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

The extreme f(x)=x^2+9/(x^2) is Minimum(-sqrt(3),6),Minimum(sqrt(3),6)

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extreme f(x)=x^2+9/(x^2)

extreme points

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

Minimum (- 3, 6), Minimum (3, 6)

Solution steps

f^{'} (x) = 2 x - \frac{18}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{2} + \frac{9}{x ^{2}} : 6

Minimum (- 3, 6)

Plug

x = 3

into

x^{2} + \frac{9}{x ^{2}} : 6

Minimum (3, 6)

Minimum (- 3, 6), Minimum (3, 6)

f (x) = - x^{4} - 8 x^{3} - 18 x^{2} + 7

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

2 - cos (x)

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

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

What is the extreme f(x)=x^2+9/(x^2) ?
The extreme f(x)=x^2+9/(x^2) is Minimum(-sqrt(3),6),Minimum(sqrt(3),6)