What is the extreme f(x)=((x+8))/(x^2) ?

The extreme f(x)=((x+8))/(x^2) is Minimum(-16,-1/32)

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extreme f(x)=((x+8))/(x^2)

extreme points

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

Minimum (- 16, - \frac{1}{32})

Solution steps

f^{'} (x) = \frac{- x - 16}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 16,

Increasing

: - 16 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 16

into

\frac{x + 8}{x ^{2}} : - \frac{1}{32}

Minimum (- 16, - \frac{1}{32})

- x^{3} + 3 x + 7

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

f (x, y) = e^{- 5 x - 5 y}

f (x, y) = 2 y - \frac{1}{5} x y

f (a, p) = x^{3} + y^{3} - 24 x y

What is the extreme f(x)=((x+8))/(x^2) ?
The extreme f(x)=((x+8))/(x^2) is Minimum(-16,-1/32)