AP

4 sign-chart patterns behind the first derivative test

A tutor-level walkthrough of the AP Calculus first derivative test for local extrema: sign charts, FRQ phrasing, and the errors that quietly cost a point or two.

5 June 202621 min
Author: Gökhan İnceReviewed by: Selin Yıldız

The AP Calculus first derivative test is the most efficient tool candidates have for classifying local maxima and minima of a differentiable function. Where a graph or a derivative expression is in hand, the test turns a sketch-and-hope classification into a deterministic one-line argument: read the sign of f'(x) on either side of a critical point, and the behaviour of the function follows. AP examiners reward that argument every year, on both the multiple-choice section of the AB exam and the free-response section, and the same logic carries through into BC-only items that mix in parametrics, polar graphs, or implicitly defined curves.

The trap is that "the first derivative test" sounds like a single rule, when in practice it is a cluster of three small habits: locating critical points, building a sign chart, and translating the chart into a sentence the grader accepts. The rest of this article walks through each habit, names the calculator moves that speed it up, contrasts it with the second derivative test, and shows the precise language a grader is looking for on a free-response solution. By the end you should be able to take any "find and classify the local extrema" prompt, plan a three-minute solution, and produce a justification that survives a strict reader.

What the first derivative test actually says

The textbook statement is short: if f is continuous at a point c and f'(x) changes sign as x crosses c, then f has a local extremum at c, and the direction of the sign change tells you whether it is a maximum or a minimum. The phrasing is deceptively simple because every clause carries weight. Continuity at c is not optional: a corner or a cusp can produce a sign change in the derivative that does not correspond to an actual extremum of the function. Differentiability is also assumed on either side of c so that the sign of f'(x) is well defined there, although f itself need not be differentiable at c.

For AP purposes, the test is applied in two modes. In the first mode, you are given f'(x) as a closed-form expression and asked to find and classify the local extrema of f. The work is mechanical: factor, find where f'(x) = 0 or is undefined, and inspect the sign of each factor. In the second mode, you are given a graph of f'(x) and asked to deduce the behaviour of f. The graph version is the one that quietly separates a 4 from a 5, because candidates routinely read derivative graphs as if they were function graphs and end up calling a maximum a minimum.

There is a subtle distinction worth flagging now and returning to later. The first derivative test classifies local extrema, not global ones. An AP prompt will almost always use the word "local" in the stem, and a careful answer mirrors that word. Saying that f has "a maximum at x = 2" without the qualifier "local" invites a deduction on the FRQ, because the question did not ask whether 2 is the largest value of f on its domain.

Finally, the test only tells you what happens at points where the derivative is zero or fails to exist. Points where the derivative is positive on both sides cannot be extrema, no matter how dramatic the graph looks. That single sentence eliminates a category of wrong answers students love to write.

Building a sign chart the way a grader expects

A sign chart is the artefact a grader expects to see, even if it is not explicitly requested. The cleanest version is a horizontal number line, the critical points marked in increasing order, the sign of f'(x) in each open interval, and a short label under each critical point that names the classification. A sign chart does two jobs at once: it forces you to check your factorisation, and it gives the reader a one-glance summary of the argument.

For a polynomial derivative, the work goes through three steps. Factor f'(x) into linear factors. List the zeros in increasing order, including any where f'(x) is undefined. Pick a test point in each interval, evaluate the sign of each factor, and multiply. The signs across the row of factors give the sign of f'(x) in that interval, and a change of sign across a critical point is the trigger for the first derivative test. For a derivative like f'(x) = (x + 2)(x − 1)²(x − 3), the squared factor does not change sign at x = 1, so 1 is not a local extremum despite being a critical point. That kind of detail is exactly the place where a sign chart earns its keep, because a quick sketch in your head will miss the double root.

When f'(x) is not a polynomial, the chart still works. For a rational derivative like f'(x) = (x² − 4) / (x − 2), simplification gives f'(x) = x + 2 for x ≠ 2, and the original critical point at x = 2 is a vertical asymptote of the derivative, not a zero. The sign chart for the simplified expression shows f positive to the right of 2 and negative to the left, so f has a local minimum at x = 2, even though f' is not defined there. For AP exam questions, that means you should keep the unsimplified form long enough to record the point at which the derivative is undefined, then simplify for the sign test.

A practical tip: if a critical point makes two or more factors zero, count its multiplicity mentally. Odd multiplicities flip the sign of f'(x), so the test applies in the usual way. Even multiplicities leave the sign unchanged, so the test reports no extremum. This shortcut saves time on the multiple-choice section, where the answer is one of five choices and you can rule out candidates by multiplicity alone.

Reading a graph of f'(x) without flipping the answer

The graph-of-derivative item is a recurring AP item type, and it punishes two habits. The first is treating a maximum of f' as a maximum of f. The second is treating a zero of f' with a horizontal tangent as a sign change. Neither is true. What f' tells you about f is its sign, not its value, and a horizontal tangent on the graph of f' means the rate of change of f' is zero, which has nothing to do with whether f has a local extremum.

The correct reading is mechanical. Find the x-intercepts of the graph of f'. Those are the candidate critical points of f. For each candidate, look at the sign of f' immediately to the left and to the right. If the sign goes from positive to negative, f has a local maximum; if it goes from negative to positive, f has a local minimum. If the sign is the same on both sides, f has no local extremum at that x, even if the graph of f' touches the x-axis and turns around. That last case is the doubled-root case in disguise, and it shows up on multiple-choice items as a tempting distractor.

A useful sanity check: where f' is positive, f is increasing; where f' is negative, f is decreasing. After you classify each critical point of f, sketch the rough behaviour of f in your margin. If your classification says "local maximum at x = 3" but your sketch shows f going up through 3, you have flipped a sign. The sketch takes ten seconds and catches most sign-reading errors before they hit the bubble sheet.

For the free-response section, the language of a graph-based answer is the same as for an algebraic one. "Because f'(x) changes from positive to negative at x = 3, f has a local maximum at x = 3" is the gold-standard sentence. Drop the "because" clause and the answer becomes a claim rather than an argument, and a strict grader will dock a point for missing justification.

First derivative test versus second derivative test: when each one is faster

The second derivative test classifies a critical point c by evaluating f''(c). If f''(c) is positive, f is concave up at c and the critical point is a local minimum; if negative, concave down and a local maximum; if zero, the test is inconclusive. For polynomial f' whose factors are easy to read, the second derivative test is faster, because no sign chart is required. For a derivative like f'(x) = x³ − 3x, the critical points are x = −1 and x = 1, f''(x) = 3x² − 3, and the second derivative test classifies both in one line.

The first derivative test, by contrast, is more general. It works for any critical point, including those where f' is not defined and those where the second derivative test is inconclusive. It is also the only test that gives useful information at points where f' changes sign without crossing zero, which is rare on a typical AP item but appears on graph questions where f' has a vertical asymptote at the critical point.

On an AP multiple-choice item, choose the second derivative test when the algebra is clean and the second derivative is easy to compute. Choose the first derivative test when the second derivative is messy, when the critical point is a point of non-differentiability, or when you are working from a graph. On a free-response item, default to the first derivative test even when the second derivative test would work, because the sign chart is itself a piece of evidence the grader can award partial credit for, whereas a single number substituted into f'' looks like guesswork if the sign comes out wrong.

A common AP trap is the second derivative test returning zero. In that case the test gives no information, and the only path to the answer is the first derivative test. A useful habit is to glance at the second derivative first: if it is easy to evaluate and clearly non-zero, use it; if it is zero or ugly, build a sign chart and use the first derivative test. That single decision rule will save you from a category of "the second derivative test said inconclusive, so I left the answer blank" failures.

Need help reaching your target score?

Book a free 15-minute call with an advisor to map out a personalised study plan.

Free consultation

FRQ language that earns the point

AP free-response items in this family typically have two or three points. The first point is for finding the critical points. The second is for the classification. The third, when present, is for a justification. The justification is where candidates lose credit they thought they had earned, because they state a classification without saying why.

The exact phrasing the rubric rewards is a sentence of the form "f has a local maximum at x = c because f'(x) changes from positive to negative at x = c." Variations that earn full credit include "f' > 0 for x < c and f' < 0 for x > c, so f has a local maximum at x = c" and the equivalent for a minimum. Phrasings that lose credit include "f'(c) = 0 so c is a maximum" (this is a sufficient condition only in conjunction with a sign change), "the graph looks like a peak at c" (graphical hand-waving without a derivative argument), and "by the first derivative test, c is a maximum" (this is a citation, not an argument; the rubric wants the sign change stated).

For non-differentiable critical points, the rubric still wants the sign change stated, with the additional acknowledgement that f' is not defined at c. A sentence like "f'(x) < 0 for x < c and f'(x) > 0 for x > c, and f is continuous at c, so f has a local minimum at x = c" is the gold standard for a corner or cusp extremum. For a graph-based item, replace the algebraic inequalities with readings of the graph: "the graph of f' is below the x-axis to the left of c and above the x-axis to the right of c, so f has a local minimum at x = c."

One more FRQ habit to lock in: state the conclusion in the same word the prompt used. If the prompt asks for "local extrema," the answer must say "local maximum at x = 2" or "local minimum at x = 3," not "maximum" or "turning point." If the prompt asks for "absolute extrema on [a, b]," the first derivative test classifies the candidates, but the rubric also wants the endpoints evaluated, and the answer must include the word "absolute." Mismatched vocabulary is a small point loss that candidates rarely see coming.

Calculator moves that speed the sign chart

On the AP Calculus exam, the calculator section is where the first derivative test becomes a one-minute operation rather than a three-minute one. The two calculator moves that matter are: finding the zeros of f'(x), and evaluating f'(x) at a test point in each open interval. Both are available on every approved graphing calculator without programming, and both should be on the muscle-memory list of any candidate who wants a 5.

To find the zeros, graph y = f'(x) on the calculator and use the zero or root finder. For a polynomial derivative, the calculator returns the zeros to several decimal places; record them to the precision the prompt requires. If the prompt is a multiple-choice item, the answer choices are usually exact values, so the decimal root must be matched to the algebraic form. A common error is to record the decimal root and forget that the algebraic factorisation might collapse a near-zero root into a clean rational number.

To evaluate at a test point, pick a value strictly between two consecutive critical points. Use the table feature of the calculator, set the independent variable to the test point, and read the value of f' there. The sign of that value is what you write on the sign chart. The habit here is to record both the test point and the value, because on a free-response solution the grader can follow a numerical sign evaluation as easily as an algebraic one.

A useful calculator trick is to store f'(x) in the Y= editor and then trace along the graph. As you move the cursor from left to right, the sign of the y-coordinate changes exactly at the zeros, and you can read the sign of each interval directly off the screen. This trick is faster than the table feature for a derivative with many critical points, and it has the side benefit of catching the doubled-root case visually, because the graph of f' will touch the x-axis and bounce back without crossing it.

One last calculator habit: after you have classified each critical point, ask the calculator to evaluate f at the critical point. The y-coordinate is the local extreme value, and on a free-response item that asks for the extreme value, the sign chart alone is not enough. The rubric separates the classification point from the value point, and a missing y-coordinate is a lost point that a sign chart cannot recover.

Common pitfalls and how to avoid them

The first derivative test looks short on paper and gets candidates into five predictable difficulties. Naming them in advance is half the defence against each.

  • Forgetting the "local" qualifier. A local extremum is not a global one, and a rubric that uses the word "local" will not award full credit for an unqualified answer. Mirror the prompt's vocabulary in your conclusion sentence.
  • Misclassifying a doubled root. When f'(x) has a squared factor, the sign of f' does not change at the corresponding critical point, so there is no local extremum there. The fastest check is the multiplicity rule: even multiplicity means no sign change means no extremum.
  • Confusing the graph of f with the graph of f'. On graph-based items, read the sign of f' on either side of the critical point, not the value of f. A maximum of f' is not a maximum of f, and a horizontal tangent on the graph of f' is not a sign change.
  • Skipping the justification. On free-response items, a bare claim of "local maximum at x = 2" loses the justification point. State the sign change explicitly: f' goes from positive to negative, or vice versa, and name the open intervals on which the sign holds.
  • Ignoring points where f' is undefined. Critical points include points where f' does not exist, provided f itself is continuous there. A sign chart that lists only the zeros of f' will miss cusp and corner extrema, which appear every exam cycle as a distractor or as the actual answer.

A habit that catches most of these in one motion is the post-classification sanity sketch. After the sign chart is complete, draw a small picture of f in the margin that is consistent with the sign of f'. If the picture disagrees with the classification, something is wrong, and the picture is usually the easier thing to trust. For most candidates reading this, that ten-second sketch is the single highest-leverage habit to install before the exam.

Worked example: a typical AP-style prompt

Consider a free-response prompt of the form: "Let f be a differentiable function with f'(x) = (x + 1)(x − 2)². Find and classify the local extrema of f." The solution runs through the sign chart, the sign-change rule, and the FRQ language in one pass.

Step one is to identify the critical points. f'(x) = 0 at x = −1 and x = 2. The factor (x − 2)² has multiplicity two, which means the sign of f' will not change at x = 2. The factor (x + 1) has multiplicity one, so a sign change is possible at x = −1. Step two is to set up the sign chart. The critical points split the real line into three intervals: (−∞, −1), (−1, 2), and (2, ∞). Pick test points, say x = −2, x = 0, and x = 3. At x = −2, f'(−2) = (−1)(−4)² = −16, negative. At x = 0, f'(0) = (1)(−2)² = 4, positive. At x = 3, f'(3) = (4)(1)² = 4, positive. The sign chart reads: negative on (−∞, −1), positive on (−1, 2), positive on (2, ∞). Step three is the classification. At x = −1, f' changes from negative to positive, so f has a local minimum at x = −1. At x = 2, f' does not change sign, so f has no local extremum at x = 2. Step four is the FRQ sentence: "f has a local minimum at x = −1 because f'(x) changes from negative to positive at x = −1, and f has no local extremum at x = 2 because f'(x) does not change sign there."

Compare that to the second derivative test on the same problem. f''(x) = derivative of (x + 1)(x − 2)². By the product rule, f''(x) = (x − 2)² + (x + 1) · 2(x − 2) = (x − 2)[(x − 2) + 2(x + 1)] = (x − 2)(3x). At x = −1, f''(−1) = (−3)(−3) = 9, positive, which confirms a local minimum. At x = 2, f''(2) = 0, so the second derivative test is inconclusive, and we are back to the sign chart to rule out an extremum. The first derivative test is the more decisive tool here, and that pattern is the reason the rubric tends to award the same point to either test, provided the sign change is stated.

Notice the precision of the wording. The sentence names the sign of f' on each side of the critical point, names the classification that follows, and addresses the inconclusive critical point at x = 2 directly. A grader reading that sentence can award the classification point, the justification point, and the no-extremum point without any inference. The cost of writing the sentence well is small, and the cost of writing it poorly is two or three points on a single FRQ.

Conclusion and next steps

The first derivative test is a one-page topic in the textbook and a six-page topic in the AP grading notes, because the test only does its job when the sign chart, the vocabulary, and the justification line up. Candidates who treat the test as a single rule to memorise tend to lose points to the doubled-root case, the non-differentiable critical point, or the missing "local" qualifier. Candidates who treat the test as a habit of reading the sign of f' on either side of every critical point tend to pick up the points they missed in the multiple-choice section and the free-response section alike.

The next study block is a sign-chart drill. Pick ten f'(x) expressions covering polynomials with repeated factors, rational functions with vertical asymptotes, and graphs of f' with horizontal tangents at the x-axis. Time yourself: a sign chart that takes more than 90 seconds per critical point is too slow for a 5. TestPrep Europe's diagnostic assessment is a natural starting point for candidates building a sharper first-derivative-test preparation plan.

Quick reference: first derivative test at a glance

The table below summarises the sign-change rules, the typical signal in a graph of f', and the FRQ sentence that earns the point. Use it as a checklist while reviewing practice items.

Sign of f' to the left of cSign of f' to the right of cGraph of f' near cClassification of f at cFRQ justification
PositiveNegativeCrosses x-axis from above to belowLocal maximum"f'(x) changes from positive to negative at x = c, so f has a local maximum at x = c."
NegativePositiveCrosses x-axis from below to aboveLocal minimum"f'(x) changes from negative to positive at x = c, so f has a local minimum at x = c."
PositivePositiveTouches x-axis and bounces back aboveNo local extremum"f'(x) does not change sign at x = c, so f has no local extremum at x = c."
NegativeNegativeTouches x-axis and bounces back belowNo local extremum"f'(x) does not change sign at x = c, so f has no local extremum at x = c."
Undefined on one sideOpposite sign on the other sideVertical asymptote of f' at cLocal extremum if f is continuous at c"f is continuous at c, f'(x) < 0 for x < c, and f'(x) > 0 for x > c, so f has a local minimum at x = c."

Frequently asked questions

What is the first derivative test on the AP Calculus exam?
It is the rule that a differentiable function f has a local extremum at a critical point c if and only if f&apos;(x) changes sign as x crosses c, with the direction of the sign change determining whether the extremum is a maximum or a minimum.
When should I use the first derivative test instead of the second derivative test?
Use the first derivative test when the second derivative is zero, when the critical point is a point of non-differentiability, when the algebra of f&apos;&apos; is messy, or when you are reading the answer off a graph of f&apos;. The second derivative test is faster only when f&apos;&apos; is easy to evaluate and clearly non-zero.
Does a doubled root of f&apos;(x) ever give a local extremum?
No. When f&apos;(x) has a factor of even multiplicity at a critical point c, the sign of f&apos; does not change across c, so the first derivative test reports no local extremum. The most common error on AP items is to classify a doubled root as a maximum or minimum, which loses the classification point.
How do I justify a local extremum on a free-response answer?
State the sign of f&apos; on each open interval around the critical point, then name the resulting classification in the same word the prompt used. A sentence of the form &quot;f&apos;(x) changes from positive to negative at x = c, so f has a local maximum at x = c&quot; is the rubric-standard justification.
Can the first derivative test classify absolute extrema?
Only in the limited sense that it classifies the local extrema that are candidates for the absolute extremum on a closed interval. To finish the classification of an absolute extremum, you must also evaluate f at the endpoints of the interval and compare, and the FRQ answer must use the word &quot;absolute&quot; rather than &quot;local.&quot;

Start your exam preparation

Explore our 1-to-1 tutoring and small-group course options with expert instructors. First-lesson money-back guarantee.

Free consultation

Comments

Be the first to comment on this article.

Leave a comment

Your comment will appear after approval.

Article rating (optional)