Why Candidate Checks Need Both Calculus and a Calculator
A calculator can evaluate a function quickly at a list of inputs. But comparing values is useful only after the list is complete: an omitted endpoint or interior critical point could change the absolute maximum or minimum.
For a continuous function on a closed interval, the absolute extrema can occur at the endpoints or at interior points where the derivative is zero or does not exist. Calculus identifies those candidates; a calculator can then help evaluate and compare their function values. The numerical comparison is evidence about which candidate wins, while the calculus argument explains why the candidates cover all possibilities.
The Closed-Interval Candidate Rule
Suppose \(f\) is continuous on \([a,b]\), where \(a<b\). The Extreme Value Theorem guarantees that \(f\) attains both an absolute maximum and an absolute minimum on that interval. A candidate for either extremum is an endpoint, or an interior point where \(f'(x)=0\) or \(f'(x)\) does not exist.
This gives a practical sequence of tasks: confirm continuity on the whole interval, include both endpoints, find every interior critical number, and evaluate \(f\) at each candidate. If there are finitely many candidates, comparing their values identifies the absolute extrema. A tie means the extremum occurs at more than one candidate.
Why the Rule Works
The argument combines the Extreme Value Theorem with Fermat’s theorem. The Extreme Value Theorem says there is at least one point \(c\in[a,b]\) where the absolute maximum occurs, and at least one point where the absolute minimum occurs. Consider either such point.
If it is \(a\) or \(b\), it is an endpoint candidate. If it is inside the interval and \(f\) is not differentiable there, it is a critical-number candidate by definition. If it is inside the interval and \(f\) is differentiable there, it is a local extremum as well as an absolute extremum. Fermat’s theorem then gives \(f'(c)=0\). Thus every absolute extremum occurs among the stated candidates.
This argument proves that the candidate list is sufficient. It does not say that every candidate is an extremum; evaluation and comparison are still needed. Nor does it guarantee that there are only finitely many critical numbers. If there are infinitely many, a finite calculator table by itself cannot represent the whole candidate set.
A Calculator-Assisted Workflow
Record the closed interval and verify that the function is continuous throughout it, as required by the theorem.
Include both endpoints. Solve \(f'(x)=0\) and find interior points where \(f'\) does not exist, retaining only points in the interval where \(f\) is defined.
Use the derivative equation and the interval restrictions to justify that no interior critical number has been omitted. A graph or numerical solver can help locate candidates, but a search display alone may not certify completeness.
Enter each candidate into the original function, not the derivative. Keep sufficient calculator precision while comparing.
The largest candidate value is the absolute maximum value; the smallest is the absolute minimum value. State the input where each occurs, including every tie.
When a candidate is obtained numerically, it is an approximation. Keep the distinction clear between an exact candidate, such as \(\pi/6\), and a rounded decimal entered to represent it. Displayed calculator values are usually rounded approximations too. For a close comparison, retain more digits than the final report requires, and verify that rounding has not reversed the ordering.
Worked Example: Comparing Polynomial Candidates
Worked Example: Absolute Extrema of a Cubic
Find the absolute maximum and minimum of \(f(x)=x^3-3x\) on \([-2,2]\), and check that the candidate list is complete.
The function is a polynomial, so it is continuous on the entire closed interval. Its derivative is
Thus \(f'(x)=0\) at \(x=-1\) and \(x=1\). Both are inside \([-2,2]\). The derivative is defined everywhere, and the factored equation has no other roots, so the complete candidate list is \(-2,-1,1,2\).
Evaluate the original function at each candidate:
A calculator confirms these exact values. The largest is \(2\), attained at \(x=-1\) and \(x=2\); the smallest is \(-2\), attained at \(x=-2\) and \(x=1\).
Check by substitution: \(f(-1)=2\), \(f(2)=2\), \(f(-2)=-2\), and \(f(1)=-2\), matching the reported extrema and ties.
Worked Example: Trigonometric Candidates and Calculator Precision
Worked Example: A Trigonometric Function on \([0,2\pi]\)
Find the absolute extrema of \(g(x)=x+2\cos x\) on \([0,2\pi]\). Use radians.
The function is continuous on the closed interval. Differentiate and solve for interior critical numbers:
On \([0,2\pi]\), the solutions are \(x=\pi/6\) and \(x=5\pi/6\). These are all the solutions in the interval: in one full revolution, sine equals \(1/2\) at those two angles only. The derivative exists throughout, so the complete candidate list is \(0,\pi/6,5\pi/6,2\pi\).
Using the exact candidate inputs gives the following exact expressions and calculator approximations:
The decimals are approximations to evaluations at the exact inputs. If instead the calculator is given rounded decimal inputs for the angles, its outputs are approximate evaluations at those rounded inputs and need not equal the exact values. The largest candidate value is \(2\pi+2\approx8.283185\), at \(x=2\pi\). The smallest is \(5\pi/6-\sqrt3\approx0.885943\), at \(x=5\pi/6\).
Check the comparison with the exact expressions: \(2\pi+2>2\), and \(5\pi/6-\sqrt3<2\), while the other interior value is approximately \(2.255650\). The calculated ordering is consistent with all four candidate evaluations.
Worked Example: Include a Point Where the Derivative Does Not Exist
Worked Example: A Corner Point on a Closed Interval
Find the absolute extrema of \(h(x)=|x-1|+\frac{x}{5}\) on \([0,3]\).
The absolute-value function is continuous, so \(h\) is continuous on \([0,3]\). For \(x<1\), \(|x-1|=1-x\), and for \(x>1\), \(|x-1|=x-1\). Therefore,
At \(x=1\), the left-hand derivative is \(-4/5\) and the right-hand derivative is \(6/5\), so the derivative does not exist there. Neither derivative formula is zero on its interval. Thus the complete candidate list is the endpoints \(0,3\) and the interior critical number \(1\).
Evaluate each candidate:
A calculator comparison gives the same ordering. The absolute minimum is \(0.2\) at \(x=1\), and the absolute maximum is \(2.6\) at \(x=3\).
Check by substitution: the function values at the three candidates are \(1\), \(0.2\), and \(2.6\), respectively, so neither reported extremum comes from an omitted candidate in the complete list.
Common Pitfall: Comparing Only Interior Critical Points
A frequent error is to solve \(f'(x)=0\), compare only those function values, and declare the largest or smallest one to be an absolute extremum. The endpoints are candidates even when the derivative is nonzero there.
For instance, in the trigonometric example, comparing only the interior values gives approximately \(2.255650\) and \(0.885943\). One might incorrectly report the first as the absolute maximum. But the omitted endpoint has \(g(2\pi)=2\pi+2\approx8.283185\), which is larger. Including both endpoints corrects the comparison.
What to Show in a Free Response
A clear solution should make the reasoning visible, not just report calculator outputs. State the interval and continuity condition, show how the interior critical numbers were found, and explain why the list is complete. Then record the candidate inputs and corresponding function values, with approximate values identified as approximations. Finish by naming the absolute maximum and minimum values and the inputs where they occur.
If a calculator’s numerical solver supplies an approximate root of \(f'(x)=0\), do not treat the displayed root as proof that it found every root in the interval. Use the derivative equation, interval restrictions, and relevant algebraic or trigonometric reasoning to support completeness. Numerical output is especially useful for evaluating and comparing candidates once that reasoning is in place.
Check Your Understanding
For each question, distinguish the work a calculator can perform from the reasoning needed to justify the candidate list.
- What condition on \(f\) and \([a,b]\) ensures that absolute extrema exist?
- Which points must be checked when \(f\) is continuous on a closed interval?
- Why should an interior point where \(f'\) does not exist be considered?
- On \([0,2]\), suppose the only interior solution of \(f'(x)=0\) is \(x=1\). Which three function values should be compared?
- Why is a calculator’s numerical root display not, by itself, a proof that the candidate list is complete?
