Find the intervals in which the function $f$ given by $f(x)=2x^{3}-3x^{2}-36x+7$ is
$(a)$ increasing
$(b)$ decreasing

(N/A) The given function is $f(x)=2x^{3}-3x^{2}-36x+7$.
First,we find the derivative of the function:
$f^{\prime}(x) = \frac{d}{dx}(2x^{3}-3x^{2}-36x+7) = 6x^{2}-6x-36$.
Now,factorize the derivative:
$f^{\prime}(x) = 6(x^{2}-x-6) = 6(x-3)(x+2)$.
To find the critical points,set $f^{\prime}(x) = 0$:
$6(x-3)(x+2) = 0 \Rightarrow x = 3, x = -2$.
These points divide the real line into three disjoint intervals: $(-\infty, -2)$,$(-2, 3)$,and $(3, \infty)$.
We test the sign of $f^{\prime}(x)$ in each interval:
$1$. For $x \in (-\infty, -2)$,let $x = -3$: $f^{\prime}(-3) = 6(-3-3)(-3+2) = 6(-6)(-1) = 36 > 0$. Thus,$f$ is strictly increasing.
$2$. For $x \in (-2, 3)$,let $x = 0$: $f^{\prime}(0) = 6(0-3)(0+2) = 6(-3)(2) = -36 < 0$. Thus,$f$ is strictly decreasing.
$3$. For $x \in (3, \infty)$,let $x = 4$: $f^{\prime}(4) = 6(4-3)(4+2) = 6(1)(6) = 36 > 0$. Thus,$f$ is strictly increasing.
Conclusion:
$(a)$ The function is strictly increasing in $(-\infty, -2) \cup (3, \infty)$.
$(b)$ The function is strictly decreasing in $(-2, 3)$.