Question

If \( m \) and \( M \) are respectively the absolute minimum and absolute maximum values of a function \( f(x) = 2x^3 + 9x^2 + 12x + 1 \) defined on \([-3,0]\), then \( m + M \) is: 

• A. \( -7 \)
• B. \( 0 \)
• C. \( 1 \)
• D. \( 5 \)

Hint:

To find absolute extrema on a closed interval, evaluate the function at critical points and endpoints, then select the maximum and minimum values.

Answer 1

The Correct Option is A

Step 1: Compute the First Derivative To find the critical points, we first differentiate the function: \[ f(x) = 2x^3 + 9x^2 + 12x + 1 \] \[ f'(x) = \frac{d}{dx} (2x^3 + 9x^2 + 12x + 1) \] \[ = 6x^2 + 18x + 12 \]

Step 2: Solve for Critical Points Set \( f'(x) = 0 \) to find critical points: \[ 6x^2 + 18x + 12 = 0 \] Dividing by 6: \[ x^2 + 3x + 2 = 0 \] Factoring: \[ (x + 1)(x + 2) = 0 \] \[ x = -1, -2 \]

 Step 3: Evaluate \( f(x) \) at Critical and Endpoint Values We evaluate \( f(x) \) at \( x = -3, -2, -1, 0 \): \[ f(-3) = 2(-3)^3 + 9(-3)^2 + 12(-3) + 1 = 2(-27) + 9(9) + 12(-3) + 1 \] \[ = -54 + 81 - 36 + 1 = -8 \] \[ f(-2) = 2(-2)^3 + 9(-2)^2 + 12(-2) + 1 \] \[ = 2(-8) + 9(4) + 12(-2) + 1 = -16 + 36 - 24 + 1 = -3 \] \[ f(-1) = 2(-1)^3 + 9(-1)^2 + 12(-1) + 1 \] \[ = 2(-1) + 9(1) + 12(-1) + 1 = -2 + 9 - 12 + 1 = -4 \] \[ f(0) = 2(0)^3 + 9(0)^2 + 12(0) + 1 = 1 \] 

Step 4: Identify Maximum and Minimum Values \[ \text{Maximum value: } M = 1 \] \[ \text{Minimum value: } m = -8 \] 

Step 5: Compute \( m + M \) \[ m + M = -8 + 1 = -7 \]