Let f(x)=x^3-6x^2+12x-3 , then at x=2 , f(x) has:

The correct answer is Option (4) → neither a maximum nor a minimum

Differentiating:

    \(f'(x) = 3x^2 - 12x + 12\)
    \(= 3(x^2 - 4x + 4)\)
    \(= 3(x-2)^2\)

So f′(2) = 0 and x = 2 is a critical point. But \(f'(x) = 3(x-2)^2 \gt 0\) on both sides of x = 2: f′ does not change sign. By the first derivative test, x = 2 is neither a point of local maximum nor a point of local minimum; f is increasing through it (it is a point of inflexion).

Tip: Here \(f''(x) = 6x - 12\) gives f′′(2) = 0. A zero second derivative means the second derivative test fails; it does not by itself mean “neither”. Decide with the sign of f′ on the two sides.