What is the extreme f(x)=2x+8/(x^2)+5,x>0 ?

The extreme f(x)=2x+8/(x^2)+5,x>0 is Minimum(2,11)

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extreme f(x)=2x+8/(x^2)+5,x>0

extreme points

f (x) = 2 x + \frac{8}{x ^{2}} + 5, x > 0

Minimum (2, 11)

Solution steps

Find the critical points:

x = 2

Domain of

2 x + \frac{8}{x ^{2}} + 5, 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}} + 5 \Rightarrow y = 11

Minimum (2, 11)

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

f (x, y) = \frac{1}{1 - x ^{2} - y ^{2}}

f (x, y) = 14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

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

x^{2} y + y^{3} - 12 y

What is the extreme f(x)=2x+8/(x^2)+5,x>0 ?
The extreme f(x)=2x+8/(x^2)+5,x>0 is Minimum(2,11)