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

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

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

extreme points

y = \frac{2 x}{x ^{2} + 1}

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, - 1)

Plug

x = 1

into

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

Maximum (1, 1)

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

f (x, y) = 6 - 2 x - 3 y

f (x, y) = x^{3} + 3 x^{2} y + y^{3} - y

x^{3} + 3 x y^{2} - 39 x - 36 y + 36

f (x, y) = 62 x^{4} + 62 y^{4} - 4 x y + 15

f (x) = 4 y - 4 - x

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