Given the function \( f(x)=\int_{-1}^{x}\left(t^{4}-12 t^{3}\right) d t \) determine all \( x \)-values, if any, for which \( f \) has a relative maximum.

To determine the \( x \)-values for which the function \( f(x) \) has a relative maximum, we will follow these steps: 1. **Find the derivative \( f'(x) \)** using the Fundamental Theorem of Calculus. 2. **Set the derivative equal to zero** to find critical points. 3. **Use the second derivative test** to determine if these critical points correspond to relative maxima. ### Step 1: Find the derivative \( f'(x) \) According to the Fundamental Theorem of Calculus, the derivative of the integral function is given by: \[ f'(x) = \frac{d}{dx} \left( \int_{-1}^{x} (t^{4} - 12t^{3}) dt \right) = x^{4} - 12x^{3} \] ### Step 2: Set the derivative equal to zero Now, we will set \( f'(x) \) equal to zero to find the critical points: \[ x^{4} - 12x^{3} = 0 \] Factoring out \( x^{3} \): \[ x^{3}(x - 12) = 0 \] This gives us the critical points: \[ x^{3} = 0 \quad \Rightarrow \quad x = 0 \] \[ x - 12 = 0 \quad \Rightarrow \quad x = 12 \] ### Step 3: Use the second derivative test Next, we need to find the second derivative \( f''(x) \): \[ f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(x^{4} - 12x^{3}) = 4x^{3} - 36x^{2} \] Now we will evaluate \( f''(x) \) at the critical points \( x = 0 \) and \( x = 12 \): 1. **At \( x = 0 \)**: \[ f''(0) = 4(0)^{3} - 36(0)^{2} = 0 \] Since \( f''(0) = 0 \), the second derivative test is inconclusive at this point. 2. **At \( x = 12 \)**: \[ f''(12) = 4(12)^{3} - 36(12)^{2} \] Calculating \( f''(12) \): \[ = 4(1728) - 36(144) = 6912 - 5184 = 1728 \] Since \( f''(12) > 0 \), this indicates that \( f(x) \) has a relative minimum at \( x = 12 \). ### Conclusion The only critical point we found was \( x = 0 \), which was inconclusive, and \( x = 12 \), which is a relative minimum. Therefore, there are no \( x \)-values for which \( f \) has a relative maximum. Thus, the final answer is: **There are no \( x \)-values for which \( f \) has a relative maximum.**