In AP Calculus, a corner is a point where a function is continuous but not differentiable because the slope approaching from the left differs from the slope approaching from the right, like the sharp point of y = |x| at x = 0.
A corner is a sharp turn in a graph. The function is continuous there (no break, no jump), but the slope coming in from the left is different from the slope leaving on the right. Since the two one-sided derivatives disagree, the derivative doesn't exist at that point. The classic example is the absolute value function. For f(x) = |x - 2|, the slope is -1 for every x left of 2 and +1 for every x right of 2, so at x = 2 the graph snaps from one straight line to another and f'(2) does not exist.
Here's the intuitive test. If you can't draw exactly one tangent line at the point, the function isn't differentiable there. At a corner, you could lay a tangent line along the left piece or along the right piece, and they're different lines. Corners are the main reason "continuous" and "differentiable" are not the same thing. Every differentiable function is continuous, but a corner is living proof that the reverse fails.
Corners live in Topic 5.1, Using the Mean Value Theorem, in Unit 5 (Analytical Applications of Differentiation). Learning objective 5.1.A asks you to justify conclusions by applying the MVT, and essential knowledge FUN-1.B.1 spells out the two hypotheses you must check first. The function has to be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). A corner is the exam's favorite way to break the second hypothesis quietly. The function looks perfectly fine, it's continuous everywhere, but that one sharp point inside the interval kills differentiability, so the MVT guarantees nothing. Recognizing corners is how you avoid applying the MVT (or Rolle's Theorem) when you're not allowed to.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryDifferentiable (Unit 5)
A corner is the textbook example of a point that is continuous but not differentiable. If a question asks whether a function is differentiable on an interval, scan for absolute values and piecewise joins where the slopes don't match.
Absolute Value Function (Unit 5)
Functions like f(x) = |x - 2| are corner factories. The corner sits wherever the expression inside the absolute value equals zero, which is exactly where AP questions plant the trap.
Using the Mean Value Theorem (Unit 5)
The MVT requires differentiability on the open interval. One corner inside (a, b) means the hypotheses fail and you cannot use the theorem to guarantee a point c, even if such a point happens to exist anyway.
Rolle's Theorem (Unit 5)
Rolle's Theorem is the special case of the MVT where f(a) = f(b), and it needs the same differentiability hypothesis. f(x) = |x| on [-1, 1] has equal endpoint values but no point where f'(c) = 0, precisely because of the corner at x = 0.
Corners show up almost entirely as hypothesis-checkers. A common multiple-choice stem gives you several functions and asks which one is differentiable over an interval like (0, 2), and the wrong answers usually hide a corner from an absolute value or a piecewise function whose pieces meet at different slopes. Another standard setup hands you f(x) = |x - 2| on an interval like [-1, 5] or [0, 4] and asks whether the MVT applies, or which hypothesis fails and why. Your job is to (1) find where the corner is, (2) check whether it sits inside the open interval, and (3) write the justification in CED language, something like "f is continuous on [0, 4] but not differentiable at x = 2, so the Mean Value Theorem cannot be applied." No released FRQ has used the word "corner" verbatim, but FRQ justifications that invoke the MVT must state both hypotheses, and a corner is the standard reason the differentiability hypothesis fails.
Both are points where a function is continuous but not differentiable, and both break the MVT's differentiability hypothesis the same way. The difference is the slope behavior. At a corner, the one-sided slopes are two different finite numbers (like -1 and +1 for |x|). At a cusp, the slopes blow up, approaching +infinity from one side and -infinity from the other, like x^(2/3) at x = 0. A corner looks like a sharp angle; a cusp looks like a needle point with vertical tangent behavior.
A corner is a point where a function is continuous but not differentiable because the left-hand and right-hand slopes are different finite values.
The standard example is f(x) = |x - 2|, which has a corner at x = 2 where the slope jumps from -1 to +1.
A corner inside the open interval (a, b) breaks the differentiability hypothesis of the Mean Value Theorem, so the MVT cannot be applied (FUN-1.B.1).
Corners prove that continuity does not imply differentiability, even though differentiability always implies continuity.
A corner has two different finite one-sided slopes, while a cusp has one-sided slopes that approach positive and negative infinity.
On MVT justification questions, you must check both hypotheses, and naming the corner point is how you explain why differentiability fails.
A corner is a point where a graph is continuous but not differentiable because the slope from the left and the slope from the right are different finite numbers. The sharp point of y = |x| at x = 0 is the classic example.
Yes. The graph has no break or jump at a corner, so the function is continuous there. What fails is differentiability, since the two one-sided derivatives don't match.
No, not if the corner is inside the open interval (a, b). The MVT requires differentiability on (a, b), so for f(x) = |x - 2| on [0, 4], the corner at x = 2 means the MVT's hypotheses fail and the theorem guarantees nothing.
At a corner the one-sided slopes are two different finite values, like -1 and +1 for |x|. At a cusp the one-sided slopes shoot off to +infinity and -infinity, like x^(2/3) at x = 0. Both make the function non-differentiable at that point.
Set the expression inside the absolute value equal to zero. For f(x) = |x - 2|, solving x - 2 = 0 gives x = 2, which is exactly where the corner sits and where the derivative does not exist.