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

Concept: A function \( f(x) \) is strictly increasing in intervals where its first derivative is strictly positive, meaning \( f'(x) > 0 \).

Step 1: Find the first derivative of the function \( f'(x) \). Differentiating with respect to \( x \) using the power rule: \[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) \] \[ f'(x) = 6x^2 - 6x - 36 \]

Step 2: Factorize the derivative expression to find the critical points. Set up the inequality for a strictly increasing condition: \[ 6x^2 - 6x - 36 > 0 \] Divide by the common scalar factor \( 6 \): \[ x^2 - x - 6 > 0 \] Splitting the middle term to factorize: \[ (x - 3)(x + 2) > 0 \] The critical points are \( x = -2 \) and \( x = 3 \).

Step 3: Apply the wavy curve method to determine the signs across intervals. Plotting the critical points on a number line creates three distinct intervals:
• For \( x \in (-\infty, -2) \): Both factors are negative, so their product is positive (\( > 0 \)).
• For \( x \in (-2, 3) \): One factor is positive and one is negative, so their product is negative (\( < 0 \)).
• For \( x \in (3, \infty) \): Both factors are positive, so their product is positive (\( > 0 \)). Since we need \( f'(x) > 0 \), the function is strictly increasing on: \[ x \in (-\infty, -2) \cup (3, \infty) \]