The problem is asking to find the absolute maximum and minimum values of the function

$f(x) = \frac{x^2 - 4}{x^2 + 4}$

on the interval $[-5, 5]$.

Step-by-Step Process:

1. Check the endpoints of the interval: Evaluate $f(x)$ at $x = -5$ and $x = 5$.

   $f(-5) = \frac{(-5)^2 - 4}{(-5)^2 + 4} = \frac{25 - 4}{25 + 4} = \frac{21}{29}$ $f(5) = \frac{(5)^2 - 4}{(5)^2 + 4} = \frac{25 - 4}{25 + 4} = \frac{21}{29}$ So, the values at the endpoints are $f(-5) = f(5) = \frac{21}{29}$.

2. Find the critical points: We take the derivative of $f(x)$ to find critical points.

   Using the quotient rule: $f'(x) = \frac{(2x(x^2 + 4)) - (x^2 - 4)(2x)}{(x^2 + 4)^2}$ Simplifying: $f'(x) = \frac{2x(x^2 + 4) - 2x(x^2 - 4)}{(x^2 + 4)^2}$ [ f'(x) = \frac{2x[(x^2 + 4) - (x^2 - 4)]}{(x^2 + 4)^2} = \frac{2x[8]}{(x^2 + 4)^2} = \frac{16x}{(x^2 + 4)^2} ]

   Set $f'(x) = 0$: $\frac{16x}{(x^2 + 4)^2} = 0$ This gives $x = 0$ as the only critical point.

3. Evaluate at the critical point: Evaluate $f(x)$ at $x = 0$: $f(0) = \frac{0^2 - 4}{0^2 + 4} = \frac{-4}{4} = -1$

4. Conclusion:

   • At $x = 0$, $f(0) = -1$ is the absolute minimum.
   • At the endpoints $x = -5$ and $x = 5$, $f(x) = \frac{21}{29}$, which is the absolute maximum.

Final Answer:

• The absolute maximum is $\frac{21}{29}$.
• The absolute minimum is $-1$.

Let me know if you'd like further details or if you have any other questions!

Here are some related questions you might find interesting:

1. How do you find critical points for other types of rational functions?
2. What are the conditions for the existence of absolute extrema on closed intervals?
3. How does the quotient rule for derivatives work in general?
4. What happens if the function is not continuous over the given interval?
5. How would you handle optimization problems on an open interval?

Tip: When analyzing extrema on a closed interval, always check both critical points and endpoints, as absolute extrema can occur at either!