Question

The function \( f(x) = 2x^3 - 15x^2 + 36x + 6 \) is strictly decreasing in the interval:

• A. \( (2, 3) \)
• B. \( (-\infty, 2) \)
• C. \( (3, 4) \)
• D. \( (-\infty, 3) \cup (4, \infty) \)
• E. \( (-\infty, 2) \cup (3, \infty) \)

Hint:

To quickly find where a quadratic $(x-a)(x-b) < 0$ (assuming $a < b$), remember it's always the region "between" the roots. For $> 0$, it's the region "outside" the roots.

Answer 1

The Correct Option is A

Concept: A function \( f(x) \) is strictly decreasing in an interval where its first derivative \( f'(x) \) is strictly less than zero (\( f'(x) < 0 \)).

Step 1: Find the first derivative \( f'(x) \).
\[ f(x) = 2x^3 - 15x^2 + 36x + 6 \] Differentiating with respect to \( x \): \[ f'(x) = 6x^2 - 30x + 36 \]

Step 2: Set the derivative to be less than zero.
\[ 6x^2 - 30x + 36 < 0 \] Divide the entire inequality by 6: \[ x^2 - 5x + 6 < 0 \]

Step 3: Factorize and find the interval.
Factor the quadratic expression: \[ (x - 2)(x - 3) < 0 \] For the product of two terms to be negative, one must be positive and the other negative. This occurs when \( x \) is between the roots: \[ 2 < x < 3 \] Thus, the function is strictly decreasing in the interval \( (2, 3) \).