To ascertain if \( f(x) = x^3 - 6x^2 + 12x - 3 \) possesses a maximum or minimum at \( x = 2 \), calculus, specifically the first and second derivative tests, is employed.
1. First Derivative Computation: The first derivative, \( f'(x) \), is calculated.
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12 \]
2. Identification of Critical Points: \( f'(x) \) is set to zero to solve for \( x \).
\[ 3x^2 - 12x + 12 = 0 \]
Dividing the equation by 3 yields:
\[ x^2 - 4x + 4 = 0 \]
This equation factors as a perfect square:
\[ (x - 2)^2 = 0 \]
Consequently, \( x = 2 \) is identified as a critical point.
3. Second Derivative Calculation: The second derivative, \( f''(x) \), is determined to classify the critical point.
\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12 \]
The value of \( f''(x) \) is evaluated at the critical point \( x = 2 \).
\[ f''(2) = 6(2) - 12 = 0 \]
4. Determination of Nature of Critical Point: The second derivative test proves inconclusive as \( f''(2) = 0 \). However, the presence of a double root in the first derivative at \( x = 2 \) indicates a point where the function's graph exhibits a horizontal tangent. Further analysis, such as the first derivative test or examining sign changes in \( f'(x) \) around \( x = 2 \), is recommended. Given the cubic nature of \( f(x) \), a transition from increasing to decreasing or vice versa is expected at \( x = 2 \), suggesting a minimum.
Therefore, \( f(x) \) has a minimum at \( x = 2 \).