The function f(x) = x^2 - 4x;\ x ∈ [0, 4] attains minimum value at:

The correct answer is Option (3) → x = 2

The critical point:

    \( f'(x) = 2x - 4 = 0 \)

    \( x = 2 \)

On a closed interval, compare f at the critical point and at the end points:

    \( f(0) = 0 \)
    \( f(2) = 4 - 8 = -4 \)
    \( f(4) = 16 - 16 = 0 \)

The least value is −4, attained at x = 2.

Shortcut: \(f(x) = (x - 2)^2 - 4\), which is least when the square is zero, at x = 2.