What is the extreme f(x)= x/(x^2+1)[-1.2] ?

The extreme f(x)= x/(x^2+1)[-1.2] is Maximum(-1,0.6),Minimum(1,-0.6)

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extreme f(x)= x/(x^2+1)[-1.2]

extreme points

f (x) = \frac{x}{x ^{2} + 1} [- 1.2]

Maximum (- 1, 0.6), Minimum (1, - 0.6)

Solution steps

Find the critical points:

x = - 1, x = 1

Domain of

\frac{x}{x ^{2} + 1} [- 1.2] : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = - 1

into

\frac{x}{x ^{2} + 1} [- 1.2] \Rightarrow y = 0.6

Maximum (- 1, 0.6)

Plug the extreme point

x = 1

into

\frac{x}{x ^{2} + 1} [- 1.2] \Rightarrow y = - 0.6

Minimum (1, - 0.6)

Maximum (- 1, 0.6), Minimum (1, - 0.6)

\frac{( x ^{2} - 1 )}{( x - 3 ) 6}

f (x) = v (x) = x (7 - x) (15 - x), 0 < x < 7

f (x) = 6 x^{2} - 8 x + 3

f (x, y) = - 25 x^{2} - 25 y^{2} + 300 x + 350 y + 190

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

What is the extreme f(x)= x/(x^2+1)[-1.2] ?
The extreme f(x)= x/(x^2+1)[-1.2] is Maximum(-1,0.6),Minimum(1,-0.6)