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

The extreme f(x)=(x^2-9)/(x^2+9) is Minimum(0,-1)

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

extreme points

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

Minimum (0, - 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2} - 9}{x ^{2} + 9} : - 1

Minimum (0, - 1)

f (x) = 5 - 2 x^{2}

f (x) = - 5 x^{4} + 15 x^{3} - 17

f (x) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

f (x) = \frac{e ^{x}}{x - 1}

What is the extreme f(x)=(x^2-9)/(x^2+9) ?
The extreme f(x)=(x^2-9)/(x^2+9) is Minimum(0,-1)