What is the extreme f(x)=((x+4))/(x^2) ?

The extreme f(x)=((x+4))/(x^2) is Minimum(-8,-1/16)

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extreme f(x)=((x+4))/(x^2)

extreme points

f (x) = \frac{( x + 4 )}{x ^{2}}

Minimum (- 8, - \frac{1}{16})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 8,

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 8

into

\frac{x + 4}{x ^{2}} : - \frac{1}{16}

Minimum (- 8, - \frac{1}{16})

f (x) = 2 x^{3} - 6 x^{2} + 9 x - 2

f (x, y) = 8 y^{2} - 8 x^{2} - 8 y + 3 x y

f (x) = 2 x^{3} - 24 x + 2

y = \frac{( ln ( x ) ) ^{2}}{x}

What is the extreme f(x)=((x+4))/(x^2) ?
The extreme f(x)=((x+4))/(x^2) is Minimum(-8,-1/16)