Answer :

To find the absolute maximum value of the function [tex]\( f(x) = 1 - 16x - \frac{16}{x} \)[/tex] on the interval [tex]\( (0, \infty) \)[/tex], we will follow these steps:

1. Find the first derivative of the function [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = -16 + \frac{16}{x^2} \][/tex]

2. Set the first derivative equal to zero to find the critical points:
[tex]\[ f'(x) = -16 + \frac{16}{x^2} = 0 \][/tex]
Solving this equation, we get:
[tex]\[ \frac{16}{x^2} = 16 \][/tex]
[tex]\[ x^2 = 1 \][/tex]
[tex]\[ x = \pm 1 \][/tex]
Since we are considering the interval [tex]\( (0, \infty) \)[/tex], we discard [tex]\( x = -1 \)[/tex] and keep [tex]\( x = 1 \)[/tex].

3. Verify the nature of the critical point (maximum or minimum) using the second derivative test:
[tex]\[ f''(x) = - \frac{32}{x^3} \][/tex]
Evaluate the second derivative at [tex]\( x = 1 \)[/tex]:
[tex]\[ f''(1) = - \frac{32}{1^3} = -32 \][/tex]
Since [tex]\( f''(1) < 0 \)[/tex], the function has a local maximum at [tex]\( x = 1 \)[/tex].

4. Evaluate the function at the critical point [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1 - 16(1) - \frac{16}{1} = 1 - 16 - 16 = -31 \][/tex]

Thus, the absolute maximum value of the function [tex]\( f(x) \)[/tex] on the interval [tex]\( (0, \infty) \)[/tex] is [tex]\( \boxed{-31} \)[/tex].