Question

Find the interval in which the function \( f(x) = 2x^3 - 3x^2 - 36x + 7 \) is strictly increasing:

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

Hint:

Always check whether the question specifies "strictly increasing" (\( f'(x) > 0 \)) or simply "increasing" (\( f'(x) \ge 0 \)) to avoid choosing an option with incorrect bracket notations.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A function is strictly increasing if its derivative $f'(x)>0$.
Step 2: Detailed Explanation:
$f(x) = 2x^3 - 3x^2 - 36x + 7$.
$f'(x) = 6x^2 - 6x - 36$.
Set $6x^2 - 6x - 36>0$.
Divide by 6: $x^2 - x - 6>0$.
$(x-3)(x+2)>0$.
Using the wavy curve method, the expression is positive for $x \in (-\infty, -2) \cup (3, \infty)$.
Step 3: Final Answer:
The interval is $(-\infty, -2) \cup (3, \infty)$.