Question

The interval in which the function \( f(x) = 4x^2 + x \) is decreasing is:

• A. \( \left( -\frac{1}{2}, \frac{1}{2} \right) \)
• B. \( \left[ -\frac{1}{2}, \frac{1}{2} \right] \)
• C. \( (-1, 1) \)
• D. \( [-1, 1] \)

Hint:

To determine where a function is increasing or decreasing: - Compute the first derivative \( f'(x) \). - Solve \( f'(x)>0 \) for increasing intervals. - Solve \( f'(x)<0 \) for decreasing intervals. - Ensure to check critical points and denominators for undefined regions.

Answer 1

The Correct Option is A

Step 1: Compute the first derivative \( f'(x) \). Given: \[ f(x) = \frac{4x^2 + 1}{x} \] Using the quotient rule: \[ \left( \frac{g(x)}{h(x)} \right)' = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \] where \( g(x) = 4x^2 + 1 \) and \( h(x) = x \), we compute their derivatives: \[ g'(x) = 8x, \quad h'(x) = 1 \] Applying the quotient rule: \[ f'(x) = \frac{(8x \cdot x) - (4x^2 + 1) \cdot 1}{x^2} \] \[ f'(x) = \frac{8x^2 - 4x^2 - 1}{x^2} \] \[ f'(x) = \frac{4x^2 - 1}{x^2} \] Step 2: Find where \( f'(x) \) is negative. \[ \frac{4x^2 - 1}{x^2}<0 \] Since \( x^2 \) in the denominator is always positive, the inequality simplifies to: \[ 4x^2 - 1<0 \] \[ 4x^2<1 \] \[ x^2<\frac{1}{4} \] \[ -\frac{1}{2}<x<\frac{1}{2} \] Thus, \( f(x) \) is decreasing in the interval: \[ \left( -\frac{1}{2}, \frac{1}{2} \right) \]