求微积分中cusp与corner的正式严谨定义及函数实例辨析

Great question—these terms can be surprisingly tricky because different textbooks sometimes use slightly varying definitions, but let's break down the standard rigorous ones used in most calculus analysis contexts.

First, a foundational rule for both points: the function must be continuous at the point in question—discontinuities never qualify as cusps or corners.

Corner #

A point (x = a) is a corner of (f(x)) if:

• (f) is continuous at (x = a),
• The left-hand derivative (f'_-(a)) exists (is a finite real number),
• The right-hand derivative (f'_+(a)) exists (is a finite real number),
• And (f'-(a) \neq f'+(a)).

In plain language: the function has two distinct, finite tangent lines approaching the point from left and right, creating a sharp angle but not a vertical spike. A classic example is (f(x) = |x|) at (x=0)—the left derivative is (-1), the right derivative is (1), both finite but unequal.

Quick correction on your example (f(x) = |x^2 - 1|) at (x=1): let’s compute the one-sided derivatives. For (x > 1), (f(x) = x^2 - 1), so (f'(x) = 2x) and (f'+(1) = 2). For (x < 1), (f(x) = 1 - x^2), so (f'(x) = -2x) and (f'-(1) = -2). Both are finite and unequal, so this is actually a corner, not a cusp—easy mix-up, no worries!

Cusp #

A point (x = a) is a cusp of (f(x)) if:

• (f) is continuous at (x = a),
• The limits of the derivative as (x) approaches (a) from the left and right are infinite, but have opposite signs:
  • Either (\lim_{x \to a^-} f'(x) = +\infty) and (\lim_{x \to a^+} f'(x) = -\infty),
  • Or (\lim_{x \to a^-} f'(x) = -\infty) and (\lim_{x \to a^+} f'(x) = +\infty).

Geometrically, this means the function’s graph pinches into a sharp point where the tangent lines on both sides approach vertical, but one side shoots upward toward infinity and the other shoots downward.

Now let’s analyze your (g(x) = \sqrt[3]{x^2} = x^{2/3}). First, compute the derivative: (g'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}).

• As (x \to 0^+), (\sqrt[3]{x}) approaches (0^+), so (g'(x) \to +\infty).
• As (x \to 0^-), (\sqrt[3]{x}) approaches (0^-), so (g'(x) \to -\infty).

Since (g(x)) is continuous at (x=0) (and (g(0)=0)), this fits the strict definition of a cusp.

Quick Distinction Cheat Sheet #

• Corner: Finite, unequal left/right derivatives → sharp angle with two finite slopes.
• Cusp: Infinite left/right derivatives with opposite signs → vertical "pinch" where slopes blow up in opposite directions.

内容的提问来源于stack exchange，提问作者S.H.W