Given the function \( f(x)=\int_{2}^{x}\left(t^{3}+18 t^{2}+81 t\right) d t \), 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 the second derivative of \( f(x) \). From the Fundamental Theorem of Calculus, the first derivative is: \[ f'(x) = \frac{d}{dx} \left( \int_{2}^{x} \left(t^{3} + 18t^{2} + 81t\right) dt \right) = x^{3} + 18x^{2} + 81x. \] Next, we find the second derivative: \[ f''(x) = \frac{d}{dx}(f'(x)) = 3x^{2} + 36x + 81. \] To find the points of inflection, we set the second derivative equal to zero: \[ 3x^{2} + 36x + 81 = 0. \] Dividing the entire equation by 3 simplifies it to: \[ x^{2} + 12x + 27 = 0. \] Now, we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \): \[ x = \frac{-12 \pm \sqrt{12^{2} - 4 \cdot 1 \cdot 27}}{2 \cdot 1} = \frac{-12 \pm \sqrt{144 - 108}}{2} = \frac{-12 \pm \sqrt{36}}{2}. \] Calculating the square root yields: \[ x = \frac{-12 \pm 6}{2}. \] Thus, we find two possible values: 1. \( x = \frac{-6}{2} = -3 \) 2. \( x = \frac{-18}{2} = -9 \) To confirm these are points of inflection, we check the sign of \( f''(x) \) around these values: - For \( x < -9 \): Choose \( x = -10 \) → \( f''(-10) = 3(-10)^{2} + 36(-10) + 81 = 300 - 360 + 81 = 21 \) (positive). - For \( -9 < x < -3 \): Choose \( x = -5 \) → \( f''(-5) = 3(-5)^{2} + 36(-5) + 81 = 75 - 180 + 81 = -24 \) (negative). - For \( x > -3 \): Choose \( x = 0 \) → \( f''(0) = 81 \) (positive). Since \( f''(x) \) changes sign at \( x = -9 \) and \( x = -3 \), both of these points are indeed points of inflection. Thus, the final answer is that \( f \) has points of inflection at: \[ \boxed{-9 \text{ and } -3}. \]