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To solve for the absolute maximum and absolute minimum values of the function \( f(x) = \frac{x}{x^2 - x + 25} \) on the interval \([0, 15]\), we will follow these steps:

### Step 1: Analyze the function

The given function is:

\[
f(x) = \frac{x}{x^2 - x + 25}
\]

We need to find the critical points, evaluate the function at those points, and check the values at the endpoints of the interval.

### Step 2: Find the critical points

To find the critical points, we first need to compute the derivative of \( f(x) \) using the quotient rule. The quotient rule states that for \( f(x) = \frac{g(x)}{h(x)} \), the derivative is:

\[
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}
\]

For our function:

- \( g(x) = x \) and \( g'(x) = 1 \)
- \( h(x) = x^2 - x + 25 \) and \( h'(x) = 2x - 1 \)

Applying the quotient rule:

\[
f'(x) = \frac{1(x^2 - x + 25) - x(2x - 1)}{(x^2 - x + 25)^2}
\]

Simplifying the numerator:

\[
f'(x) = \frac{x^2 - x + 25 - x(2x - 1)}{(x^2 - x + 25)^2} = \frac{x^2 - x + 25 - (2x^2 - x)}{(x^2 - x + 25)^2} = \frac{x^2 - x + 25 - 2x^2 + x}{(x^2 - x + 25)^2}
\]

\[
f'(x) = \frac{-x^2 + 25}{(x^2 - x + 25)^2}
\]

Now, set the derivative equal to zero to find the critical points:

\[
\frac{-x^2 + 25}{(x^2 - x + 25)^2} = 0
\]

The numerator must be zero for the derivative to be zero:

\[
-x^2 + 25 = 0
\]

\[
x^2 = 25
\]

\[
x = \pm 5
\]

However, since we are only interested in the interval \([0, 15]\), the only critical point within this interval is \( x = 5 \).

### Step 3: Evaluate \( f(x) \) at the critical points and endpoints

Now, we evaluate the function at the critical point \( x = 5 \) and at the endpoints \( x = 0 \) and \( x = 15 \):

- At \( x = 5 \):

\[
f(5) = \frac{5}{5^2 - 5 + 25} = \frac{5}{25 - 5 + 25} = \frac{5}{45} = \frac{1}{9}
\]

- At \( x = 0 \):

\[
f(0) = \frac{0}{0^2 - 0 + 25} = 0
\]

- At \( x = 15 \):

\[
f(15) = \frac{15}{15^2 - 15 + 25} = \frac{15}{225 - 15 + 25} = \frac{15}{235} = \frac{3}{47}
\]

### Step 4: Conclusion

Now, we compare the function values at these points:

- \( f(0) = 0 \)
- \( f(5) = \frac{1}{9} \)
- \( f(15) = \frac{3}{47} \)

The absolute maximum value is \( f(5) = \frac{1}{9} \), and the absolute minimum value is \( f(0) = 0 \).

### Summary

- **Absolute maximum**: \( \frac{1}{9} \) at \( x = 5 \)
- **Absolute minimum**: \( 0 \) at \( x = 0 \)

Would you like any further clarification or details on any step? 

Here are 5 related questions to consider:
1. How do we apply the quotient rule for derivatives?
2. What happens to a function's behavior at critical points?
3. How do we determine concavity and behavior at critical points using the second derivative test?
4. How would we handle optimization if the interval were different or unbounded?
5. How can the derivative's sign tell us about increasing or decreasing behavior?

**Tip:** Always check endpoints when looking for absolute extrema on a closed interval.

Find the absolute maximum and absolute minimum values of the function f(x) = x / (x^2 - x + 25) on the interval [0, 15].

Learn how to find the absolute maximum and minimum values of f(x) = x / (x^2 - x + 25) using the quotient rule and critical points. Evaluate the function at critical points and endpoints to determine the extrema on the interval [0, 15].

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