To determine the local maximum and local minimum of the function [tex]\( f(x) = 2x + \frac{8}{x} \)[/tex], we will follow these steps:
1. Find the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx} \left(2x + \frac{8}{x}\right)
= 2 - \frac{8}{x^2}
\][/tex]
2. Set the first derivative to zero and solve for [tex]\( x \)[/tex] to find the critical points:
[tex]\[
2 - \frac{8}{x^2} = 0
\][/tex]
[tex]\[
2 = \frac{8}{x^2}
\][/tex]
[tex]\[
2x^2 = 8
\][/tex]
[tex]\[
x^2 = 4
\][/tex]
[tex]\[
x = \pm 2
\][/tex]
So, the critical points are [tex]\( x = 2 \)[/tex] and [tex]\( x = -2 \)[/tex].
3. Find the second derivative [tex]\( f''(x) \)[/tex]:
[tex]\[
f''(x) = \frac{d}{dx} \left(2 - \frac{8}{x^2}\right)
= \frac{16}{x^3}
\][/tex]
4. Evaluate the second derivative at each critical point to determine the nature of each point.
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
f''(-2) = \frac{16}{(-2)^3}
= \frac{16}{-8}
= -2
\][/tex]
Since [tex]\( f''(-2) < 0 \)[/tex], [tex]\( x = -2 \)[/tex] is a point of local maximum.
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
f''(2) = \frac{16}{2^3}
= \frac{16}{8}
= 2
\][/tex]
Since [tex]\( f''(2) > 0 \)[/tex], [tex]\( x = 2 \)[/tex] is a point of local minimum.
5. Calculate the value of the function at these critical points to find the local extrema:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = 2(-2) + \frac{8}{-2}
= -4 - 4
= -8
\][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 2(2) + \frac{8}{2}
= 4 + 4
= 8
\][/tex]
Therefore, the function [tex]\( f(x) = 2x + \frac{8}{x} \)[/tex] has:
- A local maximum at [tex]\( x = -2 \)[/tex] with value [tex]\( -8 \)[/tex]
- A local minimum at [tex]\( x = 2 \)[/tex] with value [tex]\( 8 \)[/tex]
Hence, the function has a local maximum at [tex]\( x = -2 \)[/tex] with value [tex]\( -8 \)[/tex] and a local minimum at [tex]\( x = 2 \)[/tex] with value [tex]\( 8 \)[/tex].
