What is the extreme f(x)=(9(x^2-3))/(x^3) ?

The extreme f(x)=(9(x^2-3))/(x^3) is Minimum(-3,-2),Maximum(3,2)

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extreme f(x)=(9(x^2-3))/(x^3)

extreme points

f (x) = \frac{9 ( x ^{2} - 3 )}{x ^{3}}

Minimum (- 3, - 2), Maximum (3, 2)

Solution steps

f^{'} (x) = \frac{9 ( - x ^{2} + 9 )}{x ^{4}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{9 ( x ^{2} - 3 )}{x ^{3}} : - 2

Minimum (- 3, - 2)

Plug

x = 3

into

\frac{9 ( x ^{2} - 3 )}{x ^{3}} : 2

Maximum (3, 2)

Minimum (- 3, - 2), Maximum (3, 2)

y = x^{4} - 2 x^{2} + 1

f (x) = 8 sin (2 x)

f (x) = - 20 x^{2} - 30 y^{2} + 560 x + 360 y + 180

\frac{3 x + 5}{5 x ^{2} + 55 x + 120}

f (x) = x^{4} e^{- 7 x}

What is the extreme f(x)=(9(x^2-3))/(x^3) ?
The extreme f(x)=(9(x^2-3))/(x^3) is Minimum(-3,-2),Maximum(3,2)