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

Answer 1

Answer: (B)

a minimum

Answer 2

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 $x = 2$:

• For $x < 2$, $f'(x) > 0$.
• For $x > 2$, $f'(x) < 0$.

This indicates that $f(x)$ decreases after $x = 2$, implying that $x = 2$ is a minimum point.