What is the extreme f(x)=(1-x^2)/((1+x^2)^2) ?

The extreme f(x)=(1-x^2)/((1+x^2)^2) is Minimum(-sqrt(3),-1/8),Maximum(0,1),Minimum(sqrt(3),-1/8)

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extreme f(x)=(1-x^2)/((1+x^2)^2)

extreme points

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

Minimum (- 3, - \frac{1}{8}), Maximum (0, 1), Minimum (3, - \frac{1}{8})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}} : - \frac{1}{8}

Minimum (- 3, - \frac{1}{8})

Plug

x = 0

into

\frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}} : 1

Maximum (0, 1)

Plug

x = 3

into

\frac{1 - x ^{2}}{( 1 + x ^{2} ) ^{2}} : - \frac{1}{8}

Minimum (3, - \frac{1}{8})

Minimum (- 3, - \frac{1}{8}), Maximum (0, 1), Minimum (3, - \frac{1}{8})

f (x) = x^{2} - 12 x - 13

f (x) = 15 x^{\frac{2}{3}} - 10 x

x^{4} + 4 x^{3} - 2

f (x) = 3 x^{2} ln (x)

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

What is the extreme f(x)=(1-x^2)/((1+x^2)^2) ?
The extreme f(x)=(1-x^2)/((1+x^2)^2) is Minimum(-sqrt(3),-1/8),Maximum(0,1),Minimum(sqrt(3),-1/8)