QPaperGen

Differentiate the function:

\[f'(x) = \dfrac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36\]

Factor the derivative:

\[f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\]

Set \(f'(x) = 0\) to find critical points:
\[6(x + 2)(x - 3) = 0 \Rightarrow x = -2,,3\]

Now analyze the sign of \(f'(x)\) in the intervals:
\[\]

In \((-\infty, -2)\):
Both \((x + 2) < 0\) and \((x - 3) < 0 \Rightarrow f'(x) > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \((-\infty, -2)\)
\[\]

In \((-2, 3)\):
\((x + 2) > 0\) and \((x - 3) < 0 \Rightarrow f'(x) < 0\)
\(\Rightarrow f(x)\) is strictly decreasing on \((-2, 3)\)
\[\]

In \((3, \infty)\):
Both \((x + 2) > 0\) and \((x - 3) > 0 \Rightarrow f'(x) > 0\)
\(\Rightarrow f(x)\) is strictly increasing on \((3, \infty)\)
\[\]

Final Answer:

\(f(x)\) is strictly increasing on \((-\infty, -2)\) and \((3, \infty)\) and\[\]

\(f(x)\) is strictly decreasing on \((-2, 3)\)