What are the critical points of f(x) = x³ - 3x + 2?

Question

Accepted Answer

x = 1 and x = -1. The correct answer is x = 1 and x = -1. Critical points occur where f'(x) = 0 or is undefined. Here f'(x) = 3x² - 3 = 3(x² - 1) = 3(x - 1)(x + 1). Setting this equal to zero gives x = 1 and x = -1. Since f' is a polynomial, it is defined everywhere, so these are the only critical points.

The value x = 0 is not a critical point because f'(0) = -3 ≠ 0; the derivative is nonzero at x = 0, so the function is neither at a maximum nor minimum there. The value x = 3 is obtained by confusing the coefficient in the derivative equation with a solution; it does not satisfy 3x² - 3 = 0. The values x = 2 and x = -2 do not satisfy the derivative equation either; substituting gives f'(±2) = 3(4) - 3 = 9 ≠ 0.

Finding critical points is a prerequisite for optimization problems, which are among the most heavily tested applications of derivatives on the CLEP exam. The process always starts with computing f'(x), setting it equal to zero, and solving. Students should then use the first or second derivative test to classify each critical point as a local maximum, local minimum, or neither.

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