Question

The function \( f(x) = 2x^3 - 3x^2 - 36x + 28 \) is increasing in

• A. \( (-\infty,-1] \cup [3,\infty) \)
• B. \( (-\infty,-2] \cup [3,\infty) \)
• C. \( (-\infty,-2] \cup [5,\infty) \)
• D. \( (-\infty,-5] \cup [3,\infty) \)
• E. \( (-\infty,-2] \cup [8,\infty) \)

Hint:

Factor derivative completely, then use sign chart.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A function is increasing on an interval where its first derivative is non-negative (\(f'(x) \ge 0\)). To find these intervals, we find the critical points by setting the first derivative to zero, and then we test the intervals between these points.
Step 2: Key Formula or Approach:
1. Find the first derivative, \(f'(x)\). 2. Find the critical points by solving \(f'(x) = 0\). 3. Analyze the sign of \(f'(x)\) in the intervals defined by the critical points. 4. The function is increasing where \(f'(x) \ge 0\).
Step 3: Detailed Explanation:
1. Find the first derivative. The function is \(f(x) = 2x^3 - 3x^2 - 36x + 28\). \[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 28) \] \[ f'(x) = 6x^2 - 6x - 36 \] 2. Find the critical points. Set the derivative to zero: \[ 6x^2 - 6x - 36 = 0 \] Divide the entire equation by 6 to simplify: \[ x^2 - x - 6 = 0 \] Factor the quadratic equation. We need two numbers that multiply to -6 and add to -1. These are -3 and 2. \[ (x - 3)(x + 2) = 0 \] The critical points are \(x = 3\) and \(x = -2\). 3. Analyze the sign of f'(x). The derivative \(f'(x) = 6(x-3)(x+2)\) is a parabola opening upwards. It will be positive outside its roots and negative between them.
Interval \((-\infty, -2)\): Let's test \(x = -3\). \(f'(-3) = 6(-3-3)(-3+2) = 6(-6)(-1) = 36\) (Positive, so f is increasing).
Interval \((-2, 3)\): Let's test \(x = 0\). \(f'(0) = 6(0-3)(0+2) = 6(-3)(2) = -36\) (Negative, so f is decreasing).
Interval \((3, \infty)\): Let's test \(x = 4\). \(f'(4) = 6(4-3)(4+2) = 6(1)(6) = 36\) (Positive, so f is increasing).
4. Identify the increasing intervals. The function is increasing where \(f'(x) \ge 0\). This occurs in the intervals \((-\infty, -2]\) and \([3, \infty)\). The union of these intervals is the answer.
Step 4: Final Answer:
The function is increasing in \((-\infty, -2] \cup [3, \infty)\).