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: Find the first derivative.
For $f(x) = x^3 - 6x^2 + 12x - 3$ we differentiate term by term.
\[ f'(x) = 3x^2 - 12x + 12 \]

Step 2: Factor it.
Pull out the $3$ and notice a perfect square.
\[ f'(x) = 3(x^2 - 4x + 4) = 3(x-2)^2 \]

Step 3: Look at the sign of the slope.
A square is never negative, so $3(x-2)^2 \ge 0$ everywhere. The slope is positive just before $x = 2$ and positive just after it. It only touches zero at $x = 2$ without flipping sign.

Step 4: Decide the nature.
Since the slope does not change from plus to minus or minus to plus, the curve keeps rising through $x = 2$. It just flattens for a moment. So there is no high point and no low point here.
\[ \boxed{\text{neither a maximum nor a minimum}} \]