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:

For any upward-facing parabola like $x^2 - 5x - 24$, the values are always strictly negative strictly between the two roots.

Answer 1

The Correct Option is B

Step 1: Recall the test.
A function is strictly decreasing where its first derivative is negative, $f'(x) < 0$.

Step 2: Differentiate.
\[ f(x) = 2x^3 - 15x^2 - 144x - 7 \] \[ f'(x) = 6x^2 - 30x - 144 \]
Step 3: Simplify.
Take out 6.
\[ f'(x) = 6(x^2 - 5x - 24) \]
Step 4: Factor.
\[ x^2 - 5x - 24 = (x - 8)(x + 3) \] So $f'(x) = 6(x - 8)(x + 3)$.

Step 5: Find where it is negative.
The product $(x-8)(x+3)$ is negative between its roots $-3$ and $8$.

Step 6: Conclusion.
So $f$ is strictly decreasing on $(-3, 8)$. \[ \boxed{(-3,\ 8) \text{ (Option 2)}} \]