Local Comparisons and Global Comparisons
A function can rise and fall several times across its domain. At some points it may be higher than nearby values, and at others lower than nearby values. Those are local comparisons: they concern a small neighborhood around a point.
An absolute comparison is broader. It asks whether a function value is at least as large as, or at least as small as, every value the function takes anywhere in its domain. A point can be a local maximum without being an absolute maximum, and a function can have several points where the same absolute maximum or minimum is attained.
The domain matters throughout. When comparing values, only inputs belonging to the specified domain count. This also affects local extrema at endpoints: a neighborhood of an endpoint is considered only where it overlaps the domain.
Precise Definitions
An absolute maximum or minimum must be attained at a point in the domain. A value that the function approaches but never takes is not an absolute extremum.
The local definition restricts the comparison to inputs sufficiently near \(c\), while the absolute definition compares with every input in \(D\). These definitions use non-strict inequalities. If the inequalities are strict for all other relevant points, the extremum is called a strict local or strict absolute extremum, respectively.
An absolute maximum is automatically a local maximum: its comparison with every point of \(D\) includes the points in any sufficiently small neighborhood. The same reasoning applies to an absolute minimum. The reverse implication does not follow, because a point can beat all nearby values while losing to a value farther away.
Worked Example: Several Extrema on a Closed Interval
Worked Example: Several Extrema on a Closed Interval
Let \(f(x)=x^3-3x\) on \(D=[-2,2]\). Find its local and absolute extrema, including all points where the absolute extrema occur.
Differentiate:
The derivative is positive on \((-2,-1)\), negative on \((-1,1)\), and positive on \((1,2)\). Thus \(f\) increases, then decreases, then increases. This gives a local maximum at \(x=-1\) and a local minimum at \(x=1\). At the left endpoint, the function increases as we move into the domain, so \(x=-2\) is a local minimum relative to \(D\). At the right endpoint, the function increases as we approach it, so \(x=2\) is a local maximum relative to \(D\).
Compute the values at these points:
The monotonicity intervals show that no value on any of the three pieces exceeds the values at their adjacent endpoints. The four displayed values therefore give the largest and smallest values on the full domain. The absolute maximum is \(2\), attained at both \(x=-1\) and \(x=2\). The absolute minimum is \(-2\), attained at both \(x=-2\) and \(x=1\).
Check the classifications by substitution: \(f(-1)=f(2)=2\), and \(f(-2)=f(1)=-2\). The derivative signs confirm that the interior points \(-1\) and \(1\) are local extrema, while the direction of change into the domain confirms the endpoint classifications. In this example, the local extrema include all the absolute extrema, but each absolute value occurs at two different points.
When a Local Extremum Is Not Absolute
The endpoint of a domain can be a local extremum simply because nearby points are available on only one side. That does not make its value the greatest or least value on the whole domain.
Worked Example: A Local Minimum That Is Not Absolute
Consider \(g(x)=\sin x\) on \(D=[0,2\pi]\). Its derivative is \(g'(x)=\cos x\). It is positive on \((0,\pi/2)\), negative on \((\pi/2,3\pi/2)\), and positive on \((3\pi/2,2\pi)\). Consequently, \(x=\pi/2\) is a local maximum and \(x=3\pi/2\) is a local minimum.
The endpoints are local extrema relative to this domain as well. On the domain immediately to the right of \(0\), \(\sin x\geq 0=\sin 0\), so \(x=0\) is a local minimum. Immediately to the left of \(2\pi\), \(\sin x\leq 0=\sin(2\pi)\), so \(x=2\pi\) is a local maximum.
The values at these points are:
Since \(-1\leq\sin x\leq 1\) for every real \(x\), and the values \(1\) and \(-1\) are attained on this domain, the absolute maximum is \(1\) at \(x=\pi/2\), and the absolute minimum is \(-1\) at \(x=3\pi/2\). In particular, \(x=0\) is a local minimum but not an absolute minimum: \(g(0)=0\), whereas \(g(3\pi/2)=-1<0\). Similarly, \(x=2\pi\) is a local maximum but not an absolute maximum.
Check the comparisons: the sine bound gives \(g(x)\leq 1\) and \(g(x)\geq -1\) throughout \(D\), and substitution confirms equality at the stated points. This example shows why a neighborhood comparison alone cannot establish an absolute extremum.
Absolute Extrema May Not Exist
Having a domain and a formula does not guarantee that either absolute extremum exists. For instance, a function may increase across its domain while approaching a limiting value at an excluded endpoint. That value is not attained, so it cannot be an absolute maximum. The following example also shows that a function can have no local extrema at all.
Worked Example: No Local or Absolute Extrema on an Open Interval
Let \(p(x)=x\) on \(D=(0,1)\). First consider absolute extrema. For any \(x\in(0,1)\), the point \(x/2\) also belongs to \((0,1)\), and
Thus \(x\) cannot be an absolute minimum. Also, \((x+1)/2\in(0,1)\) and
so \(x\) cannot be an absolute maximum. Since this applies to every \(x\in D\), there is no absolute minimum or maximum.
To rule out local extrema, fix an arbitrary \(x\in(0,1)\) and an arbitrary neighborhood radius \(\delta>0\). Choose
Then \(x-\varepsilon\) and \(x+\varepsilon\) both lie in \((0,1)\), and both are within distance \(\delta\) of \(x\). Moreover,
So in every neighborhood of \(x\), there is a domain point with a smaller value and another with a larger value. Therefore \(x\) is neither a local maximum nor a local minimum. Because \(x\) and \(\delta\) were arbitrary, \(p\) has no local extrema anywhere in \((0,1)\).
As a final check, the function is strictly increasing, consistent with the comparisons above. Its values approach \(0\) and \(1\) near the two excluded endpoints, but neither value is attained on the domain.
A Reliable Way to Think About the Distinction
When a point is proposed as an extremum, first identify which comparison is being claimed. For a local extremum, the claim concerns all domain points within some positive distance of the point. For an absolute extremum, the claim concerns every point in the domain, no matter how far away.
A useful logical distinction follows directly from the definitions:
Show that the proposed point belongs to the domain and compare its value with the value at every other point in the domain.
It is enough to find one point in the domain with a larger or smaller value, as appropriate.
Find one positive radius for which the required inequality holds for every domain point inside that neighborhood.
Show that every positive-radius neighborhood contains domain points violating the local maximum or minimum comparison.
These are definitions-based checks, not promises that an extremum exists. Whether a function must attain its extreme values under particular domain and continuity assumptions is a separate question.
Common Mistake: Treating Local as Absolute
A common error is to see a local minimum and conclude that its value is the least value on the whole domain. In the sine example, the calculation “\(g(0)=0\), and \(0\) is a local minimum, so the absolute minimum is \(0\)” is wrong. The local conclusion only compares \(g(0)\) with nearby values in \([0,2\pi]\).
The correction is to make a domain-wide comparison. Since \(3\pi/2\in[0,2\pi]\) and
the value at \(0\) cannot be an absolute minimum. Indeed, the sine bound shows that \(-1\) is the absolute minimum value. A single comparison with a smaller value is enough to reject the proposed absolute minimum; proving that a value really is absolute requires ruling out all smaller values on the domain.
Why the Distinction Matters
Local extrema describe behavior near a point. They help identify where a function changes its direction of movement, and they are useful in analyzing graphs and optimization problems. Absolute extrema answer a different question: what is the greatest or least value available over the entire specified domain?
Keeping the two ideas separate prevents two opposite errors. A local extremum may not be globally best, as the endpoint minima of the sine example illustrate. Conversely, an absolute extremum must be attained, so an excluded endpoint or an unattained limiting value cannot qualify. Repeated values also matter: “the absolute maximum” may refer to the greatest value, while several distinct inputs attain that value.
Check Your Understanding
Answer each question using the domain in its statement.
- State the definition of an absolute maximum for a function \(f:D\to\mathbb{R}\).
- Why is every absolute minimum also a local minimum, but not every local minimum an absolute minimum?
- For \(f(x)=x^3-3x\) on \([-2,2]\), list all points where the absolute maximum and minimum are attained.
- For \(g(x)=\sin x\) on \([0,2\pi]\), why is \(x=0\) a local minimum but not an absolute minimum?
- For \(p(x)=x\) on \((0,1)\), explain why no \(x\) in the domain can be a local extremum. Your argument should work for every positive neighborhood radius.
- Can an absolute maximum be attained at more than one point? Give an example from this tutorial.
