What is the extreme f(x)=2x+8/(x^2)+1,x>0 ?

The extreme f(x)=2x+8/(x^2)+1,x>0 is Minimum(2,7)

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extreme f(x)=2x+8/(x^2)+1,x>0

extreme points

f (x) = 2 x + \frac{8}{x ^{2}} + 1, x > 0

Minimum (2, 7)

Solution steps

Find the critical points:

x = 2

Domain of

2 x + \frac{8}{x ^{2}} + 1, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 2

into

2 x + \frac{8}{x ^{2}} + 1 \Rightarrow y = 7

Minimum (2, 7)

f (x) = \frac{1}{x ^{2}} + \frac{1}{x ^{3}}

f (x, y) = x^{3} + y^{3} - 63 (x + y) + 12 x y

f (x, y) = x^{2} + 3 y^{2} - 4 x + 6 y

g (x) = \frac{1}{9} x^{9} - x

f (x, y) = x^{2} + y^{2} - 18 x + 12 y - 7

What is the extreme f(x)=2x+8/(x^2)+1,x>0 ?
The extreme f(x)=2x+8/(x^2)+1,x>0 is Minimum(2,7)