What is the extreme f(x)=(x^2-4)/(9-x^2) ?

The extreme f(x)=(x^2-4)/(9-x^2) is Minimum(0,-4/9)

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extreme f(x)=(x^2-4)/(9-x^2)

extreme points

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

Minimum (0, - \frac{4}{9})

Solution steps

f^{'} (x) = \frac{10 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{x ^{2} - 4}{9 - x ^{2}} : - \frac{4}{9}

Minimum (0, - \frac{4}{9})

- 2 x^{2} + 3 x + 1

f (x) = x + sin^{2} (x)

y = x + \frac{1}{x}

2 x + 1 + \frac{5}{x - 5}

f (x) = (x - y) (1 - x y)

What is the extreme f(x)=(x^2-4)/(9-x^2) ?
The extreme f(x)=(x^2-4)/(9-x^2) is Minimum(0,-4/9)