What is the extreme f(x)= x/(x^2+6x+8) ?

The extreme f(x)= x/(x^2+6x+8) is Minimum(-2sqrt(2),(3+2sqrt(2))/2),Maximum(2sqrt(2),(3-2sqrt(2))/2)

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extreme f(x)= x/(x^2+6x+8)

extreme points

f (x) = \frac{x}{x ^{2} + 6 x + 8}

Minimum (- 22, \frac{3 + 2 2}{2}), Maximum (22, \frac{3 - 2 2}{2})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 8}{( x ^{2} + 6 x + 8 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - 22,

Increasing

: - 22 < x < - 2,

Increasing

: - 2 < x < 22,

Decreasing

: 22 < x < \infty

Plug

x = - 22

into

\frac{x}{x ^{2} + 6 x + 8} : \frac{3 + 2 2}{2}

Minimum (- 22, \frac{3 + 2 2}{2})

Plug

x = 22

into

\frac{x}{x ^{2} + 6 x + 8} : \frac{3 - 2 2}{2}

Maximum (22, \frac{3 - 2 2}{2})

Minimum (- 22, \frac{3 + 2 2}{2}), Maximum (22, \frac{3 - 2 2}{2})

y = x^{2} - 2 x - 3

f (x) = 3 x^{\frac{2}{3}} - 2 x

2 x - 3 y = 12

2 - \frac{300}{x ^{2}}

- x

What is the extreme f(x)= x/(x^2+6x+8) ?
The extreme f(x)= x/(x^2+6x+8) is Minimum(-2sqrt(2),(3+2sqrt(2))/2),Maximum(2sqrt(2),(3-2sqrt(2))/2)