Question

Consider the function \( f: \mathbb{R} \to \mathbb{R} \), defined as \[ f(x) = 2x^3 - 3x^2 - 12x + 1. \] Which of the following statements is/are correct? (Here, \( \mathbb{R} \) is the set of real numbers.)

• A. \( f \) has no global maximizer
• B. \( f \) has no global minimizer
• C. \( x = -1 \) is a local minimizer of \( f \)
• D. \( x = 2 \) is a local maximizer of \( f \)

Hint:

For cubic functions, check the critical points and use the second derivative test to identify the nature of the function. Keep in mind that cubic functions do not have global maxima or minima because they go to \( \infty \) and \( -\infty \) in opposite directions.

Answer 1

The Correct Option is A, B

We are given the function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \). Let's first find the critical points by taking the derivative of \( f(x) \). 
Step 1: Find the first derivative of \( f(x) \): \[ f'(x) = 6x^2 - 6x - 12. \] 
Step 2: Set the first derivative equal to zero to find the critical points: \[ 6x^2 - 6x - 12 = 0. \] Simplifying the equation: \[ x^2 - x - 2 = 0. \] Factoring: \[ (x - 2)(x + 1) = 0. \] Thus, the critical points are \( x = 2 \) and \( x = -1 \). 
Step 3: Second derivative test to determine the nature of the critical points: 
The second derivative is: \[ f''(x) = 12x - 6. \] At \( x = -1 \), \( f''(-1) = 12(-1) - 6 = -18 \), which is less than 0, indicating a local maximum at \( x = -1 \).
At \( x = 2 \), \( f''(2) = 12(2) - 6 = 18 \), which is greater than 0, indicating a local minimum at \( x = 2 \).
Step 4: Global maximizer and minimizer 
The function \( f(x) \) is a cubic function, and cubic functions have no global maxima or minima because they tend to infinity in one direction and negative infinity in the other direction. Thus, \( f(x) \) has no global maximizer or global minimizer. Therefore, the correct answers are (A) and (B).