For the function f(x) = (x - 2)^3 + 1 , x ∈ R , the point x = 2 is

The correct answer is Option (3) → neither a point of local maximum nor a point of local minimum

Differentiating:

    \(f'(x) = 3(x - 2)^2\)

So f’(2) = 0 and x = 2 is a critical point. Here \(f''(x) = 6(x - 2)\) gives f’’(2) = 0, so the second derivative test does not decide. By the first derivative test:

    \(x \lt 2\): \(f'(x) = 3(x - 2)^2 \gt 0\)
    \(x \gt 2\): \(f'(x) = 3(x - 2)^2 \gt 0\)

The sign of f’(x) does not change at x = 2; f is increasing on both sides. So x = 2 is neither a point of local maximum nor a point of local minimum (it is a point of inflection).