Let $f\left(x\right)=x^3-3x^2+1$. What is the absolute minimum value of $f$ on the closed interval $[0,3]$?

• A. $0$
• B. $-5$
• C. $-2$
• D. $1$

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the problem. We are tasked with finding the absolute minimum value of the function f(x) = x³ - 3x² + 1 on the closed interval [0, 3]. To do this, we need to evaluate f(x) at critical points and endpoints within the interval.
2. Step 2: Find the derivative of f(x) to locate critical points. The derivative f'(x) is calculated as follows: f'(x) = d/dx(x³ - 3x² + 1) = 3x² - 6x. Set f'(x) = 0 to find critical points.
3. Step 3: Solve the equation 3x² - 6x = 0. Factorize the derivative: 3x(x - 2) = 0. This gives two critical points: x = 0 and x = 2.
4. Step 4: Evaluate f(x) at the critical points and endpoints of the interval [0, 3]. The endpoints are x = 0 and x = 3, and the critical point within the interval is x = 2. Calculate f(0), f(2), and f(3).
5. Step 5: Compare the values of f(0), f(2), and f(3) to determine the absolute minimum value. The smallest value among these is the absolute minimum of f(x) on the interval [0, 3].