Step 1: Find critical points.
To locate local maxima and minima, calculate the first derivative of \( f(x) = 2x^3 - 3x^2 - 12x + 4 \):
\[\nf'(x) = \frac{d}{dx} (2x^3 - 3x^2 - 12x + 4) = 6x^2 - 6x - 12.\n\]
Set \( f'(x) = 0 \) to identify critical points:
\[\n6x^2 - 6x - 12 = 0 \implies x^2 - x - 2 = 0.\n\]
Solve the quadratic equation:
\[\nx = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2},\n\]
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:
\[\nf''(x) = \frac{d}{dx} (6x^2 - 6x - 12) = 12x - 6.\n\]
Evaluate \( f''(x) \) at the critical points:
- At \( x = -1 \):
\[\nf''(-1) = 12(-1) - 6 = -12 - 6 = -18<0,\n\]
indicating a local maximum (since \( f''(x)<0 \)).
- At \( x = 2 \):
\[\nf''(2) = 12(2) - 6 = 24 - 6 = 18>0,\n\]
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 results in 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.