Given the function \( f(x)=\int_{1}^{x}\left(t^{4}-16 t^{3}\right) d t \) determine all \( x \)-values, if any, for which \( f \) has a point of inflection.

To determine the \( x \)-values where the function \( f(x) = \int_{1}^{x} \left(t^4 - 16t^3\right) dt \) has a point of inflection, we'll follow these steps: 1. **Find the First Derivative \( f'(x) \):** By the Fundamental Theorem of Calculus, \[ f'(x) = x^4 - 16x^3. \] 2. **Find the Second Derivative \( f''(x) \):** Differentiate \( f'(x) \) with respect to \( x \): \[ f''(x) = \frac{d}{dx}(x^4 - 16x^3) = 4x^3 - 48x^2 = 4x^2(x - 12). \] 3. **Determine Potential Points of Inflection:** Points of inflection occur where \( f''(x) = 0 \) and there is a change in the sign of \( f''(x) \). Set \( f''(x) = 0 \): \[ 4x^2(x - 12) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 12. \] 4. **Test for Sign Change Around Each Critical Point:** - **At \( x = 0 \):** - **For \( x < 0 \):** \[ f''(x) = 4x^2(x - 12) > 0 \times (-) = < 0. \] - **For \( x > 0 \):** \[ f''(x) = 4x^2(x - 12) > 0 \times (-) = < 0. \] Since there is no sign change, \( x = 0 \) is **not** a point of inflection. - **At \( x = 12 \):** - **For \( x < 12 \):** \[ f''(x) = 4x^2(x - 12) > 0 \times (-) = < 0. \] - **For \( x > 12 \):** \[ f''(x) = 4x^2(x - 12) > 0 \times (+) = > 0. \] There is a sign change from negative to positive, indicating a point of inflection at \( x = 12 \). **Conclusion:** The function \( f(x) \) has a single point of inflection at \( x = 12 \). **Answer:** Only at x = 12 does f  have a point of inflection.