A relative (local) maximum is a point where a function's value is greater than or equal to all nearby values. In AP Calculus, it occurs at a critical point where f' changes sign from positive to negative (FUN-4.A.2), or where f' = 0 and f'' < 0.
A relative maximum (also called a local maximum) is a point where a function is bigger than everything around it. Think of it as the top of one hill, not necessarily the tallest mountain on the whole graph. Formally, f has a relative maximum at x = c if f(c) ≥ f(x) for all x near c.
On the AP exam, the definition matters less than the proof. Relative maxima can only happen at critical points, places where f'(c) = 0 or f'(c) doesn't exist. But not every critical point is a max, so you confirm it one of two ways. The First Derivative Test (Topic 5.4) says f has a relative max at c if f' changes from positive to negative there (the function stops rising and starts falling). The Second Derivative Test (Topic 5.7) says if f'(c) = 0 and f''(c) < 0, the curve is concave down at c, so c is a relative max. Both tests live under the same essential knowledge: the first derivative of a function determines the location of relative extrema (FUN-4.A.2).
Relative maxima sit at the heart of Unit 5 (Analytical Applications of Differentiation), specifically Topics 5.4, 5.7, and 5.9, all under the learning objective "Justify conclusions about the behavior of a function based on the behavior of its derivatives" (5.4.A, 5.7.A, 5.9.A). That word "justify" is the whole game. The exam doesn't just ask where the max is; it asks you to write the sentence that proves it, usually "f' changes from positive to negative at x = c, so f has a relative maximum at x = c."
The term also crosses into Unit 6. When a function is defined as an accumulation function g(x) = ∫ from a to x of f(t) dt, then g' = f by the Fundamental Theorem of Calculus, so finding relative maxima of g means reading sign changes off the graph of f (Topic 6.5, FUN-5.A.3). And one more high-value fact from FUN-4.A: if a continuous function has only one critical point on an interval and it's a relative max, it's automatically the absolute max on that interval.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryFirst Derivative Test (Unit 5)
This is the most common way to prove a relative maximum. If f' switches from positive to negative at a critical point, the function climbed and then fell, so that point is a peak. The exact wording of that sign change is what earns the justification point on FRQs.
Second Derivative Test (Unit 5)
The alternate proof. If f'(c) = 0 and f''(c) < 0, the graph is concave down at c (frowning shape), so c is a relative max. Watch the trap from practice questions: if f''(c) = 0, the test is inconclusive and you have to fall back on the First Derivative Test.
Accumulation Function (Unit 6)
When g(x) = ∫ from a to x of f(t) dt, the graph you're handed of f is actually the graph of g'. So g has a relative maximum exactly where f crosses from positive to negative. This is the bridge that makes Unit 5 logic reappear on Unit 6 problems like the 2023 FRQ Q4.
Critical Point (Unit 5)
Every relative maximum happens at a critical point, but the reverse isn't true. A critical point can be a max, a min, or neither (like f(x) = x³ at x = 0). Critical points are the candidates; the derivative tests are the interview.
Multiple choice loves testing the logic of the two tests. Typical stems ask what you can conclude when f'' is zero at a critical point (answer: nothing yet, the Second Derivative Test is inconclusive) or what happens to an integrally-defined function h when its derivative changes from negative to positive (that's a relative minimum, the mirror-image trap). On the free response, relative maxima show up in graph-of-the-derivative problems like the 2023 FRQ Q4, where f' is given as line segments and a semicircle and you have to locate and justify extrema of f from sign changes. The 2021 FRQ Q5 pushed it further with implicit differentiation, asking whether a point on a curve was a relative max or min using the Second Derivative Test. In every case, the credited response requires a sign-based justification in words, not just an x-value. Saying "f' = 0 at x = c" alone never earns the point.
A relative maximum beats its neighbors; an absolute maximum beats every point on the entire interval. A function can have several relative maxima but at most one absolute maximum value, and on a closed interval the absolute max might happen at an endpoint where no relative max exists at all. One important crossover from FUN-4.A: if a continuous function has exactly one critical point on an interval and it's a relative max, that point is also the absolute max. That fact is a fast justification shortcut on optimization FRQs.
A relative maximum is a point where the function value is greater than or equal to all nearby values, even if it's not the highest point on the whole graph.
Relative maxima can only occur at critical points, where f' equals zero or is undefined, but not every critical point is a relative maximum.
First Derivative Test: f has a relative maximum at x = c if f' changes from positive to negative at c, and you must state that sign change to earn justification points.
Second Derivative Test: if f'(c) = 0 and f''(c) < 0, then c is a relative maximum; if f''(c) = 0, the test is inconclusive.
For an accumulation function g(x) = ∫ from a to x of f(t) dt, relative maxima of g occur where the graph of f crosses from positive to negative, because f is g's derivative.
If a continuous function has only one critical point on an interval and it's a relative maximum, it is automatically the absolute maximum on that interval.
It's a point where a function's value is greater than or equal to all nearby values, like the top of one hill on the graph. It occurs at a critical point where f' changes from positive to negative, or where f' = 0 and f'' < 0.
No. A critical point is only a candidate. It could be a relative max, a relative min, or neither (f(x) = x³ has a critical point at x = 0 that's neither). You need the First or Second Derivative Test to classify it.
A relative max only has to beat nearby points; an absolute max beats every point on the interval. On a closed interval, the absolute max can occur at an endpoint, which is never a relative max. The CED notes one shortcut: a single critical point that's a relative max on an interval is also the absolute max there.
You can't tell. When the second derivative is zero at a critical point, the Second Derivative Test is inconclusive, and you have to check the sign change of f' instead. This is a classic multiple-choice trap.
By the Fundamental Theorem of Calculus, g' = f, so look at where the graph of f crosses from positive to negative. That x-value is a relative maximum of g. The 2023 FRQ Q4 tested exactly this skill with a graph of f' made of line segments and a semicircle.