The function f(x) = x^3 - 3x^2 - 9x + 4 has a local maximum at

The correct answer is Option (1) → \(x = -1\)

The critical points are where f’(x) = 0:

    \(f'(x) = 3x^2 - 6x - 9\)
    \(f'(x) = 3(x^2 - 2x - 3)\)
    \(f'(x) = 3(x - 3)(x + 1)\)

So x = −1 or x = 3. The second derivative test:

    \(f''(x) = 6x - 6\)
    \(f''(-1) = -12 \lt 0\)
    \(f''(3) = 12 \gt 0\)

So f has a local maximum at x = −1 and a local minimum at x = 3.

Why the other options are incorrect

• Option (2) \(x = 3\) — Incorrect. Here \(f''(3) = 12 \gt 0\), so x = 3 is the point of local minimum.

• Option (3) \(x = 1\) — Incorrect. \(f'(1) = 3 - 6 - 9 = -12 \ne 0\), so x = 1 is not a critical point.

• Option (4) \(x = -3\) — Incorrect. \(f'(-3) = 27 + 18 - 9 = 36 \ne 0\), so x = −3 is not a critical point.