Question

In the closed interval [0, 3], the minimum value of the function \( f \) given below is: \[ f(x) = 2x^3 - 9x^2 + 12x \]

• A. \(0\)
• B. \(4\)
• C. \(5\)
• D. \(9\)

Hint:

To find the minimum or maximum value of a function in a given interval, always check the function's values at the critical points and endpoints of the interval.

Answer 1

The Correct Option is A

To find the minimum value of the function, we need to calculate its critical points and evaluate the function at the endpoints of the interval. 
Step 1: First, find the first derivative of the function \( f(x) \): \[ f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x) = 6x^2 - 18x + 12 \] 
Step 2: Set \( f'(x) = 0 \) to find the critical points: \[ 6x^2 - 18x + 12 = 0 \] Divide through by 6: \[ x^2 - 3x + 2 = 0 \] Factor the quadratic equation: \[ (x - 1)(x - 2) = 0 \] Thus, \( x = 1 \) and \( x = 2 \) are the critical points. 
Step 3: Evaluate \( f(x) \) at the critical points and the endpoints \( x = 0 \) and \( x = 3 \): \( f(0) = 2(0)^3 - 9(0)^2 + 12(0) = 0 \)
\( f(1) = 2(1)^3 - 9(1)^2 + 12(1) = 2 - 9 + 12 = 5 \)
\( f(2) = 2(2)^3 - 9(2)^2 + 12(2) = 16 - 36 + 24 = 4 \)
\( f(3) = 2(3)^3 - 9(3)^2 + 12(3) = 54 - 81 + 36 = 9 \)
Step 4: The minimum value of \( f(x) \) on the interval [0, 3] occurs at \( x = 0 \), and the minimum value is \( f(0) = 0 \).