Why a Closed-Interval Check Must Include Nondifferentiable Points
Finding absolute extrema on a closed interval requires more than solving \(f'(x)=0\). A function can attain a highest or lowest value at an interior point where its derivative does not exist. A sharp cusp is one familiar example; a vertical tangent is another.
The candidate method is a way to make sure such points are not missed. It does not declare every candidate to be an extremum. Instead, it collects the points that might be extrema, evaluates the function at them, and compares the resulting values.
The Candidate Rule and Its Conditions
A point \(c\) in the domain of \(f\) is called a critical number if \(c\) is an interior point of the domain and either \(f'(c)=0\) or \(f'(c)\) does not exist. This tutorial focuses on the second possibility: an interior point where the function is defined but its derivative fails to exist.
A closed-interval candidate check uses critical numbers in \((a,b)\) together with the endpoints \(a\) and \(b\). Continuity on \([a,b]\) matters: by the Extreme Value Theorem, it guarantees that \(f\) attains both an absolute maximum and an absolute minimum somewhere on the interval. The candidate rule then identifies where those values can occur.
A nondifferentiable point is a candidate, not automatically an extremum. A cusp may be a minimum or maximum, while a vertical tangent can occur where the function is increasing straight through the point. The function values must still be compared, and the local behavior can help explain the result.
Why Interior Extrema Are on the Candidate List
The key fact is Fermat’s theorem: if a function has a local maximum or local minimum at an interior point and is differentiable there, its derivative must be zero. Consequently, any interior absolute extremum that does not have derivative zero must be a point where the derivative does not exist.
Here is the argument for a local maximum. For sufficiently small positive \(h\), \(f(c+h)\leq f(c)\), so the difference quotient \(\frac{f(c+h)-f(c)}{h}\) is at most zero. For sufficiently small negative \(h\), the numerator is still at most zero but the denominator is negative, so the quotient is at least zero. If the two-sided limit defining \(f'(c)\) exists, its value must be both at most zero and at least zero. It is therefore zero. For a local minimum, the inequalities reverse, and the same conclusion follows.
This argument explains why points where the derivative does not exist must not be discarded. Fermat’s theorem only forces the derivative to be zero when it exists. It makes no such claim when the derivative fails to exist.
For a continuous function on \([a,b]\), an absolute extremum exists. If it occurs at an endpoint, the endpoint is already on the list. If it occurs in the interior, then either the derivative exists and equals zero, or the derivative does not exist. These are exactly the interior cases covered by the candidate rule.
Recognizing Nondifferentiable Interior Candidates
A corner or cusp often produces different one-sided slopes. A vertical tangent can arise when the difference quotients become unbounded rather than approaching a finite derivative. In either case, the point belongs on the candidate list if it lies inside the interval and the function is defined there.
It is useful to distinguish where a function is defined from where it is differentiable. A point at which the function itself is undefined cannot be an extremum of that function. In the examples here, the nondifferentiable points are in the domain, and the functions remain continuous on their entire closed intervals.
Worked Example: A Cusp That Is an Absolute Minimum
Find the absolute extrema of \(f(x)=|x-1|\) on \([0,3]\). The function is continuous on this closed interval. Away from \(x=1\), write it as \(1-x\) for \(x<1\) and \(x-1\) for \(x>1\). Thus
At \(x=1\), the difference quotient from the left is \(-1\), while the difference quotient from the right is \(1\). Since these one-sided limits differ, \(f'(1)\) does not exist. There are no points where the derivative is zero, so the candidate list consists of \(0\), \(1\), and \(3\).
The smallest candidate value is \(0\), attained at \(x=1\); the largest is \(2\), attained at \(x=3\). Therefore the absolute minimum is \(f(1)=0\), and the absolute maximum is \(f(3)=2\). As a check, the formula \(|x-1|\) is nonnegative everywhere and equals zero at \(x=1\), while its endpoint values are \(1\) and \(2\), consistent with the comparison.
A Vertical Tangent Can Be a Candidate Without Being an Extremum
A vertical tangent is a useful reminder that a nondifferentiable point is only a possible location for an extremum. Consider \(f(x)=\sqrt[3]{x}\). At zero, the difference quotient is positive and grows without bound as \(h\) approaches zero, so there is no finite derivative there. Nevertheless, the function increases through zero rather than turning around.
Worked Example: A Vertical Tangent That Is Not an Extremum
Find the absolute extrema of \(f(x)=\sqrt[3]{x}\) on \([-8,8]\). The function is continuous on this closed interval. For \(x\neq 0\),
This derivative is positive wherever it exists and is never zero. At zero, the difference quotient is
As \(h\) approaches zero from either side, \(\frac{1}{|h|^{2/3}}\) grows without bound. Thus \(f'(0)\) does not exist as a finite real number. The graph has a vertical tangent there in the limiting-slope sense. The candidate list is \(-8\), \(0\), and \(8\).
Comparing the values gives an absolute minimum of \(-2\) at \(x=-8\) and an absolute maximum of \(2\) at \(x=8\). The candidate \(x=0\), with value \(0\), is neither. The increasing behavior on both sides of zero is consistent with this result. In particular, the failure of the derivative to exist does not by itself mean that the point is a local or absolute extremum.
A Cusp Can Be an Absolute Maximum
Nondifferentiable candidates can also be absolute maxima. The following example has a cusp at an interior point, and comparison of the candidate values identifies that cusp as the maximum.
Worked Example: A Cusp That Is an Absolute Maximum
Find the absolute extrema of \(f(x)=-|x|^{2/3}\) on \([-1,1]\). The function is continuous on the interval. For \(x<0\), it is \(-(-x)^{2/3}\), and for \(x>0\), it is \(-x^{2/3}\). Differentiating on each side gives
Neither expression is zero, and both become unbounded in magnitude as \(x\) approaches zero. More directly, for \(h\neq 0\),
The quotient has no finite limit at zero, so \(f'(0)\) does not exist. The candidates are \(-1\), \(0\), and \(1\), with values
The largest value is \(0\), attained at \(x=0\), and the smallest value is \(-1\), attained at both endpoints. Hence the absolute maximum is \(f(0)=0\), and the absolute minimum is \(-1\) at \(x=-1\) and \(x=1\). This agrees with \(f(x)=-|x|^{2/3}\leq 0\) throughout the interval.
A Reliable Candidate-Check Procedure
For a continuous function on a closed interval, use a consistent procedure. In particular, do not stop after solving \(f'(x)=0\): first identify where the derivative does not exist as well.
Confirm that \(f\) is continuous on the entire closed interval \([a,b]\). This ensures that absolute extrema exist.
Locate every \(x\in(a,b)\) where \(f'(x)=0\) and every point in \((a,b)\) where \(f'(x)\) does not exist.
Include \(a\) and \(b\), even when the derivative exists at them. The interior derivative test does not replace endpoint checks.
Compute \(f(x)\) at every candidate. The largest of these values is the absolute maximum value, and the smallest is the absolute minimum value.
If two or more candidates share the largest or smallest value, each corresponding point is an absolute extremum. Also, candidate lists need not always be finite: if a particular problem produces infinitely many candidates, the instruction to compare values still describes the mathematical requirement, though a finite table may not be a practical method.
Common Mistake: Solving Only \(f'(x)=0\)
A frequent error is to assume every interior extremum must occur where the derivative equals zero. That is true only if the derivative exists at the extremum. For \(f(x)=|x-1|\) on \([0,3]\), solving \(f'(x)=0\) produces no candidates at all: the derivative is \(-1\) on the left and \(1\) on the right. If one checks only the endpoints, one gets \(f(0)=1\) and \(f(3)=2\), and incorrectly reports \(1\) as the minimum.
The correction is to test where the derivative does not exist. At \(x=1\), the left-hand and right-hand slopes differ, so \(f'(1)\) does not exist. Including this candidate gives \(f(1)=0\), the actual absolute minimum. The error was not in the endpoint calculations; it was an incomplete candidate list.
Why the Distinction Matters
The candidate rule has two logically separate parts. The differentiable interior extrema are found through the condition \(f'(x)=0\), justified by Fermat’s theorem. The nondifferentiable interior extrema must be checked separately because that theorem does not apply to them. The endpoints are included because an extremum on a closed interval can occur there without satisfying an interior derivative condition.
After building the complete list, evaluate the function rather than relying only on a sketch or on a label such as “cusp.” A cusp can be a minimum or maximum depending on how the function behaves. A vertical tangent may be an extremum or may pass through a point without turning. The candidate comparison determines which values are globally largest and smallest on the interval.
Check Your Understanding
Use the closed-interval candidate rule to answer these questions.
- For a continuous function on \([a,b]\), which interior points belong on the candidate list?
- Why does Fermat’s theorem not rule out an extremum at a point where the derivative does not exist?
- For \(f(x)=|x-2|\) on \([0,4]\), list the candidates and compare their function values.
- For \(f(x)=\sqrt[3]{x}\) on \([-1,1]\), explain why zero is a candidate but not an absolute extremum.
- If two different candidates have the same smallest function value, what does that imply about the absolute minimum?
