Let `f(x)=x^3-6x^2+12x-3` . Then at x=2 f(x) has

Options

A. a maximum
B. a minimum
C. both a maximum and a minimum
D. neither a maximum nor a minimum

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 - 6x^2 + 12x - 3 \) at \( x = 2 \) to determine whether it has a maximum, minimum, both, or neither.

### Step-by-Step Solution:

1. **Differentiate the Function**:
We start by finding the first derivative of the function \( f(x) \).
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12 \]

**Hint**: The first derivative helps us find critical points where the function may have maxima or minima.

2. **Evaluate the First Derivative at \( x = 2 \)**:
Now, we substitute \( x = 2 \) into the first derivative.
\[ f'(2) = 3(2^2) - 12(2) + 12 = 3(4) - 24 + 12 = 12 - 24 + 12 = 0 \]

**Hint**: A critical point occurs when the first derivative equals zero.

3. **Differentiate Again to Find the Second Derivative**:
Next, we find the second derivative of the function.
\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12 \]

**Hint**: The second derivative helps us determine the concavity of the function at the critical points.

4. **Evaluate the Second Derivative at \( x = 2 \)**:
We now substitute \( x = 2 \) into the second derivative.
\[ f''(2) = 6(2) - 12 = 12 - 12 = 0 \]

**Hint**: If the second derivative is also zero, we cannot conclude whether we have a maximum or minimum.

5. **Conclusion**:
Since both \( f'(2) = 0 \) and \( f''(2) = 0 \), we cannot determine if \( x = 2 \) is a maximum, minimum, or neither using the second derivative test. Therefore, we conclude that at \( x = 2 \), \( f(x) \) has neither a maximum nor a minimum.

**Final Answer**: The correct option is **4: neither maximum nor minimum**.