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].
