Find stationary points of f(x) = x^3 - 3x and classify using f' and f''. Which statement is correct?

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Find stationary points of f(x) = x^3 - 3x and classify using f' and f''. Which statement is correct?

The main idea is that stationary points occur where the derivative is zero, and we use the second derivative to tell whether each point is a maximum or a minimum. Compute the derivative: f'(x) = 3x^2 - 3 = 3(x^2 - 1). Setting it to zero gives x^2 = 1, so x = -1 and x = 1 are stationary points. Classify them with the second derivative: f''(x) = 6x. At x = -1, f''(-1) = -6 < 0, so a local maximum. At x = 1, f''(1) = 6 > 0, so a local minimum. Thus the correct description is that there are stationary points at x = -1 and x = 1, with x = -1 a maximum and x = 1 a minimum. The other possibilities fail because there are no stationary points at 0 or at ±√2, and there are indeed two stationary points.

The main idea is that stationary points occur where the derivative is zero, and we use the second derivative to tell whether each point is a maximum or a minimum.

Compute the derivative: f'(x) = 3x^2 - 3 = 3(x^2 - 1). Setting it to zero gives x^2 = 1, so x = -1 and x = 1 are stationary points.

Classify them with the second derivative: f''(x) = 6x. At x = -1, f''(-1) = -6 < 0, so a local maximum. At x = 1, f''(1) = 6 > 0, so a local minimum.

Thus the correct description is that there are stationary points at x = -1 and x = 1, with x = -1 a maximum and x = 1 a minimum. The other possibilities fail because there are no stationary points at 0 or at ±√2, and there are indeed two stationary points.

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