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The problem asks you to determine the maximum and minimum values of the function:

\[
f(x) = \frac{4x - 10}{2x^2 - 6x + 27}
\]

on the interval \([-2, 5]\). 

### Solution Overview
To find the maximum and minimum values on a closed interval, we need to:
1. **Compute the derivative** of \(f(x)\).
2. **Find critical points** by setting the derivative equal to zero or undefined.
3. **Evaluate the function at critical points and the endpoints** \(-2\) and \(5\).
4. Compare all these values to find the largest and smallest.

---

### Step 1: Compute the Derivative \(f'(x)\)
We use the **quotient rule** for differentiation:

\[
\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}
\]

where:
- \(u(x) = 4x - 10\), so \(u'(x) = 4\)
- \(v(x) = 2x^2 - 6x + 27\), so \(v'(x) = 4x - 6\)

Applying the quotient rule:

\[
f'(x) = \frac{4(2x^2 - 6x + 27) - (4x - 10)(4x - 6)}{(2x^2 - 6x + 27)^2}
\]

Simplify the numerator:

\[
4(2x^2 - 6x + 27) = 8x^2 - 24x + 108
\]
\[
(4x - 10)(4x - 6) = 16x^2 - 40x - 24x + 60 = 16x^2 - 64x + 60
\]

Now:

\[
8x^2 - 24x + 108 - (16x^2 - 64x + 60) = -8x^2 + 40x + 48
\]

Thus:

\[
f'(x) = \frac{-8x^2 + 40x + 48}{(2x^2 - 6x + 27)^2}
\]

---

### Step 2: Find Critical Points
We need to set \(f'(x) = 0\), which means the numerator must be zero:

\[
-8x^2 + 40x + 48 = 0
\]

Divide by \(-8\):

\[
x^2 - 5x - 6 = 0
\]

Solve using the quadratic formula:

\[
x = \frac{5 \pm \sqrt{25 + 24}}{2} = \frac{5 \pm \sqrt{49}}{2} = \frac{5 \pm 7}{2}
\]

Thus:

\[
x = 6 \quad \text{or} \quad x = -1
\]

---

### Step 3: Evaluate \(f(x)\) at Critical Points and Endpoints
We evaluate the function \(f(x)\) at \(x = -2\), \(x = -1\), \(x = 5\), and any valid critical points in the interval.

- \(f(-2) = \frac{4(-2) - 10}{2(-2)^2 - 6(-2) + 27} = \frac{-8 - 10}{8 + 12 + 27} = \frac{-18}{47}\)
- \(f(-1) = \frac{4(-1) - 10}{2(-1)^2 - 6(-1) + 27} = \frac{-4 - 10}{2 + 6 + 27} = \frac{-14}{35} = -\frac{2}{5}\)
- \(f(5) = \frac{4(5) - 10}{2(5)^2 - 6(5) + 27} = \frac{20 - 10}{50 - 30 + 27} = \frac{10}{47}\)

---

### Step 4: Compare Values
Now we compare the function values at the points:

- \(f(-2) = \frac{-18}{47} \approx -0.383\)
- \(f(-1) = -\frac{2}{5} = -0.4\)
- \(f(5) = \frac{10}{47} \approx 0.213\)

Thus:
- The **minimum value** is \(f(-1) = -0.4\).
- The **maximum value** is \(f(5) \approx 0.213\).

---

### Final Answer
- Maximum value: \(f(5) \approx 0.213\)
- Minimum value: \(f(-1) = -0.4\)

---

Do you need further details on any of these steps?

---

Here are five related questions for further exploration:
1. How does the quotient rule work in calculus?
2. Why do we evaluate endpoints in optimization problems on closed intervals?
3. How can we confirm whether critical points are maxima, minima, or saddle points?
4. What is the significance of the second derivative in optimization?
5. How would the solution change if the function were defined on an open interval?

**Tip:** When solving optimization problems, always check both the critical points *and* endpoints to ensure the correct extrema are found.

Bestäm eventuella största och minsta värden av funktionen f(x) = (4x - 10) / (2x² - 6x + 27) på intervallet [-2, 5].

Solve an optimization problem involving the function f(x) = (4x - 10) / (2x² - 6x + 27) on the interval [-2, 5]. Learn how to apply the quotient rule, find critical points, and evaluate function values at endpoints to determine maxima and minima.

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