To determine the absolute maximum value of $f(x) = x^3 - 6x^2 + 12x + 9$ on the closed interval $[1, 3]$, follow these steps:

Step 1: Find the derivative of $f(x)$

We compute $f'(x)$ to determine the critical points: $f'(x) = \frac{d}{dx}[x^3 - 6x^2 + 12x + 9]$ $f'(x) = 3x^2 - 12x + 12$

Step 2: Solve $f'(x) = 0$ to find critical points

Set $f'(x) = 0$: $3x^2 - 12x + 12 = 0$ Divide through by 3: $x^2 - 4x + 4 = 0$ Factorize: $(x - 2)(x - 2) = 0 \implies x = 2$

Thus, $x = 2$ is the only critical point in the interval $[1, 3]$.

Step 3: Evaluate $f(x)$ at endpoints and critical points

We now compute $f(x)$ at $x = 1$, $x = 2$, and $x = 3$.

1. At $x = 1$: $f(1) = (1)^3 - 6(1)^2 + 12(1) + 9$ $f(1) = 1 - 6 + 12 + 9 = 16$

2. At $x = 2$ (critical point): $f(2) = (2)^3 - 6(2)^2 + 12(2) + 9$ $f(2) = 8 - 24 + 24 + 9 = 17$

3. At $x = 3$: $f(3) = (3)^3 - 6(3)^2 + 12(3) + 9$ $f(3) = 27 - 54 + 36 + 9 = 18$

Step 4: Compare the values

From the calculations:

• $f(1) = 16$
• $f(2) = 17$
• $f(3) = 18$

The absolute maximum value occurs at $x = 3$, and the maximum value is: $\boxed{18}$

Final Answer:

The absolute maximum value of $f(x)$ on the interval $[1, 3]$ is 18, and it occurs at $x = 3$.

Would you like further clarifications or help with a similar problem? Here are 5 related questions:

1. How do you find absolute minimum values on a closed interval?
2. What is the difference between critical points and extrema?
3. How do you handle endpoints when finding absolute extrema?
4. How do derivatives help in identifying critical points?
5. Can you apply the same process for trigonometric or exponential functions?

Tip: Always check endpoints when finding absolute maxima or minima on a closed interval!