering explains the theorem and can make a classification easier to interpret.
Check Your Understanding
Use the definitions and theorem to answer the following questions. State the interval you are considering whenever you apply strict monotonicity.
- If \(f\) is strictly decreasing on an open interval, can it have a strict local maximum at an interior point? Explain why.
- For \(f(x)=x^3\), why does \(f'(0)=0\) not imply a local extremum at \(0\)?
- Show that \(f(x)=1/x\) is strictly decreasing on \((0,\infty)\) by comparing \(f(u)\) and \(f(v)\) when \(0<u<v\). What does the theorem then imply?
- Why does strict increase on \([0,1]\) not rule out an absolute maximum at \(x=1\)?
- The function \(x^2\) is strictly monotone on each of \((-\infty,0)\) and \((0,\infty)\). Why does that not rule out a local minimum at \(0\)?
