What is the extreme f(x)=(4x)/(x^2+1),(-4,0) ?

The extreme f(x)=(4x)/(x^2+1),(-4,0) is Minimum(-1,-2)

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extreme f(x)=(4x)/(x^2+1),(-4,0)

extreme points

f (x) = \frac{4 x}{x ^{2} + 1}, (- 4, 0)

Minimum (- 1, - 2)

Solution steps

Find the critical points:

x = - 1

Domain of

\frac{4 x}{x ^{2} + 1}, (- 4, 0) : - 4 < x < 0

Find intervals:

Decreasing

: - 4 < x < - 1,

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 1

into

\frac{4 x}{x ^{2} + 1} \Rightarrow y = - 2

Minimum (- 1, - 2)

f (x) = x^{2} + y^{2} - 6 y + 9

f (x) = 8 x^{\frac{3}{4}} - x

f (x) = - 5 x^{6} ln (x)

E (a, b) = (2 a + b)^{2} - (a + b) (a - b)

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

What is the extreme f(x)=(4x)/(x^2+1),(-4,0) ?
The extreme f(x)=(4x)/(x^2+1),(-4,0) is Minimum(-1,-2)