Questions: Find the absolute maximum and minimum values of the function f(x) = 5/(x-4) + 9 ln(x-4) on the interval [18/3, 7].

Find the absolute maximum and minimum values of the function \( f(x) = \frac{5}{x-4} + 9\ln(x-4) \) on the interval \([6, 7]\).

Find the derivative of \( f(x) \).

The derivative of \( f(x) \) is \( f'(x) = -\frac{5}{(x-4)^2} + \frac{9}{x-4} \).

Set the derivative equal to zero to find critical points.

Set \( f'(x) = 0 \):
\[ -\frac{5}{(x-4)^2} + \frac{9}{x-4} = 0 \]
\[ \frac{9}{x-4} = \frac{5}{(x-4)^2} \]
\[ 9(x-4)^2 = 5(x-4) \]
Divide both sides by \( (x-4) \) (since \( x \neq 4 \)):
\[ 9(x-4) = 5 \]
\[ 9x - 36 = 5 \]
\[ 9x = 41 \]
\[ x = \frac{41}{9} \]

Check if the critical point is within the interval \([6, 7]\).

The critical point \( x = \frac{41}{9} \approx 4.556 \) is not within the interval \([6, 7]\).

Evaluate \( f(x) \) at the endpoints of the interval.

Evaluate at \( x = 6 \):
\[ f(6) = \frac{5}{6-4} + 9\ln(6-4) = \frac{5}{2} + 9\ln(2) \]
Evaluate at \( x = 7 \):
\[ f(7) = \frac{5}{7-4} + 9\ln(7-4) = \frac{5}{3} + 9\ln(3) \]

Compare the values to determine the absolute maximum and minimum.

Since \( f(7) > f(6) \), the absolute maximum value is \( f(7) = \frac{5}{3} + 9\ln(3) \) and the absolute minimum value is \( f(6) = \frac{5}{2} + 9\ln(2) \).

Absolute maximum value: \(\boxed{\frac{5}{3} + 9\ln(3)}\)
Absolute minimum value: \(\boxed{\frac{5}{2} + 9\ln(2)}\)

Absolute maximum value: \(\boxed{\frac{5}{3} + 9\ln(3)}\)
Absolute minimum value: \(\boxed{\frac{5}{2} + 9\ln(2)}\)