To find the absolute maximum and minimum values of $f(x) = 16 + 4x - x^2$ on the interval $[0, 5]$, we can use the following steps:

1. Differentiate the function to find the critical points within the interval.
2. Evaluate $f(x)$ at the critical points and endpoints.
3. Compare these values to determine the absolute maximum and minimum.

Step 1: Find the Critical Points

We start by finding $f'(x)$: $f(x) = 16 + 4x - x^2$ $f'(x) = 4 - 2x$ Setting $f'(x) = 0$ to find critical points: $4 - 2x = 0$ $x = 2$ So, $x = 2$ is a critical point within the interval $[0, 5]$.

Step 2: Evaluate $f(x)$ at the Endpoints and Critical Points

We need to evaluate $f(x)$ at $x = 0$, $x = 5$, and $x = 2$.

1. At $x = 0$: $f(0) = 16 + 4(0) - (0)^2 = 16$

2. At $x = 5$: $f(5) = 16 + 4(5) - (5)^2 = 16 + 20 - 25 = 11$

3. At $x = 2$: $f(2) = 16 + 4(2) - (2)^2 = 16 + 8 - 4 = 20$

Step 3: Identify the Absolute Maximum and Minimum

• The highest value is $f(2) = 20$, so the absolute maximum on $[0, 5]$ is $20$.
• The lowest value is $f(5) = 11$, so the absolute minimum on $[0, 5]$ is $11$.

Final Answer

• Absolute maximum value: $20$ at $x = 2$
• Absolute minimum value: $11$ at $x = 5$

Would you like additional details or have questions on any of the steps?

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Here are 5 questions to further explore the topic:

1. How would we find critical points if the interval were different?
2. What if $f(x)$ were a cubic polynomial—would we follow the same steps?
3. Why do we need to check endpoints when finding absolute extrema on a closed interval?
4. How can we confirm if a critical point is a maximum or minimum using the second derivative?
5. What role does the shape of a parabola play in identifying extrema for quadratic functions?

Tip: Always check both critical points and endpoints in closed intervals to find absolute extrema.