Question

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

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Check endpoints and critical points for absolute extrema on closed intervals.

Answer 1

The Correct Option is C

Step 1: Find critical points by differentiating: \[ y = x^{3} - 3x + 2 \] \[ y' = 3x^{2} - 3 = 3(x^{2} - 1) \] Set derivative to zero: \[ 3(x^{2} - 1) = 0 \implies x^{2} = 1 \implies x = \pm 1 \] Only \(x=1\) lies in the interval \([0, 2]\). Step 2: Evaluate function values at critical points and endpoints: \[ y(0) = 0^{3} - 3(0) + 2 = 2 \] \[ y(1) = 1 - 3 + 2 = 0 \] \[ y(2) = 8 - 6 + 2 = 4 \] Step 3: The absolute maximum in \([0,2]\) is the highest among these values: \[ \max\{2, 0, 4\} = 4 \]