Question

Suppose the function $f$ is continuous at the point $P(c, f(c))$. Then, the graph of $f$ has a cusp at $P$ if

• A. $\lim_{x \to c^-} f'(x) = +\infty$ and $\lim_{x \to c^+} f'(x) = +\infty$
• B. $\lim_{x \to c^-} f'(x) = -\infty$ and $\lim_{x \to c^+} f'(x) = -\infty$
• C. $\lim_{x \to c^-} f'(x) = +\infty$ and $\lim_{x \to c^+} f'(x) = -\infty$ (or vice versa)
• D. $\lim_{x \to c^-} f'(x) = 0$ and $\lim_{x \to c^+} f'(x) = 1$

Hint:

A classic example of a cusp is $f(x) = x^{2/3}$ at $x=0$. Here, $f'(x) = \frac{2}{3x^{1/3}}$, which goes to $-\infty$ as $x \to 0^-$ and $+\infty$ as $x \to 0^+$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A cusp is a point on a curve where the function is continuous, but has a sharp turn where the tangent lines on either side approach verticality with opposite directions.

Step 2: Detailed Explanation:
For a function $f(x)$ continuous at $x = c$, let us analyze the behavior of its derivative $f'(x)$ as $x$ approaches $c$:
1. If $\lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x) = \pm\infty$, the graph has a vertical tangent line at $x = c$, but not a cusp.
2. If the one-sided derivatives approach infinity with opposite signs: \[ \lim_{x \to c^-} f'(x) = +\infty \text{ and } \lim_{x \to c^+} f'(x) = -\infty \] or: \[ \lim_{x \to c^-} f'(x) = -\infty \text{ and } \lim_{x \to c^+} f'(x) = +\infty \] then the curve rises sharply on one side and falls sharply on the other side, forming a sharp point or "cusp" at $P(c, f(c))$.
At this point, the left-hand and right-hand tangents both become vertical, but they point in opposite directions, making the function non-differentiable at $c$.

Step 3: Final Answer:
The graph has a cusp if the one-sided limits of the derivative approach infinity with opposite signs.