This question is designed to be answered with a calculator.Let f(x) = (1 − x2)2/3. Which type of non-differentiable point exists at x = 1?

At x = 1, the expression inside the power is 1 − 1² = 0, so f(1) = 0. Near x = 1, 1 − x² behaves like −2(x − 1), making f(x) behave like a positive constant times |x − 1|^(2/3). The slopes become unbounded with opposite signs on the two sides, which gives a cusp, not a vertical tangent.

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