Let the function be defined by $f\left(x\right)=x^3-3x^2+2$. At what value(s) of $x$ does $f\left(x\right)$ have a relative maximum?

• A. $x=0$
• B. $x=1$
• C. $x=2$
• D. $x=-1$

Answer: (B)

Verified step by step guidance

1. Step 1: To find the relative maximum of the function f(x) = x^3 - 3x^2 + 2, start by finding the first derivative f'(x). The derivative of f(x) is calculated using the power rule: f'(x) = 3x^2 - 6x.
2. Step 2: Set the first derivative f'(x) equal to zero to find the critical points. Solve the equation 3x^2 - 6x = 0. Factorize the equation: 3x(x - 2) = 0. This gives the critical points x = 0 and x = 2.
3. Step 3: Use the second derivative test to determine whether each critical point corresponds to a relative maximum, minimum, or neither. Compute the second derivative f''(x) by differentiating f'(x): f''(x) = 6x - 6.
4. Step 4: Evaluate f''(x) at each critical point. For x = 0, f''(0) = 6(0) - 6 = -6 (negative, indicating a relative maximum). For x = 2, f''(2) = 6(2) - 6 = 6 (positive, indicating a relative minimum).
5. Step 5: Conclude that the function f(x) has a relative maximum at x = 0. Verify the behavior of the function around x = 0 to confirm the result.