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

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

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

extreme points

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

Minimum (- 2, - \frac{1}{2}), Maximum (0, \frac{1}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, - \frac{1}{2})

Plug

x = 0

into

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

Maximum (0, \frac{1}{2})

Minimum (- 2, - \frac{1}{2}), Maximum (0, \frac{1}{2})

f (x) = - x^{4} + 4 x^{3} + 8 x^{2}

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

(x^{2} + y^{2}) e^{y^{2} - x^{2}}

f (x) = 2 x^{3} + 3 x^{2}

f (x) = x^{2} e^{- 8 x}

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