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 the function \( y = -x^4 + 4x^3 \) on the interval \([-1, 4]\), follow these steps: 1. **Find the Critical Points:** - First, compute the derivative of \( y \): \[ y' = \frac{d}{dx}(-x^4 + 4x^3) = -4x^3 + 12x^2 = -4x^2(x - 3) \] - Set the derivative equal to zero to find critical points: \[ -4x^2(x - 3) = 0 \implies x = 0 \text{ (double root)} \quad \text{or} \quad x = 3 \] - So, the critical points within the interval \([-1, 4]\) are \( x = 0 \) and \( x = 3 \). 2. **Evaluate \( y \) at the Critical Points and Endpoints:** - **At \( x = -1 \):** \[ y(-1) = -(-1)^4 + 4(-1)^3 = -1 - 4 = -5 \] - **At \( x = 0 \):** \[ y(0) = -0^4 + 4(0)^3 = 0 \] - **At \( x = 3 \):** \[ y(3) = -(3)^4 + 4(3)^3 = -81 + 108 = 27 \] - **At \( x = 4 \):** \[ y(4) = -(4)^4 + 4(4)^3 = -256 + 256 = 0 \] 3. **Determine the Maximum Value:** - The values of \( y \) at the critical points and endpoints are \(-5\), \(0\), \(27\), and \(0\). - The **absolute maximum** value of \( y \) on the interval \([-1, 4]\) is **27**. **Answer:** E. 27