For f'(x) = x^3 - 3x, on which intervals is f increasing?

For f'(x) = x^3 - 3x, on which intervals is f increasing?

• A. (-√3, 0) and (√3, ∞)
• B. (-∞, -√3)
• C. (0, √3)
• D. (-∞, ∞)

Answer: (A)

The function increases where its derivative is positive. For f'(x) = x^3 − 3x, factor as x(x−√3)(x+√3), so the critical points are −√3, 0, and √3. The sign of f' on the intervals between these points is determined by testing values: at x = −2, f' is negative; at x = −1, f' is positive; at x = 1, f' is negative; at x = 2, f' is positive. Therefore, f'(x) > 0 on (−√3, 0) and (√3, ∞). Hence f increases on those intervals.

The function increases where its derivative is positive. For f'(x) = x^3 − 3x, factor as x(x−√3)(x+√3), so the critical points are −√3, 0, and √3. The sign of f' on the intervals between these points is determined by testing values: at x = −2, f' is negative; at x = −1, f' is positive; at x = 1, f' is negative; at x = 2, f' is positive. Therefore, f'(x) > 0 on (−√3, 0) and (√3, ∞). Hence f increases on those intervals.