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

The extreme f(x)=((2x^2))/(9-x^2) is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{36 x}{( 9 - x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{9 - x ^{2}} : 0

Minimum (0, 0)

f (x) = - \frac{1}{x - 6}

f (x, y) = 18 x^{2} - 3 y^{2} - 36 x - 128 y - 110

f (x) = ln (3 x^{2} + 3 x - 10)

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

y = x^{2} - 8 x + 7

What is the extreme f(x)=((2x^2))/(9-x^2) ?
The extreme f(x)=((2x^2))/(9-x^2) is Minimum(0,0)