What is the extreme f(x)=((2x^2+4x))/((2+x^2)) ?

The extreme f(x)=((2x^2+4x))/((2+x^2)) is Minimum(1-sqrt(3),1-sqrt(3)),Maximum(1+sqrt(3),1+sqrt(3))

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extreme f(x)=((2x^2+4x))/((2+x^2))

extreme points

f (x) = \frac{( 2 x ^{2} + 4 x )}{( 2 + x ^{2} )}

Minimum (1 - 3, 1 - 3), Maximum (1 + 3, 1 + 3)

Solution steps

f^{'} (x) = \frac{- 4 x ^{2} + 8 x + 8}{( 2 + x ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3 + 1,

Increasing

: - 3 + 1 < x < 1 + 3,

Decreasing

: 1 + 3 < x < \infty

Plug

x = 1 - 3

into

\frac{2 x ^{2} + 4 x}{2 + x ^{2}} : 1 - 3

Minimum (1 - 3, 1 - 3)

Plug

x = 1 + 3

into

\frac{2 x ^{2} + 4 x}{2 + x ^{2}} : 1 + 3

Maximum (1 + 3, 1 + 3)

Minimum (1 - 3, 1 - 3), Maximum (1 + 3, 1 + 3)

f (x) = x^{3} + 8 x + 12

7 - 6 x - x^{3}

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

f (x) = x^{2} - 14 x + 4

f (x) = x^{2} - 14 x + 6

What is the extreme f(x)=((2x^2+4x))/((2+x^2)) ?
The extreme f(x)=((2x^2+4x))/((2+x^2)) is Minimum(1-sqrt(3),1-sqrt(3)),Maximum(1+sqrt(3),1+sqrt(3))