Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The sign of f′ tells you which way f is going, and a change of sign tells you where it turned around. That is the whole first derivative test. Build the sign chart from the factored derivative, mark the zeros, and read the pattern. Positive then negative means a relative maximum. Negative then positive means a relative minimum. No change at all means neither, and that case is the one students forget. The same chart answers two different questions that the exam asks separately: on which intervals the function increases or decreases, and what type of extremum sits at each critical point. Two sentences carry the credit — the interval statement with its reason, and the sign pattern that names the type.
Five of the nine released exams since 2017 contain a part that requires a sign pattern, and every one of those parts asks for the direction of the function rather than for a number.
Figure 1 — The direction of the sign change decides the type. A zero of f′ with no sign change produces neither a maximum nor a minimum.
f′ changes +→− at c⟹f(c) is a relative maximum
f′ changes −→+ at c⟹f(c) is a relative minimum
Both implications run in one direction only, and the direction matters. A sign change guarantees the extremum. An extremum at a critical point guarantees the sign change, but only when f is differentiable on both sides. The test never claims that every zero of f′ is a turning point, because f(x)=x3 has f′(0)=0 and no turning point at all.
Two words, used precisely, cover every interval question.
| Statement | Meaning | How it is proved |
|---|---|---|
| f is increasing on an interval | larger input gives larger output | f′(x)>0 on that interval |
| f is decreasing on an interval | larger input gives smaller output | f′(x)<0 on that interval |
| f is increasing at a point x=k | the local trend at k is upward | f′(k)>0 |
| f is decreasing at a point x=k | the local trend at k is downward | f′(k)<0 |
Intervals and points are graded with different sentences.
An interval answer needs open intervals and a reason attached to each one. A point answer needs only the sign of the derivative at that input, which is why the released exams accept "decreasing at x=2 because m′(2)=−1.5" as a complete response. Mixing the two forms is a writing error, and it costs credit even when the sign work is right.
Three moves, in this order.
The third move is where speed comes from. Testing the far left and the far right and then counting multiplicity changes is faster than testing every interval, and it is the same amount of work on a five-interval chart. A factor with an even multiplicity touches the axis without crossing it, so the sign survives that zero untouched.
The exam has asked this directly. When f′(c)=0, three things are possible, and the derivative alone cannot tell you which.
The third case is the one that separates a memorised rule from an understood one. It is also the reason the rubric asks for the sign pattern rather than the equation f′(c)=0.
Borrowed idea: A sign chart is a map of one question, and the question is whether the curve goes up or down. Every interval question on this topic is that map read aloud. Nothing more. So the work has two halves. First draw the map, then translate it into the sentence the rubric wants. Students who can draw the chart still lose points here, because they translate badly. They write the zeros instead of the intervals, and they name the critical point instead of the type. The chart was never the answer. The sentence is.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "on what intervals is f increasing" | a formula for f′ | sign of f′ per interval | factor, chart, write open intervals |
| "does f have a relative minimum at x=6" | a graph of f′ | sign of f′ on each side of 6 | report the pattern, then the type |
| "is m increasing at x=2" | a numeric m′(2) | one sign, one point | one sentence, no chart needed |
| "increasing and concave up" | two conditions at once | sign of f′ and sign of f′′ | intersect the two interval lists |
| "find the x-coordinates of the extrema" | a factored f′ | zeros plus sign change | classify, then list |
① Factor first, always. An unfactored cubic hides its multiplicities, and multiplicity is what decides whether the sign flips.
② Draw the chart even when the question is a single point. The chart costs fifteen seconds and prevents the off-by-one error of reading a zero as a sign change.
③ Answer in the grammar of the question. Intervals get intervals. Types get types. Points get one sentence with the sign attached.
Core Archetype (Simplest Form): Let f′(x)=(x−1)(x−4). Find the open intervals on which f is increasing, and classify each critical point.
Standard Execution:
- Zeros. f′(x)=0 at x=1 and x=4.
- Signs. Testing x=0 gives f′>0; testing x=2 gives f′<0; testing x=5 gives f′>0.
- Intervals. f is increasing on (−∞,1) and on (4,∞), and decreasing on (1,4).
- Types. At x=1 the sign goes + to −, so f(1) is a relative maximum. At x=4 the sign goes − to +, so f(4) is a relative minimum.
Exam Variation (Context / Parameter Shift): Let f be a differentiable function with f′(x)=(x+3)(x−1)2 on the interval [−4,3]. Find the open intervals on which f is decreasing, and determine whether f has a relative extremum at x=1.
Solution: The factor (x−1)2 is never negative, so the sign of f′ is the sign of x+3. On (−4,−3) we have f′<0, and on (−3,1) and (1,3) we have f′>0. So f is decreasing only on (−4,−3). At x=1 the derivative is zero but the sign does not change, so f has neither a relative maximum nor a relative minimum there; the graph has a horizontal tangent and continues to increase.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f be a differentiable function with f′(x)=(x−4)(x+1). On which open intervals is f decreasing?
MCQ 2
Let f′(x)=x3(x−2). What type of extremum does f have at x=0?
MCQ 3
The function m is differentiable and m′(2)=−1.5. Which statement about m at x=2 is correct?
FRQ 4
Let f be a differentiable function defined on the closed interval [−4,3], with f′(x)=(x+3)(x−1)2.
(a) Find all values of x in the open interval (−4,3) at which f has a critical point.
(b) On what open intervals is f increasing? On what open intervals is f decreasing? Justify your answer.
(c) For each critical point found in part (a), determine whether f has a relative maximum, a relative minimum, or neither at that point.
(d) A student claims that f must have a relative extremum at every critical point. Use this function to explain why the claim is false, and describe what the graph of f does at x=1.
| What to build | What to write | |
|---|---|---|
| "on what intervals is f increasing" | sign chart of f′ | open intervals where f′>0, with the sign as the reason |
| "relative maximum, minimum, or neither" | sign of f′ on both sides of the zero | the pattern, then the type |
| "is f increasing at x=k" | the single value f′(k) | one sentence with that sign |
| "increasing and concave up" | sign of f′ and sign of f′′ | the overlap of the two interval lists |
| "find the x-coordinates of the extrema" | zeros plus sign changes | the inputs that survive the test |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
The sign of f′ tells you which way f is going, and a change of sign tells you where it turned around. That is the whole first derivative test. Build the sign chart from the factored derivative, mark the zeros, and read the pattern. Positive then negative means a relative maximum. Negative then positive means a relative minimum. No change at all means neither, and that case is the one students forget. The same chart answers two different questions that the exam asks separately: on which intervals the function increases or decreases, and what type of extremum sits at each critical point. Two sentences carry the credit — the interval statement with its reason, and the sign pattern that names the type.
Five of the nine released exams since 2017 contain a part that requires a sign pattern, and every one of those parts asks for the direction of the function rather than for a number.
Figure 1 — The direction of the sign change decides the type. A zero of f′ with no sign change produces neither a maximum nor a minimum.
f′ changes +→− at c⟹f(c) is a relative maximum
f′ changes −→+ at c⟹f(c) is a relative minimum
Both implications run in one direction only, and the direction matters. A sign change guarantees the extremum. An extremum at a critical point guarantees the sign change, but only when f is differentiable on both sides. The test never claims that every zero of f′ is a turning point, because f(x)=x3 has f′(0)=0 and no turning point at all.
Two words, used precisely, cover every interval question.
| Statement | Meaning | How it is proved |
|---|---|---|
| f is increasing on an interval | larger input gives larger output | f′(x)>0 on that interval |
| f is decreasing on an interval | larger input gives smaller output | f′(x)<0 on that interval |
| f is increasing at a point x=k | the local trend at k is upward | f′(k)>0 |
| f is decreasing at a point x=k | the local trend at k is downward | f′(k)<0 |
Intervals and points are graded with different sentences.
An interval answer needs open intervals and a reason attached to each one. A point answer needs only the sign of the derivative at that input, which is why the released exams accept "decreasing at x=2 because m′(2)=−1.5" as a complete response. Mixing the two forms is a writing error, and it costs credit even when the sign work is right.
Three moves, in this order.
The third move is where speed comes from. Testing the far left and the far right and then counting multiplicity changes is faster than testing every interval, and it is the same amount of work on a five-interval chart. A factor with an even multiplicity touches the axis without crossing it, so the sign survives that zero untouched.
The exam has asked this directly. When f′(c)=0, three things are possible, and the derivative alone cannot tell you which.
The third case is the one that separates a memorised rule from an understood one. It is also the reason the rubric asks for the sign pattern rather than the equation f′(c)=0.
Borrowed idea: A sign chart is a map of one question, and the question is whether the curve goes up or down. Every interval question on this topic is that map read aloud. Nothing more. So the work has two halves. First draw the map, then translate it into the sentence the rubric wants. Students who can draw the chart still lose points here, because they translate badly. They write the zeros instead of the intervals, and they name the critical point instead of the type. The chart was never the answer. The sentence is.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| "on what intervals is f increasing" | a formula for f′ | sign of f′ per interval | factor, chart, write open intervals |
| "does f have a relative minimum at x=6" | a graph of f′ | sign of f′ on each side of 6 | report the pattern, then the type |
| "is m increasing at x=2" | a numeric m′(2) | one sign, one point | one sentence, no chart needed |
| "increasing and concave up" | two conditions at once | sign of f′ and sign of f′′ | intersect the two interval lists |
| "find the x-coordinates of the extrema" | a factored f′ | zeros plus sign change | classify, then list |
① Factor first, always. An unfactored cubic hides its multiplicities, and multiplicity is what decides whether the sign flips.
② Draw the chart even when the question is a single point. The chart costs fifteen seconds and prevents the off-by-one error of reading a zero as a sign change.
③ Answer in the grammar of the question. Intervals get intervals. Types get types. Points get one sentence with the sign attached.
Core Archetype (Simplest Form): Let f′(x)=(x−1)(x−4). Find the open intervals on which f is increasing, and classify each critical point.
Standard Execution:
- Zeros. f′(x)=0 at x=1 and x=4.
- Signs. Testing x=0 gives f′>0; testing x=2 gives f′<0; testing x=5 gives f′>0.
- Intervals. f is increasing on (−∞,1) and on (4,∞), and decreasing on (1,4).
- Types. At x=1 the sign goes + to −, so f(1) is a relative maximum. At x=4 the sign goes − to +, so f(4) is a relative minimum.
Exam Variation (Context / Parameter Shift): Let f be a differentiable function with f′(x)=(x+3)(x−1)2 on the interval [−4,3]. Find the open intervals on which f is decreasing, and determine whether f has a relative extremum at x=1.
Solution: The factor (x−1)2 is never negative, so the sign of f′ is the sign of x+3. On (−4,−3) we have f′<0, and on (−3,1) and (1,3) we have f′>0. So f is decreasing only on (−4,−3). At x=1 the derivative is zero but the sign does not change, so f has neither a relative maximum nor a relative minimum there; the graph has a horizontal tangent and continues to increase.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
Let f be a differentiable function with f′(x)=(x−4)(x+1). On which open intervals is f decreasing?
MCQ 2
Let f′(x)=x3(x−2). What type of extremum does f have at x=0?
MCQ 3
The function m is differentiable and m′(2)=−1.5. Which statement about m at x=2 is correct?
FRQ 4
Let f be a differentiable function defined on the closed interval [−4,3], with f′(x)=(x+3)(x−1)2.
(a) Find all values of x in the open interval (−4,3) at which f has a critical point.
(b) On what open intervals is f increasing? On what open intervals is f decreasing? Justify your answer.
(c) For each critical point found in part (a), determine whether f has a relative maximum, a relative minimum, or neither at that point.
(d) A student claims that f must have a relative extremum at every critical point. Use this function to explain why the claim is false, and describe what the graph of f does at x=1.
| What to build | What to write | |
|---|---|---|
| "on what intervals is f increasing" | sign chart of f′ | open intervals where f′>0, with the sign as the reason |
| "relative maximum, minimum, or neither" | sign of f′ on both sides of the zero | the pattern, then the type |
| "is f increasing at x=k" | the single value f′(k) | one sentence with that sign |
| "increasing and concave up" | sign of f′ and sign of f′′ | the overlap of the two interval lists |
| "find the x-coordinates of the extrema" | zeros plus sign changes | the inputs that survive the test |
Strengthen this topic with AP-style practice, or continue to next topic.