91Ó°ÊÓ

In calculus, the first derivative of a function gives us important information about the function's rate of change. For the given function, \( f(x) = x + \frac{4}{x} \), we need to find its first derivative to understand where the function's slope is zero or undefined.

The first derivative \( f'(x) \) of our function is calculated using differentiation rules. By employing the power rule for \( x \) and the quotient rule for \( \frac{4}{x} \), we get:

\[ f'(x) = 1 - \frac{4}{x^2} \]
Finding this derivative is crucial because setting it to zero will help us find the critical points, which is our next step.

Critical points occur where the first derivative is zero or undefined. These points can indicate potential maxima, minima, or inflection points of the function. To find the critical points for our function \( f(x) = x + \frac{4}{x} \), we set the first derivative to zero and solve for \( x \):

\[ 1 - \frac{4}{x^2} = 0 \]
Rearranging the equation gives:
\[ \frac{4}{x^2} = 1 \]
By multiplying both sides by \( x^2 \) and solving for \( x \), we get:
\[ 4 = x^2 \]
\[ x = \pm 2 \]
Here, we have potential critical points at \( x = -2 \) and \( x = 2 \). However, since our interval of interest is \([-8, -1]\), we disregard \( x = 2 \) because it doesn't lie within this interval. Thus, only \( x = -2 \) is a valid critical point for the problem.

Once we have our critical points, determining the absolute maximum and minimum values involves evaluating the original function at these points and the endpoints of the interval. Here, we calculate \( f(x) \) at the critical point \( x = -2 \), and at the interval endpoints \( x = -8 \) and \( x = -1 \).

Evaluations:

• For \( x = -2 \): \[ f(-2) = -2 + \frac{4}{-2} = -4 \]
• For \( x = -8 \): \[ f(-8) = -8 + \frac{4}{-8} = -8.5 \]
• For \( x = -1 \): \[ f(-1) = -1 + \frac{4}{-1} = -5 \]

By comparing these values, we find:

• Absolute Minimum: \( f(-8) = -8.5 \) at \( x = -8 \)
• Absolute Maximum: \( f(-2) = -4 \) at \( x = -2 \)

Interval evaluation is the final step to identify which critical points and interval boundaries give the absolute maximum and minimum values. Here, we carefully evaluate \( f(x) \) within the specified interval \([-8, -1]\). This involves ensuring that we include the endpoints of the interval as part of our calculations.

We found earlier that our function is evaluated at key points:

• \( f(-8) = -8.5 \)
• \( f(-2) = -4 \)
• \( f(-1) = -5 \)

By comparing the function's values at these points, we conclude:

• The function achieves its absolute minimum value of \( -8.5 \) at \( x = -8 \).
• The absolute maximum value of \( -4 \) occurs at \( x = -2 \).

This interval evaluation verifies the critical points and confirms the absolute extrema within the defined interval.