Answer :
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].
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].