Why Make a Candidate Table?
A function can have its absolute maximum or minimum at an endpoint, at a point where its derivative is zero, or at an interior point where its derivative does not exist. A candidate table keeps these possibilities together so that none is overlooked.
The method is especially useful when the question asks for the absolute extrema of a function on a closed interval. It does not require deciding whether each critical point is a local maximum or minimum. Instead, it compares the function values at every point that could contain an absolute extremum.
The Candidate-Table Method
The method applies most directly when \(f\) is continuous on \([a,b]\) and the candidate list is finite. Continuity matters: the Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval attains both an absolute maximum and an absolute minimum. The derivative helps identify where those attained extrema can occur.
The procedure is:
Confirm the domain includes the whole closed interval and that \(f\) is continuous there.
On \((a,b)\), solve \(f'(x)=0\) and find points where \(f'(x)\) does not exist, retaining only points where \(f\) is defined.
The endpoints \(a\) and \(b\) are candidates whether or not the derivative is zero there.
Compute \(f(x)\) at every candidate. The largest listed value is the absolute maximum value; the smallest is the absolute minimum value.
A candidate is not automatically an absolute extremum. It is a point that must be checked. The table makes the final decision by comparing the actual function values.
Why the Candidate List Is Complete
The key fact about interior extrema is Fermat’s theorem: if a function has a local maximum or minimum at an interior point and is differentiable there, its derivative at that point is zero. Here is the reason for the maximum case.
Suppose \(c\) is an interior local maximum and \(f\) is differentiable at \(c\). For sufficiently small positive \(h\), \(c+h\) remains near \(c\), and the local maximum condition gives \(f(c+h)\leq f(c)\). Thus
For sufficiently small negative \(h\), the numerator is still nonpositive, but the denominator is negative, so
Because \(f'(c)\) exists, the difference quotients from both sides approach the same limit. The right-hand quotients are nonpositive and the left-hand quotients are nonnegative, so their common limit must be zero. Therefore \(f'(c)=0\). The minimum case follows by reversing the inequalities.
This conclusion combines the Extreme Value Theorem with Fermat’s theorem. The Extreme Value Theorem ensures there are extrema to find. Fermat’s theorem rules out a differentiable interior extremum with a nonzero derivative. If an interior extremum is not differentiable, its point is still included as a candidate. The endpoints must be included separately because the two-sided derivative argument does not apply there.
Worked Example: A Polynomial with an Endpoint Tie
Worked Example: A Polynomial with an Endpoint Tie
Find the absolute extrema of \(f(x)=x^3-3x^2+1\) on \([0,3]\).
The polynomial is continuous on \([0,3]\), so its absolute extrema exist. Differentiate:
Set the derivative equal to zero:
The point \(x=0\) is an endpoint, and \(x=2\) is an interior critical point. Include the other endpoint \(x=3\). Evaluate each:
| Candidate \(x\) | Reason for inclusion | Value \(f(x)\) |
|---|---|---|
| \(0\) | Left endpoint | \(1\) |
| \(2\) | \(f'(2)=0\) | \(-3\) |
| \(3\) | Right endpoint | \(1\) |
The largest candidate value is \(1\), attained at both \(x=0\) and \(x=3\). The smallest is \(-3\), attained at \(x=2\). Therefore the absolute maximum value is \(1\), and the absolute minimum value is \(-3\). Checking the table confirms that no candidate has been left out: the endpoints and the only interior solution of \(f'(x)=0\) are all included.
Worked Example: A Nondifferentiable Interior Candidate
Worked Example: A Nondifferentiable Interior Candidate
Find the absolute extrema of \(g(x)=|x-1|\) on \([0,2]\).
The absolute-value function is continuous on the interval. For \(x<1\), \(g(x)=1-x\), so \(g'(x)=-1\). For \(x>1\), \(g(x)=x-1\), so \(g'(x)=1\). At \(x=1\), the left and right derivatives differ, so \(g'(1)\) does not exist. Thus the interior candidate is \(x=1\). Add the endpoints and evaluate:
| Candidate \(x\) | Reason for inclusion | Value \(g(x)\) |
|---|---|---|
| \(0\) | Left endpoint | \(1\) |
| \(1\) | \(g'(1)\) does not exist | \(0\) |
| \(2\) | Right endpoint | \(1\) |
The smallest value is \(0\), so the absolute minimum occurs at \(x=1\). The largest value is \(1\), attained at \(x=0\) and \(x=2\). As a check, the endpoint values match, while the interior value is zero because the absolute value is taken at its vertex.
Worked Example: A Rational Function
Worked Example: A Rational Function
Find the absolute extrema of \(h(x)=x+\frac{4}{x}\) on \([1,4]\).
Since \(x\) is never zero on this interval, \(h\) is continuous on \([1,4]\) and differentiable in its interior. Differentiate:
Solve \(h'(x)=0\):
Only \(x=2\) lies in the interior \((1,4)\). The endpoints are \(1\) and \(4\). Evaluate:
| Candidate \(x\) | Reason for inclusion | Value \(h(x)\) |
|---|---|---|
| \(1\) | Left endpoint | \(5\) |
| \(2\) | \(h'(2)=0\) | \(4\) |
| \(4\) | Right endpoint | \(5\) |
The absolute minimum value is \(4\), attained at \(x=2\). The absolute maximum value is \(5\), attained at both endpoints. The derivative check gives \(h'(2)=1-\frac{4}{2^2}=1-1=0\), confirming that the interior candidate satisfies the condition used to find it.
Common Pitfall: Leaving Out the Endpoints
A common error is to solve \(f'(x)=0\) and treat those solutions as the entire candidate list. Endpoints are not required to have derivative zero, so this can miss an absolute extremum.
Worked Example: Correcting an Incomplete Candidate List
Consider \(p(x)=x^2\) on \([-1,2]\). A mistaken approach finds \(p'(x)=2x\), solves \(2x=0\) to get \(x=0\), and then reports \(p(0)=0\) as the absolute maximum. This conclusion does not follow: it compares only one candidate and even labels its value incorrectly.
The correct list includes the endpoints \(-1\) and \(2\), as well as the interior critical point \(0\). Their values are
Thus the absolute minimum value is \(0\) at \(x=0\), and the absolute maximum value is \(4\) at \(x=2\). The correction is to include both endpoints before comparing values.
How to Read the Final Comparison
The table reports function values, not just candidate locations. To state an answer completely, give both the extremum value and where it occurs. If the largest or smallest value appears in multiple rows, the absolute extremum occurs at every corresponding candidate point.
The table method is a finite comparison only when the relevant candidate list is finite. If there are infinitely many critical points or nondifferentiable points, listing a few of them is not enough to justify the result; an additional argument is needed to control all possible candidates. For the usual finite-list problems, the logic is direct: continuity guarantees absolute extrema, the interior-extremum result places them among the candidates, and comparing every candidate value identifies them.
Check Your Understanding
For each question, focus on the candidate list and the reason the comparison is valid.
- Why must the two endpoints of a closed interval be included even when the derivative is nonzero there?
- What interior points should be included in a candidate table?
- For \(f(x)=x^2-4x\) on \([0,5]\), find the candidate points and their function values. Which are the absolute extrema?
- For \(q(x)=|x|\) on \([-2,1]\), why must \(x=0\) be included, and what are the absolute extrema?
- What does the Extreme Value Theorem contribute to the candidate-table method?
