AP Calculus AB and BC

Corner vs Cusp

Q: Is x to the two thirds a corner or a cusp at zero?

A cusp. Its derivative \frac{2}{3x^{1/3}} has no finite one-sided values at 0: it goes to -\infty from the left and +\infty from the right. A corner would need both of those to be actual numbers.

Q: Does a corner mean the function is discontinuous there?

Corners leave continuity intact. The graph of \lvert x \rvert is unbroken everywhere, corner included. Differentiability forces continuity, but continuity does not force differentiability, and corners and cusps are exactly where that gap shows.

The one-sided slopes separate them: a corner has two different finite slopes, while a cusp has slopes running to positive infinity on one side and negative infinity on the other. Both keep the graph continuous, and at both the derivative fails to exist, so both are critical numbers.

Corner

Use when: The one-sided derivatives both exist as finite numbers and disagree, which is what absolute values and piecewise definitions usually produce.

Cusp

Use when: The one-sided derivatives both grow without bound in opposite directions, so the two branches pinch together against a vertical tangent line.

Side by side

	Corner	Cusp
One-sided slopes	Two different finite numbers	One runs to $+\infty$ , the other to $-\infty$
Tangent line at the point	None, since the two half-tangents have different slopes	Vertical, and the same vertical line is approached from both sides
Standard example	$f(x) = \lvert x \rvert$ at $x = 0$	$f(x) = x^{2/3}$ at $x = 0$
Under a graphing zoom	Stays a fixed kink, two straight edges at set slopes	The branches keep steepening and squeeze toward vertical
Usual source	Absolute values and piecewise formulas	Even-numerator fractional powers below $1$ , such as $x^{2/3}$

Both points pass the continuity test and fail the differentiability test, so everything that separates them lives in the one-sided limits of the difference quotient. At a corner those two limits are finite and disagree: approaching $x = 0$ along $y = \lvert x \rvert$ gives slope $-1$ from the left and $+1$ from the right, and no single number can be the derivative. At a cusp there are no finite numbers to compare at all.

\lim_{x \to 0^{-}} f'(x) = -1, \ \lim_{x \to 0^{+}} f'(x) = 1 \qquad \text{versus} \qquad \lim_{x \to 0^{\pm}} \frac{2}{3x^{1/3}} = \pm\infty

Opposite directions are what make a cusp a cusp. For $f(x) = x^{2/3}$ the derivative $\frac{2}{3x^{1/3}}$ is large and negative just left of $0$ and large and positive just right of it, so the curve drops in, turns, and climbs back out against a vertical tangent. Compare $f(x) = x^{1/3}$ , whose derivative $\frac{1}{3x^{2/3}}$ runs to $+\infty$ from both sides: that graph also has no derivative at $0$ , but it passes straight through, which is a vertical tangent and not a cusp. In all three cases $x = 0$ is a critical number, because $f$ is defined there and $f'$ is not.

The reason has to match the point

A justification that says the derivative fails to exist at a cusp because the left and right slopes are different numbers has borrowed the corner argument, and there are no numbers there to be different. Say instead that both one-sided slopes are unbounded. Keep the different finite slopes wording for a corner such as $\lvert x \rvert$ at $0$ , where those slopes are genuinely $-1$ and $1$ . Graders read the reason, not only the verdict, and the two verdicts happen to agree.

Frequently asked questions

Is x to the two thirds a corner or a cusp at zero?

A cusp. Its derivative $\frac{2}{3x^{1/3}}$ has no finite one-sided values at $0$ : it goes to $-\infty$ from the left and $+\infty$ from the right. A corner would need both of those to be actual numbers.

Does a corner mean the function is discontinuous there?

Corners leave continuity intact. The graph of $\lvert x \rvert$ is unbroken everywhere, corner included. Differentiability forces continuity, but continuity does not force differentiability, and corners and cusps are exactly where that gap shows.

Is a vertical tangent the same thing as a cusp?

Not quite. A cusp has a vertical tangent, but a vertical tangent alone only needs both one-sided slopes to be infinite, and $x^{1/3}$ sends both to $+\infty$ while the graph passes smoothly through. The cusp is the case where the two signs disagree.

In the CED: Unit 2: Defining the Derivative