The absolute maximum value of $y = x^{3} - 3 x + 2 in 0 \leq x \leq 2 is$

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answer is C.

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Detailed Solution

Explanation:

We need to find the absolute maximum value of the function:

y = x³ - 3x + 2

over the interval 0 ≤ x ≤ 2.

Step 1: Differentiate the Function

To find the critical points, take the derivative:

dy/dx = 3x² - 3

Step 2: Set the Derivative Equal to Zero

Setting the derivative equal to zero:

3x² - 3 = 0
x² = 1
x = ±1

Since our interval is 0 ≤ x ≤ 2, we consider only x = 1.

Step 3: Evaluate the Function at Critical Points and Boundaries

Evaluating the function:

y(0) = 0³ - 3(0) + 2 = 2
y(1) = 1³ - 3(1) + 2 = 0
y(2) = 2³ - 3(2) + 2 = 4

Step 4: Identify the Maximum

The maximum value occurs at:

y(2) = 4

Final Answer:

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