Find the absolute maximum value of the function f(x) = 2x^3 - 3x^2 - 12x + 1 on the closed interval [-2, 3].

Answer and explanation

Correct answer: C. 8

  1. Step 1: Find critical points by differentiating: f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1).
  2. Step 2: Set f'(x) = 0 => x = -1 and x = 2. Both lie within [-2, 3].
  3. Step 3: Evaluate f(x) at critical points and interval endpoints:
  4. Step 4: f(-2) = 2(-8) - 3(4) - 12(-2) + 1 = -16 - 12 + 24 + 1 = -3.
  5. Step 5: f(-1) = 2(-1) - 3(1) - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8.
  6. Step 6: f(2) = 2(8) - 3(4) - 12(2) + 1 = 16 - 12 - 24 + 1 = -19.
  7. Step 7: f(3) = 2(27) - 3(9) - 12(3) + 1 = 54 - 27 - 36 + 1 = -8.
  8. Step 8: The absolute maximum value is 8 (occurring at x = -1).
Answer
8.
Why not the others: '1' is f(0) which is not a critical point; '-19' is the local minimum value at x = 2; '-26' is an arithmetic error evaluating endpoints.

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