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

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

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

extreme points

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

Minimum (- 1, - 1), Maximum (1, 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - 1)

Plug

x = 1

into

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

Maximum (1, 1)

Minimum (- 1, - 1), Maximum (1, 1)

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

x + \frac{32}{x ^{2}}

f (x, y) = 1 - x^{2} - 4 - y^{2}

f (x) = x - 2 cos (x), 0 \leq x \leq 2 π

f (x) = x^{4} - 20 x^{3} + 28 x^{2} - 4

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