Determine the intervals on which the function f(x) = x^3 - 3x^2 + 4 i…

The correct answer is Option (2) → Increasing on $(-\infty, 0) \cup (2, \infty)$; Decreasing on $(0, 2)$ ##

The derivative:

$ f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 4) = 3x^2 - 6x $

The derivative is equal to zero to find critical points:

$ 3x^2 - 6x = 0 $

Factor the equation: $3x(x - 2) = 0$

So, $x = 0$ and $x = 2$ are critical points.

The sign of $f'(x)$ in the intervals determined by these critical points: $(-\infty, 0), (0, 2)$ and $(2, \infty)$.

$\textbf{For $x \in (-\infty, 0)$, $x = -1$:}$

$ f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 > 0 $

$f(x)$ is increasing in $(-\infty, 0)$.

$\textbf{For $x \in (0, 2)$, $x = 1$:}$

$ f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 < 0 $

$f(x)$ is decreasing in $(0, 2)$.

$\textbf{For $x \in (2, \infty)$, $x = 3$:}$

$ f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 > 0 $

$f(x)$ is increasing in $(2, \infty)$.

Hence, $f(x)$ is increasing in $(-\infty, 0) \cup (2, \infty)$ and $f(x)$ is decreasing in $(0, 2)$.