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

decreasing on R

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

Given function: $f(x) = 4 - 3x + 3x^2 - x^3$

Derivative: $f'(x) = -3 + 6x - 3x^2 = -3(x^2 - 2x + 1) = -3(x - 1)^2$

Since $(x - 1)^2 \ge 0$ for all real $x$, $f'(x) \le 0$ for all $x \in \mathbb{R}$.

Therefore, the function is decreasing for all real $x$ (never increasing).

Final Answer:

decreasing on R