Given the function \( f(x)=\int_{0}^{x}\left(t^{3}-12 t^{2}+36 t\right) d t \), determine all intervals on which \( f \) is concave down.

To determine the intervals where the function \( f(x) \) is concave down, we need to first find the second derivative of \( f(x) \). We begin by computing the first derivative \( f'(x) \). By the Fundamental Theorem of Calculus: \[ f'(x) = x^3 - 12x^2 + 36x \] Next, we compute the second derivative \( f''(x) \): \[ f''(x) = \frac{d}{dx}(f'(x)) = 3x^2 - 24x + 36 \] Now, we need to find where \( f''(x) < 0 \). We start by solving the equation \( f''(x) = 0 \): \[ 3x^2 - 24x + 36 = 0 \] Dividing the entire equation by 3 gives: \[ x^2 - 8x + 12 = 0 \] Next, we factor the quadratic: \[ (x - 6)(x - 2) = 0 \] Thus, the roots are \( x = 2 \) and \( x = 6 \). These points divide the \( x \)-axis into several intervals: \( (-\infty, 2) \), \( (2, 6) \), and \( (6, \infty) \). We test the sign of \( f''(x) \) in each interval: 1. For \( x < 2 \) (e.g., \( x = 0 \)): \[ f''(0) = 3(0)^2 - 24(0) + 36 = 36 > 0 \] (Concave up) 2. For \( 2 < x < 6 \) (e.g., \( x = 4 \)): \[ f''(4) = 3(4)^2 - 24(4) + 36 = 48 - 96 + 36 = -12 < 0 \] (Concave down) 3. For \( x > 6 \) (e.g., \( x = 7 \)): \[ f''(7) = 3(7)^2 - 24(7) + 36 = 147 - 168 + 36 = 15 > 0 \] (Concave up) Thus, the function \( f(x) \) is concave down on the interval \( (2, 6) \).