Find the interval of increasing, decreasing, concave up and down for \[f(x) = 2{x^3} - 3{x^2} - 36x - 7\] .

Hint: The given problem revolves around the nature of the curve, its slope and concavity. The increasing and decreasing nature of the curve is judged by its first derivative as it gives an idea about the slope of the curve, Concavity of a curve is judged by the second derivative as it gives an idea about the rate of change of slope of the curve.

Complete step-by-step answer:
Consider, \[f(x) = 2{x^3} - 3{x^2} - 36x - 7\] .
So, to find the increasing or decreasing nature of a curve, we have to find the variation of slope of the curve. To find critical points, we need to find the first order derivative of f(x).
 $ f'(x) = 2(3{x^2}){\text{ - 3(2x) - 36}} $
= $ f'(x) = 6{x^2}{\text{ - 6x - 36}} $
Now equate the derivative function equal to zero to find critical points of a function where the function actually changes its nature from increasing to decreasing or vice versa.
 $ f'(x) = 6{x^2}{\text{ - 6x - 36}} = 0 $
= $ {\text{ }}{x^2}{\text{ - x - 6 = 0}} $
Factorise, by splitting of middle term
= $ {x^2} + x( - 3 + 2) - 6 = 0 $
= $ {x^2} - 3x + 2x - 6 = 0 $
= \[x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\]
= $ (x - 3)(x + 2) = 0 $
Either \[\left( {x-3} \right) = 0\] or \[\left( {x + 2} \right) = 0\] .
So Either \[x = 3\] or \[x = - 2\] .
So, critical points are $ x = - 2,3 $ .
These two values of x divide R (Real number set) into three disjoint intervals, namely $ \left( { - \infty , - 2} \right) $ , $ ( - 2,3) $ and $ (3,\infty ) $ .
Now we observe sign of the first derivative $ f'(x) $ in these intervals.
Now, $ x \in \left( { - \infty , - 2} \right) $ or \[x \in \left( { - 2,3} \right)\] or \[x \in \left( {3,\infty } \right)\]
For $ x \in \left( { - \infty , - 2} \right) $ $ f'(x) > 0 $ . Thus the given function f(x) is increasing in this interval.
For \[x \in \left( { - 2,3} \right)\] $ f'(x) < 0 $ . Thus the given function f(x) is decreasing in this interval.
For \[x \in \left( {3,\infty } \right)\] $ f'(x) > 0 $ . Thus the given function f(x) is increasing in this interval.
Now for finding the concavity of the curve, we have to find the second derivative of the function.
 $ \dfrac{{d\left( {f'(x)} \right)}}{{dx}} = 12x - 6 $
Equating it to zero for finding the point where the slope of rate of change of the given function becomes zero.
So, $ \dfrac{{d\left( {f'(x)} \right)}}{{dx}} = 12x - 6 = 0 $
 $ = x = \dfrac{1}{2} $ .
So, the second derivative of the function for $ x \in \left( { - \infty ,\dfrac{1}{2}} \right) $ is negative and for $ x \in \left( {\dfrac{1}{2},\infty } \right) $ is positive.
So, the curve is concave upwards for $ x \in \left( {\dfrac{1}{2},\infty } \right) $ and concave downwards for $ x \in \left( { - \infty ,\dfrac{1}{2}} \right) $ .

Note: The problem is a question related to the nature of curve, whether it is increasing or decreasing based on applications of derivatives. It also requires the knowledge of concavity of a curve where we have to determine whether a curve is concave upward or concave downward using algebraic analysis.