Find the intervals in which the function $f$ given by $f(x) = 4x^3 - 6x^2 - 72x + 30$ is
$(a)$ increasing
$(b)$ decreasing.

(A) We have $f(x) = 4x^3 - 6x^2 - 72x + 30$.
First,find the derivative $f'(x)$:
$f'(x) = 12x^2 - 12x - 72$
$f'(x) = 12(x^2 - x - 6)$
$f'(x) = 12(x - 3)(x + 2)$
To find the critical points,set $f'(x) = 0$:
$12(x - 3)(x + 2) = 0$
$x = 3$ or $x = -2$.
The points $x = -2$ and $x = 3$ divide the real line into three disjoint intervals: $(-\infty, -2)$,$(-2, 3)$,and $(3, \infty)$.
We test the sign of $f'(x)$ in each interval:
$1$. For $(-\infty, -2)$,choose $x = -3$: $f'(-3) = 12(-3-3)(-3+2) = 12(-6)(-1) = 72 > 0$. Thus,$f$ is increasing.
$2$. For $(-2, 3)$,choose $x = 0$: $f'(0) = 12(0-3)(0+2) = 12(-3)(2) = -72 < 0$. Thus,$f$ is decreasing.
$3$. For $(3, \infty)$,choose $x = 4$: $f'(4) = 12(4-3)(4+2) = 12(1)(6) = 72 > 0$. Thus,$f$ is increasing.

Interval	Sign of $f'(x)$	Nature of function $f$
$(-\infty, -2)$	$f'(x) > 0$	Increasing
$(-2, 3)$	$f'(x) < 0$	Decreasing
$(3, \infty)$	$f'(x) > 0$	Increasing