In AP Calculus, a function is continuous at a point x = c when three things hold: f(c) is defined, the limit of f(x) as x approaches c exists, and that limit equals f(c). A function is continuous on an interval if it's continuous at every point in the interval (EK LIM-2.B.1).
The "draw it without lifting your pencil" idea is the right intuition, but the AP exam wants the precise version. Continuity at x = c is a three-part checklist. First, f(c) must exist. Second, the limit of f(x) as x approaches c must exist, which means the left-hand and right-hand limits agree. Third, that limit must actually equal f(c). Break any one of the three and you get a discontinuity (Topic 1.10 names the types: removable, jump, and discontinuities from vertical asymptotes).
Continuity over an interval just means continuity at every point in that interval. Here's the shortcut the CED hands you (EK LIM-2.B.2): polynomial, rational, power, exponential, logarithmic, and trigonometric functions are continuous everywhere on their domains. So for most functions, continuity questions are really domain questions. The places to actually check are domain gaps (like x = 0 for 1/x) and the seam points of piecewise functions, where you have to verify the two pieces meet.
Continuity lives in Unit 1 (Topics 1.10-1.12, learning objectives 1.10.A and 1.12.A), but it never leaves the course. It's the entry fee for almost every big theorem. The Intermediate Value Theorem requires continuity on a closed interval. The Mean Value Theorem (Topic 5.1, LO 5.1.A) requires continuity on [a, b] plus differentiability on (a, b). Definite integrals and average value arguments assume it too. And in Unit 2, LO 2.4.A makes the famous one-way connection. Differentiable implies continuous, but continuous does not imply differentiable. If you can't state and check continuity, you lose justification points all over the FRQ section.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryLimit (Unit 1)
Continuity is literally defined by a limit. Saying f is continuous at c is just saying the limit as x approaches c equals f(c). Every continuity check is a limit computation in disguise.
Differentiability (Unit 2)
Differentiable implies continuous, never the reverse. f(x) = |x| is continuous everywhere but not differentiable at x = 0 because the left and right difference quotients disagree, and the cube root of x has a vertical tangent at 0. These two are the CED's go-to counterexamples.
Intermediate Value Theorem (Unit 1)
IVT only works because continuous functions can't skip values. If f is continuous on [a, b], it hits every y-value between f(a) and f(b). Table-based FRQs love asking you to invoke this, and "f is continuous" must appear in your justification.
Mean Value Theorem (Unit 5)
MVT needs continuity on the closed interval and differentiability on the open interval. FRQ stems often say "twice-differentiable," which quietly hands you continuity for free since differentiable implies continuous. Spotting that connection is the whole game.
Multiple choice tests the three-part definition directly. A classic stem gives you left-hand limit, right-hand limit, and function value at a point and asks whether the function is continuous or what type of discontinuity it has. Practice questions like "what kind of discontinuity does 1/x have at x = 0" (vertical asymptote) or "how do you fix a removable discontinuity at x = 2" (redefine the function value to equal the limit) are standard. On FRQs, continuity usually shows up as a hypothesis you must cite. The 2018 FRQ 4 gave a table for a twice-differentiable tree-height function and expected you to recognize that differentiability guarantees continuity, unlocking IVT and MVT. The 2017 FRQ 1 stated that A(h) is continuous so a Riemann sum approximation made sense. The rubric pattern is consistent. Earning theorem justification points means writing "since f is continuous on [a, b]..." before applying IVT, MVT, or the Extreme Value Theorem. Skipping that sentence costs the point even when your answer is right.
These are not interchangeable. Differentiability is the stronger condition. If f is differentiable at a point, it's automatically continuous there (LO 2.4.A). But continuity alone doesn't guarantee a derivative. f(x) = |x| is continuous at x = 0 yet has a sharp corner there, so f'(0) doesn't exist. Think of it this way. Continuous means no breaks, differentiable means no breaks AND no corners or vertical tangents. The exam loves the trap answer "continuous, therefore differentiable." That direction is always false.
Continuity at x = c requires three things: f(c) exists, the limit as x approaches c exists, and the limit equals f(c).
Polynomial, rational, power, exponential, logarithmic, and trig functions are continuous everywhere on their domains, so check the domain first.
Differentiable implies continuous, but continuous does not imply differentiable, and f(x) = |x| at x = 0 is the counterexample to memorize.
The three discontinuity types are removable (a hole you can fix by redefining one point), jump (left and right limits disagree), and vertical asymptote (the function blows up).
IVT, MVT, and EVT all require continuity, so explicitly state "f is continuous on the interval" in any FRQ justification that uses them.
For piecewise functions, check continuity at the seam by confirming the left piece, right piece, and function value all match.
A function is continuous at x = c if f(c) is defined, the limit of f(x) as x approaches c exists, and that limit equals f(c). It's continuous on an interval if this holds at every point in the interval.
Yes, |x| is continuous at x = 0 because the limit from both sides equals 0, which matches f(0). It's just not differentiable there, since the difference quotient gives -1 from the left and +1 from the right.
Differentiability is stronger. Every differentiable function is continuous, but a continuous function can fail to be differentiable at corners (like |x| at x = 0) or vertical tangents (like the cube root of x at x = 0).
Redefine the function's value at that point to equal the limit there. If f has a hole at x = 2 where the limit is 5, setting f(2) = 5 fills the hole and makes f continuous at x = 2.
Yes, when you apply IVT, MVT, or EVT. Rubrics award the justification point for verifying hypotheses, so write something like "because H is twice-differentiable, H is continuous on [a, b], so by the MVT..." The 2018 exam's tree-height FRQ tested exactly this move.
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