A local (relative) minimum is a point where a function's value is lower than at all nearby points, even if it isn't the lowest value on the whole domain. On the AP Calc exam, local minima occur only at critical points, and you justify them with the First or Second Derivative Test.
A local minimum (the CED also calls it a relative minimum) is a point where a function dips lower than everything immediately around it. Picture a valley on a graph. It's the bottom of that valley, but there might be deeper valleys somewhere else on the function. That's the whole "local" part. It only has to beat its neighbors, not the entire function.
The CED gives you the machinery for finding these points. Per FUN-1.C.3, every local extremum happens at a critical point, which is a point where f′ equals zero or doesn't exist (FUN-1.C.2). But the reverse isn't true. A critical point might be a local min, a local max, or neither (think of f(x) = x³ at x = 0, where the derivative is zero but the graph just flattens and keeps climbing). So finding candidates is step one. Step two is justifying which candidates are actually local minima, and that's where the First Derivative Test (f′ changes from negative to positive) or the Second Derivative Test (f′ = 0 and f″ > 0) comes in.
Local minima live in Unit 5: Analytical Applications of Differentiation, specifically Topics 5.2, 5.3, and 5.7. They support learning objectives 5.2.A (justify conclusions using the Extreme Value Theorem and critical points), 5.3.A (justify conclusions about a function from its derivatives), and 5.7.A (use the second derivative to classify critical points). Notice the word that keeps repeating in those objectives. Justify. The AP exam doesn't just want you to find a local minimum. It wants you to prove it, with a sentence like "f has a relative minimum at x = 2 because f′ changes sign from negative to positive at x = 2." Unit 5 is where calculus stops being computation and starts being argument, and local minima are one of the most common things you'll be asked to argue about.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryCritical Point (Unit 5)
Critical points are where local minima are allowed to live. FUN-1.C.3 says all local extrema occur at critical points, but not every critical point is an extremum. So solving f′(x) = 0 gives you a list of suspects, not convictions. You still need a derivative test to classify each one.
First Derivative Test (Unit 5)
This is the workhorse justification for a local minimum. If f′ switches from negative to positive at a critical point, the function was falling and then started rising, so that point has to be the bottom of a valley. This test always works, even when the Second Derivative Test is inconclusive.
Absolute Minimum (Unit 5)
An absolute minimum has to beat every value on the interval, not just the neighbors. The Candidates Test (comparing critical points and endpoints) finds absolute extrema. One nice shortcut from the CED, if a continuous function has only one critical point on an interval and it's a local min, it's automatically the absolute min too.
Increasing Interval (Unit 5)
A local minimum is literally the seam between a decreasing interval and an increasing interval. Topic 5.3 sign charts for f′ aren't a separate skill, they ARE the First Derivative Test laid out in table form. Build the sign chart and the local minima fall right out.
Local minima show up everywhere on the AP Calc exam, in both multiple choice and FRQs. MCQ stems sound like "At which value of x does f have a relative minimum?" and often hand you a graph or table of f′ instead of f itself. You're expected to read the sign change in f′ directly. FRQs almost always ask for a justification, and the rubric is picky about it. "f′ changes from negative to positive at x = c" earns the point; "f′(c) = 0" alone does not, because a zero derivative can also mean a max or neither (that's the x³ trap). You should also be ready to use the Second Derivative Test (f′(c) = 0 and f″(c) > 0 means local min) and to distinguish relative from absolute extrema, since questions frequently ask for one and tempt you with the other.
A local minimum only needs to be lower than nearby points; an absolute (global) minimum is the single lowest value on the entire interval or domain. A function can have several local minima but at most one absolute minimum value. The tests differ too. Local minima get the First or Second Derivative Test, while absolute minima on a closed interval get the Candidates Test, where you evaluate f at every critical point and both endpoints and compare. Watch the endpoints, an absolute minimum on [a, b] can sit at an endpoint where f′ never equals zero at all.
A local minimum is a point where the function's value is lower than at all nearby points, even if a lower value exists elsewhere on the function.
Every local minimum occurs at a critical point (where f′ is zero or undefined), but not every critical point is a local minimum, so you always need a test to confirm.
First Derivative Test justification means f′ changes sign from negative to positive at the critical point.
Second Derivative Test justification means f′(c) = 0 and f″(c) > 0, which tells you the graph is concave up at the bottom of a valley.
If a continuous function has exactly one critical point on an interval and it's a local minimum, the CED says it's also the absolute minimum on that interval.
On FRQs, writing only "f′(c) = 0" does not earn the justification point; you must show the sign change or the second derivative condition.
A local (relative) minimum is a point where a function's value is lower than at all points immediately around it. It occurs at a critical point and is confirmed when f′ changes from negative to positive there, or when f′ = 0 and f″ > 0.
No. A zero derivative gives you a critical point, but that point could be a local min, a local max, or neither. The classic counterexample is f(x) = x³ at x = 0, where f′ = 0 but the function just keeps increasing. You always need the First or Second Derivative Test to classify it.
A local minimum only beats its neighbors; an absolute minimum is the single lowest value on the whole interval or domain. On a closed interval, you find absolute minima with the Candidates Test by comparing critical points and endpoints, and the answer can be an endpoint.
Write that f′ changes sign from negative to positive at the critical point (First Derivative Test), or that f′(c) = 0 and f″(c) > 0 (Second Derivative Test). Just stating f′(c) = 0 won't earn the point because that alone doesn't distinguish a min from a max.
Yes. The CED states that if a continuous function has only one critical point on an interval and that point is a relative minimum, it's also the absolute minimum on that interval. With multiple critical points, you need the Candidates Test to compare values.