To determine whether \( f(x) = x^3 - 6x^2 + 12x - 3 \) has a maximum or minimum at \( x = 2 \), we need to use calculus, specifically the first and second derivative tests.
1. First Derivative: Find \( f'(x) \).
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12 \]
2. Critical Points: Set \( f'(x) = 0 \) and solve for \( x \).
\[ 3x^2 - 12x + 12 = 0 \]
Divide the entire equation by 3:
\[ x^2 - 4x + 4 = 0 \]
This is a perfect square:
\[ (x - 2)^2 = 0 \]
Thus, \( x = 2 \) is a critical point.
3. Second Derivative: Find \( f''(x) \) to determine the nature of the critical point.
\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12 \]
Evaluate \( f''(x) \) at the critical point \( x = 2 \).
\[ f''(2) = 6(2) - 12 = 0 \]
4. Conclusion: The second derivative test is inconclusive since \( f''(2) = 0 \). However, since the first derivative has a double root at \( x = 2 \), it indicates the presence of a point where the graph of the function has a flat tangent line. To confirm, consider using the first derivative test or analyzing the sign changes in \( f'(x) \) around \( x = 2 \). In this case, the cubic nature of the function implies \( f(x) \) transitions from increasing to decreasing or vice versa, indicating a minimum point at \( x = 2 \).
Therefore, at \( x = 2 \), \( f(x) \) has a minimum.
First, compute the first derivative \( f'(x) \):
\[ f'(x) = 3x^2 - 12x + 12. \]
Set \( f'(x) = 0 \) to find critical points:
\[ 3x^2 - 12x + 12 = 0 \implies x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0 \implies x = 2. \]
Next, compute the second derivative \( f''(x) \):
\[ f''(x) = 6x - 12. \]
At \( x = 2 \):
\[ f''(2) = 6(2) - 12 = 0. \]
Since \( f''(2) = 0 \), perform the higher-order derivative test or inspect the behavior of \( f'(x) \) around \( x = 2 \):
This indicates that \( f(x) \) decreases after \( x = 2 \), implying that \( x = 2 \) is a minimum point.