The function f(x) = 4-3x+3x^2-x^3 is (Here R is set of real numbers)

The correct answer is Option (1) → decreasing on R

The sign of the derivative decides where f increases or decreases:

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

Since \((x - 1)^2 \ge 0\), \(f'(x) \le 0\) for every real x, and \(f'(x) = 0\) only at the single point x = 1.

A derivative that is negative everywhere except at an isolated point still makes f decreasing, so f is decreasing on R. In particular, it is decreasing (not increasing) on \((0, \infty)\) and decreasing on \((-\infty, 0)\).