Question

The absolute maximum value of the function \( f(x) = 2x^3 - 3x^2 - 36x + 9 \) defined on \( [-3, 3] \) is

• A. \( 36 \)
• B. \( 53 \)
• C. \( 63 \)
• D. \( 72 \)

Hint:

To find the absolute extrema, evaluate the function at both the critical points and the endpoints of the given interval.

Answer 1

The Correct Option is C

We are required to find the absolute maximum value of the function: \[ f(x) = 2x^3 - 3x^2 - 36x + 9 \quad \text{on the interval} \quad [-3, 3]. \] Step 1: Find the critical points of \( f(x) \) First, compute the derivative of \( f(x) \): \[ f'(x) = 6(x^2 - x - 6). \] Set \( f'(x) = 0 \): \[ x^2 - x - 6 = 0. \] Factor the quadratic equation: \[ (x - 3)(x + 2) = 0. \] The critical points are: \[ x = -2, \quad x = 3. \] Step 2: Evaluate \( f(x) \) at the critical points and the endpoints Now, evaluate the function \( f(x) \) at the critical points and the endpoints of the interval \( [-3, 3] \): 1. At \( x = -2 \): \[ f(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 9 = -16 - 12 + 72 + 9 = 53. \] 2. At \( x = -3 \): \[ f(-3) = 2(-3)^3 - 3(-3)^2 - 36(-3) + 9 = -54 - 27 + 108 + 9 = 36. \] 3. At \( x = 3 \): \[ f(3) = 2(3)^3 - 3(3)^2 - 36(3) + 9 = 54 - 27 - 108 + 9 = -72. \] Step 3: Determine the absolute maximum value From the above evaluations, we get: \[ f(-2) = 53, \quad f(-3) = 36, \quad f(3) = -72. \] Thus, the absolute maximum value of \( f(x) \) on the interval \( [-3, 3] \) is: \[ \boxed{53 \, \text{(Option B)}}. \]