Question

The function \( f(x) = 6x^4 - 3x^2 - 5 \) is increasing in the set:

• A. \( (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, 1) \)
• B. \( (-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty) \)
• C. \( (-\frac{1}{2}, \frac{1}{2}) \)
• D. \( (-\infty, \frac{1}{2}) \)
• E. \( (-\infty, -\frac{1}{2}) \cup (\frac{1}{2}, \infty) \)

Hint:

For increasing/decreasing functions, find critical points by setting \( f'(x) = 0 \) and analyze sign changes.

Answer 1

The Correct Option is B

Step 1: Find \( f'(x) \) \[ f'(x) = \frac{d}{dx} (6x^4 - 3x^2 - 5). \] \[ = 24x^3 - 6x. \] \[ = 6x (4x^2 - 1). \] \[ = 6x (2x - 1)(2x + 1). \] Step 2: Find critical points Setting \( f'(x) = 0 \): \[ 6x (2x - 1)(2x + 1) = 0. \] \[ x = 0, \quad x = \frac{1}{2}, \quad x = -\frac{1}{2}. \] Step 3: Determine increasing intervals Using sign analysis, \( f(x) \) is increasing in: \[ (-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty). \] Thus, the correct answer is (B) \( (-\frac{1}{2}, 0) \cup (\frac{1}{2}, \infty) \).