Question

The difference between the absolute maximum and absolute minimum values of the function \( f(x) = 2x^3 - 15x^2 + 36x - 30 \) on \( [-1, 4] \) is:

• A. 80
• B. 1
• C. 85
• D. 4

Hint:

To find absolute extrema on a closed interval, always evaluate the function at critical points within the interval and at the endpoints, then compare values to identify the maximum and minimum.

Answer 1

The Correct Option is C

We are asked to find the absolute maximum and minimum values of: \[ f(x) = 2x^3 - 15x^2 + 36x - 30 \] on the interval \( [-1, 4] \), and then find their difference. Step 1: Find the derivative of \( f(x) \) \[ f'(x) = \frac{d}{dx}\left(2x^3 - 15x^2 + 36x - 30\right) = 6x^2 - 30x + 36 \] Step 2: Find critical points by setting \( f'(x) = 0 \) \[ 6x^2 - 30x + 36 = 0 \] Divide through by 6: \[ x^2 - 5x + 6 = 0 \] Factorizing: \[ (x-2)(x-3) = 0 \] So, critical points are: \[ x = 2, 3 \] Step 3: Evaluate \( f(x) \) at critical points and at endpoints \( x = -1, 4 \) \[ f(-1) = 2(-1)^3 - 15(-1)^2 + 36(-1) - 30 = -2 - 15 - 36 - 30 = -83 \] \[ f(2) = 2(2)^3 - 15(2)^2 + 36(2) - 30 = 16 - 60 + 72 - 30 = -2 \] \[ f(3) = 2(3)^3 - 15(3)^2 + 36(3) - 30 = 54 - 135 + 108 - 30 = -3 \] \[ f(4) = 2(4)^3 - 15(4)^2 + 36(4) - 30 = 128 - 240 + 144 - 30 = 2 \] Step 4: Identify the absolute maximum and minimum values From the computed values: \[ \text{Maximum value} = 2 \] \[ \text{Minimum value} = -83 \] Step 5: Compute the difference \[ \text{Difference} = 2 - (-83) = 85 \] % Final Answer Hence, the difference is 85.