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], follow these steps:
1. Calculate the first derivative [tex]\( f'(x) \)[/tex]:
To find the critical points, we need to differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx} \left(1 - 16x - \frac{16}{x}\right).
\][/tex]
The derivative of each term is:
[tex]\[
f'(x) = 0 - 16 - \left(-\frac{16}{x^2}\right) = -16 + \frac{16}{x^2}.
\][/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 for [tex]\( x \)[/tex], we get:
[tex]\[
-16 + \frac{16}{x^2} = 0 \Rightarrow \frac{16}{x^2} = 16 \Rightarrow x^2 = 1 \Rightarrow x = 1 \text{ (since } x > 0).
\][/tex]
3. Check if the critical points lie in the interval [tex]\( (0, \infty) \)[/tex]:
The critical point [tex]\( x = 1 \)[/tex] lies within the interval [tex]\( (0, \infty) \)[/tex].
4. Evaluate the function at the critical point:
To find the function value at [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 1 - 16(1) - \frac{16}{1} = 1 - 16 - 16 = -31.
\][/tex]
Since [tex]\( f(x) \)[/tex] has only one critical point in the interval [tex]\( (0, \infty) \)[/tex] and we have evaluated the function at this point, we determine that:
The absolute maximum value of [tex]\( f(x) \)[/tex] on [tex]\( (0, \infty) \)[/tex] is [tex]\( -31 \)[/tex] at [tex]\( x = 1 \)[/tex].
Therefore, the correct choice is:
A. The absolute maximum is [tex]\( -31 \)[/tex] at [tex]\( x = 1 \)[/tex].
