Given the function \( f(x)=-x^{4}-12 x^{3} \), determine all intervals on which \( f^{\prime} \) is decreasing Answer Attempt 1 out of 2

To determine the intervals on which the derivative \( f'(x) \) of the function \( f(x) = -x^4 - 12x^3 \) is decreasing, follow these steps: ### 1. Compute the First Derivative \( f'(x) \): \[ f(x) = -x^4 - 12x^3 \] \[ f'(x) = \frac{d}{dx}(-x^4) + \frac{d}{dx}(-12x^3) \] \[ f'(x) = -4x^3 - 36x^2 \] ### 2. Compute the Second Derivative \( f''(x) \): \[ f''(x) = \frac{d}{dx}(-4x^3) + \frac{d}{dx}(-36x^2) \] \[ f''(x) = -12x^2 - 72x \] ### 3. Determine When \( f'(x) \) is Decreasing: The derivative \( f'(x) \) is decreasing where its second derivative \( f''(x) \) is negative: \[ f''(x) < 0 \] \[ -12x^2 - 72x < 0 \] #### Simplify the Inequality: \[ -12x^2 - 72x < 0 \] Divide both sides by -12 (note that this reverses the inequality sign): \[ x^2 + 6x > 0 \] #### Factor the Quadratic: \[ x(x + 6) > 0 \] #### Find Critical Points: The critical points are \( x = 0 \) and \( x = -6 \). #### Test Intervals: 1. **Interval \((-\infty, -6)\):** - Choose \( x = -7 \): \[ (-7)(-7 + 6) = (-7)(-1) = 7 > 0 \] - **Condition holds.** 2. **Interval \((-6, 0)\):** - Choose \( x = -3 \): \[ (-3)(-3 + 6) = (-3)(3) = -9 < 0 \] - **Condition does not hold.** 3. **Interval \((0, \infty)\):** - Choose \( x = 1 \): \[ (1)(1 + 6) = (1)(7) = 7 > 0 \] - **Condition holds.** ### 4. Conclusion: The derivative \( f'(x) \) is decreasing where \( f''(x) < 0 \), which corresponds to the intervals where \( x < -6 \) or \( x > 0 \). **Therefore, \( f'(x) \) is decreasing on the intervals:** \[ (-\infty, -6) \quad \text{and} \quad (0, \infty) \]