Questions: Find the absolute maximum and absolute minimum values of f(x) = (x^2 - 16) / (x^2 + 16) on the interval [-5,5]. a) Absolute maximum b) Absolute minimum

Solution Steps

To find the absolute maximum and minimum values of the function \( f(x) = \frac{x^2 - 16}{x^2 + 16} \) on the interval \([-5, 5]\), we need to follow these steps:

1. Evaluate the function at the endpoints of the interval, \( x = -5 \) and \( x = 5 \).
2. Find the critical points of the function within the interval by setting the derivative \( f'(x) \) to zero and solving for \( x \).
3. Evaluate the function at the critical points found in step 2.
4. Compare the values obtained from steps 1 and 3 to determine the absolute maximum and minimum values.

We evaluate the function \( f(x) = \frac{x^2 - 16}{x^2 + 16} \) at the endpoints of the interval \([-5, 5]\):

• At \( x = -5 \): \[ f(-5) = \frac{(-5)^2 - 16}{(-5)^2 + 16} = \frac{25 - 16}{25 + 16} = \frac{9}{41} \]

• At \( x = 5 \): \[ f(5) = \frac{(5)^2 - 16}{(5)^2 + 16} = \frac{25 - 16}{25 + 16} = \frac{9}{41} \]

Next, we find the critical points by setting the derivative \( f'(x) \) to zero. The only critical point found is:

\[ x = 0 \]

Now we evaluate the function at the critical point \( x = 0 \):

• At \( x = 0 \): \[ f(0) = \frac{0^2 - 16}{0^2 + 16} = \frac{-16}{16} = -1 \]

We have the following values from our evaluations:

• \( f(-5) = \frac{9}{41} \)
• \( f(5) = \frac{9}{41} \)
• \( f(0) = -1 \)

Now we compare these values to determine the absolute maximum and minimum:

• The absolute maximum value is \( \frac{9}{41} \).
• The absolute minimum value is \( -1 \).

Final Answer