What is the extreme f(x)=(2(x+2)^2)/(x^2) ?

The extreme f(x)=(2(x+2)^2)/(x^2) is Minimum(-2,0)

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extreme f(x)=(2(x+2)^2)/(x^2)

extreme points

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

Minimum (- 2, 0)

Solution steps

f^{'} (x) = - \frac{8 ( x + 2 )}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

\frac{2 ( x + 2 ) ^{2}}{x ^{2}} : 0

Minimum (- 2, 0)

f (x, y) = y^{3} x^{5} + y x

f (x, y) = 2 x^{3} - 2 y^{3} + 12 x y + 3

2 x^{3} - 3 x^{2} - 12 x + 8

f (x, y) = x y + e^{x y}

g (x, y) = ln (x^{2} + y^{2} - 1)

What is the extreme f(x)=(2(x+2)^2)/(x^2) ?
The extreme f(x)=(2(x+2)^2)/(x^2) is Minimum(-2,0)