The absolute minimum of a function is the single lowest output value the function attains over its entire domain (or a given closed interval). On the AP exam, you find it by comparing the function's values at critical points and endpoints, a process called the Candidates Test.
The absolute minimum (also called the global minimum) is the lowest y-value a function ever reaches on the interval you're looking at. There can be lots of relative minima, little local dips, but only one absolute minimum value. It happens at exactly one of two kinds of places: a critical point inside the interval (where the derivative is zero or undefined) or an endpoint of the interval.
That fact gives you the whole strategy. List the candidates (critical points + endpoints), plug each one into the function, and pick the smallest output. That's the Candidates Test, and it's the move the AP exam expects you to show. On a closed interval, the Extreme Value Theorem guarantees an absolute minimum exists for a continuous function, so the search always pays off.
Absolute extrema show up all over AP Calc, but this term maps most directly to Topic 6.5 and learning objective 6.5.A, where you represent accumulation functions using definite integrals. Per FUN-5.A.3, a graph of f tells you everything about g(x) = ∫ from a to x of f(t) dt, and a classic exam question hands you the graph of f' (or f) and asks for the absolute minimum of f (or g) on a closed interval. You can't just eyeball it. You have to find where the derivative changes sign, compute candidate values using signed areas under the graph, and compare them to the endpoint values. It's the spot where Unit 5 optimization logic and Unit 6 accumulation logic fuse into one problem.
Relative Minimum (Unit 5)
A relative minimum is the lowest point in its own neighborhood; the absolute minimum is the lowest point anywhere on the interval. Every interior absolute minimum is also a relative minimum, but most relative minima lose the comparison. The Candidates Test exists exactly to settle that comparison.
First Derivative (Units 4-5)
Critical points come from setting f'(x) = 0 or finding where f' is undefined. When f' changes from negative to positive, the function stops falling and starts rising, which flags a candidate for the minimum. On graph-of-the-derivative FRQs, you read those sign changes straight off the picture.
Accumulation Functions (Unit 6)
When g(x) = ∫ from a to x of f(t) dt, the Fundamental Theorem says g'(x) = f(x). So the graph of f IS the derivative graph of g, and finding g's absolute minimum means tracking where f crosses zero and adding up signed areas. This is the Topic 6.5 version of the problem.
Absolute Maximum (Units 5-6)
Same hunt, opposite winner. The Candidates Test finds both at once, since you're already comparing every critical-point and endpoint value. Exam questions often ask for one and bait you into reporting the other, so read carefully.
This is an FRQ regular. The 2017 Q3, 2022 Q3, and 2023 Q4 free-response questions all gave you the graph of f' (made of line segments and a semicircle) on a closed interval and asked you to analyze f, including finding absolute extrema. The expected work is the Candidates Test: identify critical points where f' changes sign, compute f at each candidate using a given value like f(2) = 1 plus signed areas under the f' graph (definite integrals via the FTC), evaluate the endpoints, and justify your answer by comparing all candidate values. Multiple-choice questions hit the same skill more quickly, like asking what you do after evaluating the relative minima and endpoint values (answer: compare them and pick the smallest). Justification matters. Saying 'f' changes from negative to positive' earns a relative min, but the absolute min requires showing you compared every candidate.
A relative minimum only has to beat the points right next to it; an absolute minimum has to beat every point on the interval, including the endpoints. A function can have several relative minima but at most one absolute minimum value. Endpoints are the trap: the absolute minimum can sit at an endpoint where f' never equals zero at all, which is why the Candidates Test always includes endpoints.
The absolute minimum is the single lowest value a function attains on its domain or on a given closed interval, and it can only occur at a critical point or an endpoint.
Use the Candidates Test: evaluate the function at every critical point and both endpoints, then pick the smallest value.
On a closed interval, the Extreme Value Theorem guarantees a continuous function actually has an absolute minimum.
On accumulation-function FRQs, the given graph is the derivative, so you compute candidate values as signed areas using g(x) = ∫ from a to x of f(t) dt.
A sign change in f' from negative to positive justifies a relative minimum, but the absolute minimum justification requires comparing all candidate values.
Don't skip endpoints; the absolute minimum often lands there even when no critical point does.
It's the lowest value the function reaches over its entire domain or a given closed interval. It occurs at a critical point (where the derivative is zero or undefined) or at an endpoint, and you find it by comparing function values at all of those candidates.
No. A relative minimum only beats nearby points, while the absolute minimum beats every point on the interval. A function can have multiple relative minima but only one absolute minimum value, and it might be at an endpoint rather than any relative minimum.
Yes, always. The absolute minimum on a closed interval can sit at an endpoint where the derivative never equals zero, so the Candidates Test requires evaluating the function at both endpoints along with every critical point.
Find where f' changes sign from negative to positive (those x-values are candidates), then compute f at each candidate and at the endpoints using a given function value plus signed areas under the f' graph. The 2022 FRQ Q3 and 2023 FRQ Q4 both tested exactly this setup.
On a closed interval, yes, as long as the function is continuous. That's the Extreme Value Theorem. On an open interval or for a discontinuous function, an absolute minimum might not exist at all.