To determine the absolute maximum of \(y = x^{3} - 3x + 2\) on the interval \([0, 2]\), we follow these steps:1. Find Critical Points: Differentiate the function to get \(y' = 3x^{2} - 3 = 3(x^{2} - 1)\). Set the derivative to zero: \(3(x^{2} - 1) = 0\), which yields \(x^{2} = 1\), so \(x = \pm 1\). Since only \(x=1\) is within the specified interval \([0, 2]\), it is our relevant critical point.2. Evaluate Function at Critical Points and Endpoints: Calculate the function's value at the critical point within the interval and at the interval's endpoints: * At \(x=0\) (endpoint): \(y(0) = 0^{3} - 3(0) + 2 = 2\) * At \(x=1\) (critical point): \(y(1) = 1^{3} - 3(1) + 2 = 0\) * At \(x=2\) (endpoint): \(y(2) = 2^{3} - 3(2) + 2 = 8 - 6 + 2 = 4\)3. Identify Absolute Maximum: Compare the evaluated function values: \(\max\{2, 0, 4\}\). The highest value is \(4\), which occurs at \(x=2\). Therefore, the absolute maximum of the function on the interval \([0, 2]\) is \(4\).