Question

The function \( f(x) = 2x^3 - 3x^2 - 12x + 4 \), \( x \in \mathbb{R} \) has:

• A. two points of local maximum.
• B. two points of local minimum.
• C. one local maximum and one local minimum.
• D. neither maximum nor minimum.

Hint:

For a polynomial, find critical points by setting the first derivative to zero, then use the second derivative test to classify them as local maxima (\( f''(x)<0 \)) or minima (\( f''(x)>0 \)).

Answer 1

The Correct Option is C

Step 1: Find the critical points.
To find local maxima and minima, compute the first derivative of \( f(x) = 2x^3 - 3x^2 - 12x + 4 \): \[ f'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12. \] Set \( f'(x) = 0 \) to find critical points: \[ 6x^2 - 6x - 12 = 0 \implies x^2 - x - 2 = 0. \] Solve the quadratic equation: \[ x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}, \] so \( x = 2 \) or \( x = -1 \). The critical points are \( x = -1 \) and \( x = 2 \).
Step 2: Determine the nature of critical points using the second derivative.
Compute the second derivative: \[ f''(x) = \frac{d}{dx} (6x^2 - 6x - 12) = 12x - 6. \] Evaluate \( f''(x) \) at the critical points: - At \( x = -1 \): \[ f''(-1) = 12(-1) - 6 = -12 - 6 = -18<0, \] indicating a local maximum (since \( f''(x)<0 \)). - At \( x = 2 \): \[ f''(2) = 12(2) - 6 = 24 - 6 = 18>0, \] indicating a local minimum (since \( f''(x)>0 \)).
Step 3: Analyze the number of extrema.
The function \( f(x) \) is a cubic polynomial, and its derivative \( f'(x) \) is a quadratic with two roots (\( x = -1, 2 \)). Thus, there are exactly two critical points:
\( x = -1 \): local maximum.
\( x = 2 \): local minimum.
This gives one local maximum and one local minimum.

Step 4: Verify against options.
% Option (A) Two local maxima: Incorrect, as there is only one local maximum.
% Option (B) Two local minima: Incorrect, as there is only one local minimum.
% Option (C) One local maximum and one local minimum: Correct, as found.
% Option (D) Neither maximum nor minimum: Incorrect, as extrema exist.