Why Candidate Values Must Be Compared
An absolute maximum or minimum is a statement about the function values attained across an entire interval. Finding an interior critical number or checking an endpoint identifies a possible location, but neither step alone tells us which value is largest or smallest. The final task is to evaluate the function at every candidate and compare those results.
This comparison is especially important when several locations compete. An absolute maximum can occur at an endpoint, at an interior critical number, or at more than one point. The same is true of an absolute minimum. To give a complete answer, report both the extreme value and every input where it occurs.
The Closed-Interval Candidate Principle
For a continuous function on a closed, bounded interval, the Extreme Value Theorem guarantees that an absolute maximum and an absolute minimum are attained. The earlier material on generating candidates for closed-interval extrema and on why the candidate list is complete gives the relevant candidate principle: when \(f\) is differentiable at every interior point of \([a,b]\), any absolute extremum occurs either at an endpoint or at an interior critical number. Here an interior critical number is an \(x\in(a,b)\) for which \(f'(x)=0\).
The differentiability condition matters: it ensures that an interior extremum must be a critical number in the sense just defined. If a function is not differentiable at an interior point, that point may also need to be considered. The candidate principle stated here applies directly when the function is differentiable throughout the interval's interior.
Why does comparing the values work? The candidate principle guarantees that the actual absolute maximum and minimum are represented in the list. At the same time, each listed input belongs to \([a,b]\), so none of its function values can exceed the actual absolute maximum or fall below the actual absolute minimum. Thus the greatest listed value must equal the absolute maximum, and the least listed value must equal the absolute minimum. This reasoning also explains why a tie must be reported in full: every candidate with the greatest or least value is a point where that extremum is attained.
Include both endpoints and every interior critical number covered by the hypotheses.
Substitute each candidate input into \(f\), keeping the input and its function value distinct.
The largest is the absolute maximum value; the smallest is the absolute minimum value.
List every candidate input attaining each extreme value, including ties.
Worked Comparisons
Worked Example: A maximum and minimum each occur twice
Find the absolute maximum and minimum of \(f(x)=x^3-3x\) on \([-2,2]\).
First find the interior critical numbers by differentiating and solving \(f'(x)=0\):
Thus \(f'(x)=0\) when \(x=-1\) or \(x=1\). Both lie in \((-2,2)\). The complete candidate list is \(-2,-1,1,2\). Evaluate \(f\) at each one:
The greatest candidate value is \(2\), attained at \(x=-1\) and \(x=2\). The least is \(-2\), attained at \(x=-2\) and \(x=1\). Therefore the absolute maximum value is \(2\), and the absolute minimum value is \(-2\). Substitution into the original formula confirms each listed value.
Worked Example: Equal endpoint values
Find the absolute extrema of \(f(x)=x^2-2x\) on \([-1,3]\).
The derivative and its zero are
The critical number \(1\) is interior to the interval, so the candidates are \(-1,1,3\). Evaluate the function at all three:
The largest value is \(3\), attained at both endpoints, \(x=-1\) and \(x=3\). The smallest value is \(-1\), attained at \(x=1\). The values can be checked directly in \(x^2-2x\): substituting \(-1,1,3\) gives \(3,-1,3\), respectively. Notice that the interval's right endpoint is not the only location of the maximum, even though it is the larger endpoint input.
Worked Example: Several interior critical numbers
Find the absolute maximum and minimum of \(f(x)=x^4-4x^2+1\) on \([-2,2]\).
Differentiate and factor:
Solving \(f'(x)=0\) gives \(x=0\) or \(x^2=2\), so the interior critical numbers are \(0\), \(-\sqrt{2}\), and \(\sqrt{2}\). Include the endpoints to obtain the candidates \(-2,-\sqrt{2},0,\sqrt{2},2\). Their values are
The greatest value is \(1\), attained at \(x=-2,0,2\). The least value is \(-3\), attained at \(x=-\sqrt{2}\) and \(x=\sqrt{2}\). Each value follows by substitution into the original function, so the comparison includes the endpoints and all three interior critical numbers.
Common Pitfall: Comparing Inputs Instead of Values
The inputs \(x\) indicate where to evaluate the function; their numerical order does not determine the order of the function values. For example, in the second worked example, one might incorrectly argue: “The largest candidate input is \(x=3\), so the maximum occurs there.” This overlooks the candidate \(x=-1\), where \(f(-1)=3=f(3)\). The correct comparison is between the values \(3,-1,3\), not between the inputs \(-1,1,3\). It shows that the maximum is attained at both endpoints.
A related error is to report only the extreme value, without the points where it occurs, or to name only one point when several candidates tie. State the value and all its locations: for example, “the absolute maximum value is \(3\), attained at \(x=-1\) and \(x=3\).” The value is an output of the function; the points of attainment are inputs.
Check Your Understanding
Use the closed-interval candidate method and show the candidate values in each case.
- For a function continuous on \([a,b]\) and differentiable on \((a,b)\), what inputs belong in the candidate list?
- If two candidates have the same greatest function value, how should the absolute maximum be reported?
- For \(g(x)=x^2\) on \([-2,1]\), find the interior critical number, evaluate all candidates, and report both absolute extrema and their locations.
- Why is comparing candidate inputs alone insufficient to determine absolute extrema?
- Which hypothesis in the stated candidate theorem ensures that interior extrema are represented among the points where the derivative is zero?
