Question

Let \( f(x) = x^3 - 6x^2 + 12x - 3 \), then at \( x = 2 \), \( f(x) \) has :

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

Hint:

Notice that \( f(x) = x^3 - 6x^2 + 12x - 8 + 5 = (x - 2)^3 + 5 \).
The function \( y = x^3 \) has a point of inflection at \( x = 0 \). Shifting it right by 2 and up by 5 yields \( f(x) \), which must have a point of inflection at \( x = 2 \). Points of inflection are neither maxima nor minima.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To analyze the local behavior of a function at a point, we use tools from calculus: the first, second, and sometimes third derivative tests.
A local maximum occurs when the function stops increasing and starts decreasing (\( f'(x) \) changes from \( + \) to \( - \)). A local minimum occurs when the function stops decreasing and starts increasing (\( f'(x) \) changes from \( - \) to \( + \)).
If the derivative at a point is zero (\( f'(a) = 0 \)), it is a critical point. We then use the second derivative to determine concavity. If \( f''(a) > 0 \), it's a minimum (concave up). If \( f''(a) < 0 \), it's a maximum (concave down). However, if \( f''(a) = 0 \), the point could be a point of inflection where the function's rate of change reaches a plateau but continues in the same direction.

Step 2: Key Formula or Approach:
1. Calculate the first derivative \( f'(x) \) and evaluate it at \( x = 2 \) to see if it is a critical point.
2. Calculate the second derivative \( f''(x) \) and evaluate it at \( x = 2 \).
3. If both are zero, analyze the nature of \( f'(x) \) around \( x = 2 \) or check the third derivative \( f'''(x) \). If \( f'''(a) \neq 0 \) when \( f'(a) = f''(a) = 0 \), the point is a point of inflection.

Step 3: Detailed Explanation:
Step 3.1: Finding the first derivative.
Function: \( f(x) = x^3 - 6x^2 + 12x - 3 \).
Power rule (\( \frac{d}{dx}x^n = nx^{n-1} \)):
\[ f'(x) = 3x^2 - 12x + 12 \]
At \( x = 2 \):
\[ f'(2) = 3(2)^2 - 12(2) + 12 = 3(4) - 24 + 12 = 12 - 24 + 12 = 0 \]
Since \( f'(2) = 0 \), \( x = 2 \) is a critical point.
Step 3.2: Finding the second derivative.
\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12 \]
At \( x = 2 \):
\[ f''(2) = 6(2) - 12 = 12 - 12 = 0 \]
Step 3.3: Analyzing the nature.
Since both \( f'(2) = 0 \) and \( f''(2) = 0 \), the standard second derivative test is inconclusive. Let's look at \( f'(x) \) more closely:
\[ f'(x) = 3(x^2 - 4x + 4) = 3(x - 2)^2 \]
The term \( (x-2)^2 \) is always greater than or equal to 0 for all real \( x \). This means for \( x < 2 \), \( f'(x) > 0 \) (increasing), and for \( x > 2 \), \( f'(x) > 0 \) (increasing).
Because the sign of the derivative does not change as we pass through \( x=2 \), the function is strictly increasing at that point. It momentarily levels off (horizontal tangent) but does not turn back. This is the definition of a point of inflection with a horizontal tangent. Thus, it is neither a maximum nor a minimum.

Step 4: Final Answer:
The function \( f(x) \) levels off at \( x=2 \) but continues to increase thereafter. Since there is no change in the sign of the slope, the point \( x=2 \) represents neither a maximum nor a minimum. This corresponds to Option (4).