A critical point is not an extremum: $f'(c)=0$ is a candidate, not a verdict

01The mistake

Students solve $f'(x)=0$, find the critical points, and label them as maxima or minima without testing. For $f(x)=x^3$, the derivative $3x^2$ is zero at $x=0$, and students report an extremum. There is none — the function is increasing on both sides and the origin is an inflection point with a horizontal tangent.

The tell is asking for justification rather than for an answer. On a well-behaved function the untested answer is usually right, so the error is invisible in the value. Ask why it is a maximum and a student who skipped the test has nothing to say beyond “the derivative is zero,” which is exactly the answer that does not distinguish the cases. The exam scores this justification separately for the same reason.

The second half of the misconception is missing critical points entirely (U5-CA5). Critical points also occur where $f'$ is undefined, not only where it is zero. For $f(x)=|x|$ the minimum is at 0, where the derivative does not exist — a student solving $f'(x)=0$ finds nothing and concludes there is no extremum.

And endpoints are routinely dropped in optimisation problems. On a closed interval the maximum may occur at an endpoint where the derivative is nowhere near zero. Students who equate “candidates” with “solutions of $f'=0$” will not check them, which is why so many optimisation answers are correct in method and wrong in value.

02Why it makes sense to the student

The picture we draw is always a hill or a valley. Every diagram introducing critical points shows a smooth peak with a horizontal tangent, and the tangent line is drawn precisely because it is flat at the top. Students form the association from the illustration, and the illustration only ever shows the cases where it holds.

In practice it is usually true. Most textbook functions have extrema at their critical points, so the shortcut is confirmed dozens of times. $x^3$ appears once as a curiosity rather than as the reason the test exists.

“Critical” is a loaded word. It suggests importance, a decisive location, something that has been determined — not a candidate awaiting a check. If the term were “candidate point,” a good deal of this would not happen.

And the tests are taught as extra steps rather than as the actual question. The first derivative test arrives after critical points have already been found and labelled in examples, so it reads as verification of something already known rather than as the step that determines the answer.

03The correction

Rename what $f'(c)=0$ gives you, out loud and repeatedly: a candidate. Finding critical points narrows the search; it does not settle it. Students who use the word candidate stop treating the list as an answer.

Then make the criterion the sign change, not the zero. The first derivative test: if $f'$ goes positive to negative at $c$, it is a local maximum; negative to positive, a local minimum; no sign change, neither. That third clause is the one that gets omitted, and it is the one $x^3$ requires.

Work $x^3$ at the origin every time you teach this. $f'(x) = 3x^2$ is zero at 0 and positive on both sides, so the function is increasing through the point. Horizontal tangent, no extremum. It is one line of work and it is the counterexample the entire topic is built around.

Give the complete candidate list explicitly, because two of the three sources get forgotten: points where $f' = 0$, points where $f'$ is undefined, and the endpoints of a closed interval. Students who write all three down before starting an optimisation problem stop losing marks to omission rather than to error.

A useful classroom test: “$f'(2) = 0$. Is $f$ guaranteed to have a local extremum at $x=2$? Justify.” The answer is no, with $x^3$ shifted to $x=2$ as the counterexample. Students who answer yes have the misconception; students who answer no but cannot produce a counterexample have heard the rule without owning it.

04A sample question

05What each wrong answer reveals

• A Critical point read as extremum. The justification given is the misconception stated outright: the derivative equals zero, therefore there is a minimum. No sign test was performed, because in this student's model finding the zero is the test. Work the sign of $3x^2$ on both sides in front of them — positive, positive — and the absence of a change does the arguing.
• B Correct. $f'(x)=3x^2$ is positive for all $x \neq 0$, so $f$ is increasing on both sides of the origin. There is a horizontal tangent and no extremum; the origin is a point of inflection.
• C Same error, opposite guess. Identical reasoning to A with the label flipped, which confirms that no test was run — a student who had checked the sign would not be choosing between max and min at random. If a class splits between A and C, that split is itself the evidence: they are guessing the type because their method does not produce one.
• D The second derivative test reached for, and its limitation missed. The most sophisticated wrong answer, and the student is right that $f''$ is relevant. What they have missed is that $f''(0) = 0$ here, so the second derivative test is inconclusive at exactly this point — which is the standard case where you must fall back on the first derivative test. Worth walking through, because this student is ready for it.

A and C together are the diagnosis: a class divided between them is a class guessing the label, since nothing in their method distinguishes a maximum from a minimum. D is genuinely further along and needs the specific fact that the second derivative test fails when $f''=0$. Only B ran a test. The justification question — ask why, not which — separates all four in one item.

06Try it in Mistake Master