Given the function \( f(x)=\int_{0}^{x}\left(t^{4}+8 t^{3}+16 t^{2}\right) d t \) determine all \( x \)-values, if any, for which \( f \) has a point of inflection.

To determine the points of inflection for the function \( f(x) \), we first need to find its second derivative. 1. **Finding \( f(x) \)**: First, we evaluate the integral: \[ f(x) = \int_{0}^{x} \left(t^{4} + 8t^{3} + 16t^{2}\right) dt. \] Computing this integral: \[ f(x) = \left[ \frac{t^{5}}{5} + \frac{8t^{4}}{4} + \frac{16t^{3}}{3} \right]_{0}^{x} = \frac{x^{5}}{5} + 2x^{4} + \frac{16x^{3}}{3}. \] 2. **Finding \( f'(x) \)**: Next, we differentiate \( f(x) \): \[ f'(x) = \frac{d}{dx}\left(\frac{x^{5}}{5} + 2x^{4} + \frac{16x^{3}}{3}\right) = x^{4} + 8x^{3} + 16x^{2}. \] 3. **Finding \( f''(x) \)**: We differentiate again to find the second derivative: \[ f''(x) = \frac{d}{dx}\left(x^{4} + 8x^{3} + 16x^{2}\right) = 4x^{3} + 24x^{2} + 32x. \] 4. **Setting \( f''(x) = 0 \)**: Now we solve for \( f''(x) = 0 \): \[ 4x^{3} + 24x^{2} + 32x = 0. \] Factoring out \( 4x \): \[ 4x(x^{2} + 6x + 8) = 0. \] This gives us one solution: \[ x = 0. \] For the quadratic \( x^{2} + 6x + 8 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2}. \] Thus: \[ x = -2 \quad \text{or} \quad x = -4. \] 5. **Analyzing Points of Inflection**: Points of inflection occur where \( f''(x) \) changes sign. We found potential inflection points at \( x = 0, -2, \) and \( -4 \). To confirm that these values are points of inflection, we can check the sign of \( f''(x) \) around these points: - For \( x < -4 \), let's choose \( x = -5 \): \[ f''(-5) = 4(-5)^{3} + 24(-5)^{2} + 32(-5) = -500 + 600 - 160 = -60 \quad (\text{negative}) \] - For \( -4 < x < -2 \), choose \( x = -3 \): \[ f''(-3) = 4(-3)^{3} + 24(-3)^{2} + 32(-3) = -108 + 216 - 96 = 12 \quad (\text{positive}) \] - For \( -2 < x < 0 \), choose \( x = -1 \): \[ f''(-1) = 4(-1)^{3} + 24(-1)^{2} + 32(-1) = -4 + 24 - 32 = -12 \quad (\text{negative}) \] - For \( x > 0 \), choose \( x = 1 \): \[ f''(1) = 4(1)^{3} + 24(1)^{2} + 32(1) = 4 + 24 + 32 = 60 \quad (\text{positive}) \] From this analysis, the function \( f(x) \) has points of inflection at \( x = -4 \) and \( x = -2