Question

If \( f(x) = 2x^3 - 15x^2 - 144x - 7 \), then \( f(x) \) is strictly decreasing in:

• A. \( (-8, 3) \)
• B. \( (-3, 8) \)
• C. \( (3, 8) \)
• D. \( (-8, -3) \)

Hint:

To find where a function is strictly decreasing, solve \( f'(x) = 0 \) for critical points, test intervals, and check the sign of \( f'(x) \).

Answer 1

The Correct Option is B

To determine where \( f(x) \) is strictly decreasing, we analyze the derivative \( f'(x) \).
The derivative is: \[ f'(x) = \frac{d}{dx}(2x^3 - 15x^2 - 144x - 7). \]
Step 1: Compute \( f'(x) \)
Differentiate term by term: \[ f'(x) = 6x^2 - 30x - 144. \]
Step 2: Solve \( f'(x) = 0 \)
Factorize \( f'(x) \) to find critical points: \[ 6x^2 - 30x - 144 = 0. \]
Divide through by 6: \[ x^2 - 5x - 24 = 0. \]
Factorize: \[ (x - 8)(x + 3) = 0. \]
Thus, the critical points are: \[ x = -3, \quad x = 8. \]
Step 3: Analyze the intervals
The critical points divide the real line into three intervals: \( (-\infty, -3) \), \( (-3, 8) \), and \( (8, \infty) \).
Test the sign of \( f'(x) \) in each interval:

• For \( x \in (-\infty, -3) \), choose \( x = -4 \): \[ f'(-4) = 6(-4)^2 - 30(-4) - 144 = 96 + 120 - 144 = 72 > 0. \] \( f'(x) > 0 \), so \( f(x) \) is increasing.
• For \( x \in (-3, 8) \), choose \( x = 0 \): \[ f'(0) = 6(0)^2 - 30(0) - 144 = -144 < 0. \] \( f'(x) < 0 \), so \( f(x) \) is decreasing.
• For \( x \in (8, \infty) \), choose \( x = 9 \): \[ f'(9) = 6(9)^2 - 30(9) - 144 = 486 - 270 - 144 = 72 > 0. \] \( f'(x) > 0 \), so \( f(x) \) is increasing.

Step 4: Conclusion
\( f(x) \) is strictly decreasing in the interval \( (-3, 8) \).
Final Answer: \[ \boxed{(-3, 8)} \]