20. Find the absolute maximum value of \( y \) on the interval \( [-1,4] \) if \( y=-x^{4}+4 x^{3} \). \( \begin{array}{lllll}\text { A. }-5 & \text { B. } 0 & \text { C. } 3 & \text { D. } 4 & \text { E. } 27\end{array} \)

To find the absolute maximum value of \( y = -x^4 + 4x^3 \) on the interval \([-1, 4]\), we first need to evaluate the function at the endpoints and any critical points within the interval. 1. **Endpoints**: - At \( x = -1 \): \[ y(-1) = -(-1)^4 + 4(-1)^3 = -1 - 4 = -5 \] - At \( x = 4 \): \[ y(4) = -(4)^4 + 4(4)^3 = -256 + 256 = 0 \] 2. **Critical Points**: We find critical points by taking the derivative and setting it to zero: \[ y' = -4x^3 + 12x^2 \] Setting \( y' = 0 \): \[ -4x^3 + 12x^2 = 0 \quad \Rightarrow \quad 4x^2(-x + 3) = 0 \] This gives us \( x = 0 \) and \( x = 3 \). Now, we evaluate \( y \) at these critical points: - At \( x = 0 \): \[ y(0) = -0^4 + 4(0)^3 = 0 \] - At \( x = 3 \): \[ y(3) = -(3)^4 + 4(3)^3 = -81 + 108 = 27 \] 3. **Final Evaluation**: Now we compare the values found: - \( y(-1) = -5 \) - \( y(4) = 0 \) - \( y(0) = 0 \) - \( y(3) = 27 \) The absolute maximum value on the interval \([-1, 4]\) is \( 27 \). Thus, the answer is **E. 27**.