What is the extreme f(x)=(16(x+3))/(x^4) ?

The extreme f(x)=(16(x+3))/(x^4) is Minimum(-4,-1/16)

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extreme f(x)=(16(x+3))/(x^4)

extreme points

f (x) = \frac{16 ( x + 3 )}{x ^{4}}

Minimum (- 4, - \frac{1}{16})

Solution steps

f^{'} (x) = - \frac{48 ( x + 4 )}{x ^{5}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 4

into

\frac{16 ( x + 3 )}{x ^{4}} : - \frac{1}{16}

Minimum (- 4, - \frac{1}{16})

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

w^{4} - 55 w^{2} + 1600

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

y = \frac{x}{x ^{2} + 3}

y = x^{2} - 8 x, - \infty \leq x \leq 8

What is the extreme f(x)=(16(x+3))/(x^4) ?
The extreme f(x)=(16(x+3))/(x^4) is Minimum(-4,-1/16)