A point of inflection is a point on a graph where concavity changes, either from concave up to concave down or vice versa. In AP Calculus, this happens where the second derivative f'' changes sign, which is the same place the first derivative f' switches from increasing to decreasing (or the reverse).
A point of inflection is where a curve changes its bend. Concave up looks like a cup, concave down looks like a frown, and the inflection point is the exact spot where one becomes the other. In derivative language, it's where the second derivative f''(x) changes sign.
Here's the part the AP exam actually tests. A sign change is required, not just f''(x) = 0. The classic counterexample is f(x) = x⁴, where f''(0) = 0 but the graph is concave up on both sides of x = 0, so there's no inflection point there. You can also spot inflection points from the graph of f' alone, since f'' is the slope of f'. Wherever f' switches from increasing to decreasing (or decreasing to increasing), so wherever f' has a local max or min, the original function f has a point of inflection. That graph-of-the-derivative reading is exactly what released FRQs ask for.
Points of inflection live in Unit 5 (Analytical Applications of Differentiation) and support learning objective AP Calc 5.5.A, which asks you to justify conclusions about a function's behavior based on the behavior of its derivatives. Inflection points are one of the core 'behaviors' you read off f'' the way you read extrema off f'. They're also a standard step in the full curve-sketching workflow, where you combine critical points, intervals of increase/decrease, concavity, and inflection points to draw an accurate graph. On the exam, the justification is what earns the point. Writing 'f'' changes sign at x = c' is the sentence graders are looking for, and anything vaguer usually doesn't count.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryConcavity (Unit 5)
Points of inflection are literally the boundary markers between concavity intervals. If you can build a sign chart for f'', the inflection points are just the spots where the signs flip.
Critical Point (Unit 5)
Critical points come from the first derivative and flag possible extrema; inflection points come from the second derivative and flag concavity changes. Same logic, one derivative deeper. A point can even be both, but the two jobs are separate.
Second Derivative Test (Unit 5)
The Second Derivative Test uses concavity at a critical point to classify it as a max or min. When f''(c) = 0 the test is inconclusive, and that's exactly the situation where c might be an inflection point instead of an extremum.
Candidates Test (Unit 5)
Topic 5.5 puts these side by side. The Candidates Test finds absolute extrema by checking critical points and endpoints. Inflection points are not candidates for absolute extrema, and knowing what each list is for keeps the two procedures from blurring together.
Inflection points show up on both multiple choice and FRQs, and the most common FRQ setup gives you the graph of f' rather than f itself. The 2022 FRQ (Q3) gave a graph of f' made of a semicircle and line segments and expected you to identify where f has inflection points by finding where f' changes direction. The 2025 FRQ (Q4) ran the same play with an accumulation function g defined from a graph of f, where f plays the role of g'. Practice questions also fold inflection points into curve sketching, ask you to interpret a function with no inflection points (its concavity never changes), and test whether you know inflection points are not candidates in the Candidates Test. The skill being graded is justification. You need to say f'' changes sign, or equivalently that f' changes from increasing to decreasing, at that x-value.
Critical points are about the FIRST derivative: places where f'(x) = 0 or f' doesn't exist, the only spots where local extrema can happen. Points of inflection are about the SECOND derivative: places where f'' changes sign and concavity flips. Mixing them up costs points fast. Inflection points are never candidates in the Candidates Test for absolute extrema, and a point where f''(c) = 0 is not automatically an inflection point any more than f'(c) = 0 automatically means a max or min.
A point of inflection is where a graph's concavity changes, which means the second derivative f'' changes sign there.
f''(c) = 0 alone is not enough; f'' must actually change sign at x = c, as the counterexample f(x) = x⁴ at x = 0 shows.
On a graph of f', inflection points of f occur wherever f' changes from increasing to decreasing or vice versa, which is wherever f' has a local max or min.
Inflection points are not candidates in the Candidates Test, because absolute extrema can only occur at critical points or endpoints.
On an FRQ, the full-credit justification is a sign-change statement, such as 'f'' changes from positive to negative at x = c, so f has a point of inflection there.'
It's a point where a function's concavity changes, from concave up to concave down or the reverse. Analytically, it's where the second derivative f'' changes sign.
No. f'' must actually change sign at that point. For f(x) = x⁴, f''(0) = 0 but the function is concave up on both sides of zero, so x = 0 is not an inflection point. This is one of the most common wrong answers on AP multiple choice.
Critical points come from f' and mark possible local maxes and mins. Inflection points come from f'' and mark concavity changes. They answer different questions, and only critical points and endpoints go into the Candidates Test for absolute extrema.
Look for where f' changes direction, meaning where f' has a local max or local min. At those x-values the slope of f' (which is f'') changes sign, so f has an inflection point there. The 2022 FRQ Q3 tested exactly this with a graph of f' made of a semicircle and line segments.
Its concavity never changes. The graph is either concave up on its entire domain or concave down on its entire domain, like f(x) = x², which is concave up everywhere.