The function f(x) = x^3 +3x^2 + 4x +4, x ∈ R (set of real numbers):

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

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

Derivative: $f'(x) = 3x^2 + 6x + 4 = 3(x^2 + 2x + \frac{4}{3}) = 3(x+1)^2 + 1$

$f'(x) = 3(x+1)^2 + 1 > 0$ for all $x \in \mathbb{R}$

Since derivative is always positive, $f(x)$ is strictly increasing on $\mathbb{R}$

Answer: is increasing on $\mathbb{R}$