When a Derivative Vanishes on an Interval
A point where \(f'(x)=0\) is a critical point when \(f\) is defined there and differentiable there. But a derivative can be zero at many points in a row, not just at one isolated point. When it is zero throughout an interval, the function does not merely have a collection of stationary points: it is constant across that interval.
An interval on which a function has a constant value is often called a flat interval or a plateau. Understanding a plateau matters when classifying local extrema. In particular, its interior points behave differently from an isolated critical point, and its endpoints may have different classifications from its interior.
Why a Zero Derivative on an Interval Forces Constancy
The Mean Value Theorem gives the reason. For any two distinct points \(x_1,x_2\in[a,b]\), the function is continuous on the interval between them and differentiable in its interior. The theorem guarantees a point \(c\) between \(x_1\) and \(x_2\) such that the average rate of change equals the instantaneous rate of change at \(c\). Since \(f'(c)=0\), the average rate of change must also be zero.
Because \(x_2-x_1\ne 0\), this implies \(f(x_2)-f(x_1)=0\), so \(f(x_2)=f(x_1)\). The two points were arbitrary, which means every pair of points in the interval has the same function value. That is exactly what it means for \(f\) to be constant there.
The hypotheses matter. The Mean Value Theorem requires continuity on the closed interval and differentiability in its interior. For an open interval, the same conclusion follows by applying the theorem to any two points chosen inside it: the closed interval between those points lies within the original interval.
Worked Example: Confirming a Flat Interval and Classifying Its Points
Worked Example: A Plateau of Global Minimum Values
Consider the function defined on all real numbers by
On the open interval \((0,2)\), the function is \(f(x)=0\), so \(f'(x)=0\) there. The function is continuous at both joins: each outside expression approaches \(0\), matching the value on the plateau.
It is also differentiable at the joins. At \(x=0\), the difference quotient from the left is \(\frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h\), which approaches \(0\) as \(h\to0^-\); from the right it is \(0\). Thus \(f'(0)=0\). At \(x=2\), the quotient from the left is \(0\), and from the right it is \(\frac{f(2+h)-f(2)}{h}=\frac{h^2}{h}=h\), which approaches \(0\). Thus \(f'(2)=0\) as well.
For \(x<0\), \(x^2>0\), and for \(x>2\), \((x-2)^2>0\). On \([0,2]\), \(f(x)=0\). Therefore \(f(x)\ge 0\) for all \(x\), and every point in \([0,2]\) is a global minimum point.
Using the non-strict definition of local extrema, every interior point \(c\in(0,2)\) is also a local maximum: a sufficiently small neighborhood around \(c\) stays on the plateau, so \(f(x)=f(c)\) throughout it. The same points are local minima. The endpoints \(0\) and \(2\) are local minima, but not local maxima: arbitrarily close points outside the plateau have positive function values. None of these minimum points is a strict local minimum, because nearby points on the plateau have the same value.
Check: The computed derivatives on the plateau and at its joins are all \(0\), and the piecewise values confirm that the plateau value is \(0\), the global minimum.
Classifying Points on a Plateau
A local-extremum classification compares \(f(c)\) with function values at nearby points in the domain. A point \(c\) is a local maximum if \(f(x)\le f(c)\) for all domain points sufficiently close to \(c\). It is a local minimum if \(f(x)\ge f(c)\) for all such points. If the inequalities are strict for every nearby \(x\ne c\), the extremum is strict.
Every interior point of a flat interval is both a non-strict local maximum and a non-strict local minimum. Indeed, each has a small neighborhood contained entirely within that interval, where all function values are equal. But those points are not strict local extrema: equality with neighboring points rules out the required strict inequality.
The endpoints need separate attention. A neighborhood of an endpoint of a plateau also includes points outside the plateau, and their values determine which local-extremum comparisons hold. An endpoint may be a local maximum, a local minimum, both, or neither. The equation \(f'(c)=0\) alone does not decide the classification.
Worked Example: A Rising Function with a Flat Interval
Worked Example: Different Classifications at the Two Plateau Endpoints
Define
For \(x<0\), \(g'(x)=3x^2>0\). On \((0,2)\), \(g'(x)=0\). For \(x>2\), \(g'(x)=3(x-2)^2>0\). At \(x=0\), the difference quotient from the left is \(h^3/h=h^2\to0\), while from the right it is \(0\), so \(g'(0)=0\). At \(x=2\), the quotient from the left is \(0\), and from the right it is \(h^3/h=h^2\to0\), so \(g'(2)=0\).
For \(x<0\), \(g(x)<0\), while plateau values are \(0\). Thus \(x=0\) is a local maximum: nearby values are either below \(0\) or equal to \(0\). It is not a local minimum, because values immediately to its left are less than \(g(0)\). For \(x>2\), \(g(x)>0\), while plateau values are \(0\). Thus \(x=2\) is a local minimum, but not a local maximum.
Every point in \((0,2)\) is both a non-strict local maximum and a non-strict local minimum, because a small enough neighborhood lies entirely on the plateau. Neither classification is strict. At the endpoints, the adjacent outside values decide which comparison holds.
Check: The left-hand values near \(0\) are negative and the right-hand values near \(2\) are positive, confirming the maximum at \(0\) and minimum at \(2\).
Using Derivative Signs Around a Flat Interval
Derivative signs can help describe how a function approaches or leaves a plateau, but the plateau itself has derivative zero and no increasing or decreasing behavior within it. When examining an endpoint, compare values on the outside with the plateau value. For instance, if the function rises toward a plateau, then stays level, the entry point can be a local maximum; if it leaves the plateau by rising, the exit point can be a local minimum. These conclusions depend on the function’s behavior on both sides of the point, not simply on the derivative value at the point.
Identify an interval on which \(f'(x)=0\) and verify that the function satisfies the conditions needed to apply the Mean Value Theorem.
Apply the theorem between arbitrary points of that interval to show that their function values are equal.
Every interior point has a neighborhood of equal values, making it both a non-strict local maximum and a non-strict local minimum, but neither a strict extremum.
Compare nearby values outside the plateau with its constant value; the endpoint’s derivative being zero does not settle the classification.
Worked Example: A Zero Derivative at One Point Is Not a Plateau
Worked Example: An Isolated Critical Point
Let \(h(x)=x^2\). Then \(h'(x)=2x\), so \(h'(0)=0\), but \(h'(x)\ne0\) for every \(x\ne0\). The derivative is not zero throughout any interval around \(0\).
For example, \(h(0)=0\) and \(h(1)=1\), so the function is not constant between \(0\) and \(1\). In fact, \(h(x)>0=h(0)\) for every \(x\ne0\), which shows that \(0\) is a strict local minimum (and also a global minimum).
Check: Differentiating \(x^2\) gives \(2x\), and substituting \(x=0\) gives \(0\). The distinct values \(h(0)=0\) and \(h(1)=1\) confirm that a zero derivative at a single point does not make the function constant.
A Common Mistake
A tempting but incorrect conclusion is: “Since \(f'(0)=0\), the function must be constant near \(0\).” For \(f(x)=x^2\), that reasoning would say the function is constant near \(0\), even though \(f(0)=0\) and \(f(1)=1\). The correction is to distinguish a derivative that is zero at a point from one that is zero throughout an interval. The Mean Value Theorem proves constancy only in the interval case, with its continuity and differentiability conditions satisfied.
A second error is to call every point on a plateau a strict local maximum and minimum. Interior points do meet the non-strict local-extremum inequalities, but equal values at nearby distinct points prevent either strict classification. Plateau endpoints must also be checked against values outside the flat interval.
Why Flat Intervals Matter
A derivative sign analysis is not complete if it treats every zero of the derivative as an isolated point. On a plateau, there are infinitely many derivative zeros, and the function has a whole interval of equal values. Recognizing this structure prevents misleading classifications and helps distinguish what is true at every interior point from what must be checked separately at the edges.
Check Your Understanding
Answer each question using the distinction between a zero derivative at a point and a zero derivative throughout an interval.
- State the continuity and differentiability conditions that allow the Mean Value Theorem to establish constancy on \([a,b]\) when \(f'(x)=0\) on \((a,b)\).
- Why is every interior point of a flat interval both a non-strict local maximum and a non-strict local minimum?
- Can a point at the edge of a plateau be a local maximum but not a local minimum? Explain what nearby values would make this happen.
- For \(f(x)=x^2\), explain why \(f'(0)=0\) does not imply that \(f\) is constant near \(0\).
- Why are points in the interior of a plateau not strict local extrema?
