Determine the average value of the piecewise function \( f \) defined below on the interval \( x=-1 \) to \( x=2 \). Express your answer in simplest form. \[ f(x)=\left\{\begin{array}{lll}-6 x^{2} & \text { for } & x<0 \\ -8 x+6 & \text { for } & x \geq 0\end{array}\right. \]

To determine the average value of the piecewise function \( f(x) \) on the interval \( x = -1 \) to \( x = 2 \), we first calculate the average value using the formula: \[ \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \] In this case, \( a = -1 \) and \( b = 2 \). ### Step 1: Calculate the integral of \( f(x) \). We split the integral at the point where the definition of \( f(x) \) changes, which is at \( x = 0 \): \[ \int_{-1}^2 f(x) \, dx = \int_{-1}^0 f(x) \, dx + \int_{0}^2 f(x) \, dx \] ### For \( x < 0 \) (from -1 to 0): Here, \( f(x) = -6x^2 \), so we evaluate: \[ \int_{-1}^0 -6x^2 \, dx \] Calculating this integral: \[ = -6 \int_{-1}^0 x^2 \, dx = -6 \left[ \frac{x^3}{3} \right]_{-1}^0 = -6 \left( 0 - \frac{(-1)^3}{3} \right) = -6 \left( 0 + \frac{1}{3} \right) = -2 \] ### For \( x \geq 0 \) (from 0 to 2): Here, \( f(x) = -8x + 6 \), so we evaluate: \[ \int_{0}^2 (-8x + 6) \, dx \] Calculating this integral: \[ = \left[ -4x^2 + 6x \right]_{0}^2 = \left( -4(2^2) + 6(2) \right) - \left( -4(0^2) + 6(0) \right) = (-16 + 12) - 0 = -4 \] ### Step 2: Combine the results from both intervals: Now we combine the integrals: \[ \int_{-1}^2 f(x) \, dx = -2 + (-4) = -6 \] ### Step 3: Find the average value: Now we apply the average value formula: \[ \text{Average value} = \frac{1}{2 - (-1)} \int_{-1}^2 f(x) \, dx = \frac{1}{3} \times (-6) = -2 \] Thus, the average value of the piecewise function \( f \) on the interval from \( x = -1 \) to \( x = 2 \) is \[ \boxed{-2} \]