Given the function \( f(x)=\int_{-2}^{x}\left(t^{3}+12 t^{2}+36 t\right) d t \), determine all intervals on which \( f \) is concave down.

To determine where the function \( f(x) \) is concave down, we first need to find its second derivative. We start by calculating the first derivative using the Fundamental Theorem of Calculus: \[ f'(x) = \frac{d}{dx}\left(\int_{-2}^{x} \left(t^3 + 12t^2 + 36t\right) dt\right) = x^3 + 12x^2 + 36x. \] Next, we find the second derivative: \[ f''(x) = \frac{d}{dx}(f'(x)) = \frac{d}{dx}(x^3 + 12x^2 + 36x) = 3x^2 + 24x + 36. \] Now, we need to determine where \( f''(x) < 0 \) to find the intervals where \( f(x) \) is concave down. We will first find the critical points of the second derivative by solving the equation: \[ 3x^2 + 24x + 36 = 0. \] We can simplify this equation by dividing all terms by 3: \[ x^2 + 8x + 12 = 0. \] Next, we factor the quadratic: \[ (x + 6)(x + 2) = 0. \] Thus, the critical points are \( x = -6 \) and \( x = -2 \). Now, we will test intervals around these critical points: 1. Choose a test point in \( (-\infty, -6) \): Let \( x = -7 \). - \( f''(-7) = 3(-7)^2 + 24(-7) + 36 = 147 - 168 + 36 = 15 > 0 \) (concave up) 2. Choose a test point in \( (-6, -2) \): Let \( x = -4 \). - \( f''(-4) = 3(-4)^2 + 24(-4) + 36 = 48 - 96 + 36 = -12 < 0 \) (concave down) 3. Choose a test point in \( (-2, \infty) \): Let \( x = 0 \). - \( f''(0) = 3(0)^2 + 24(0) + 36 = 36 > 0 \) (concave up) Combining these results, we find that \( f(x) \) is concave down on the interval \( (-6, -2) \).