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.