Question

Find the open interval across which the cubic polynomial function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) is classified as strictly decreasing.

• A. \( (-2, 3) \)
• B. \( (-\infty, -2) \cup (3, \infty) \)
• C. \( (-3, 2) \)
• D. \( (0, \infty) \)

Hint:

When evaluating quadratic inequalities of the form \( (x-a)(x-b) < 0 \) where \( a < b \), the solution space will always be the interior bounded range \( (a, b) \).

Answer 1

The Correct Option is A

Concept: A continuous function \( f(x) \) is strictly decreasing across an open interval if its first derivative is strictly negative (\( f'(x) < 0 \)) at every point within that interval.

Step 1: Calculate the first derivative of the polynomial function.
Apply the standard power derivative rule to each term of the polynomial: \[ f'(x) = \frac{d}{dx}\left( 2x^3 - 3x^2 - 36x + 7 \right) = 6x^2 - 6x - 36 \]

Step 2: Find the critical roots by factoring the quadratic derivative expression.
Set the derivative expression equal to zero to isolate the boundary roots: \[ 6x^2 - 6x - 36 = 0 \quad \Rightarrow \quad 6\left( x^2 - x - 6 \right) = 0 \] Factor the quadratic trinomial inside the parentheses: \[ 6(x - 3)(x + 2) = 0 \] This isolates the critical turning points at \( x = 3 \) and \( x = -2 \).

Step 3: Analyze the derivative signs using the wavy curve method.
The critical points divide the real number line into three separate intervals: \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \).

• For \( x \in (-\infty, -2) \), \( f'(x) > 0 \) (Strictly Increasing)
• For \( x \in (-2, 3) \), \( f'(x) < 0 \) (Strictly Decreasing)
• For \( x \in (3, \infty) \), \( f'(x) > 0 \) (Strictly Increasing)

Thus, the function is strictly decreasing on the open interval \( (-2, 3) \).