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

The extreme f(x)=(x+1)/(x^2+x+1) is Minimum(-2,-1/3),Maximum(0,1)

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

extreme points

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

Minimum (- 2, - \frac{1}{3}), Maximum (0, 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, - \frac{1}{3})

Plug

x = 0

into

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

Maximum (0, 1)

Minimum (- 2, - \frac{1}{3}), Maximum (0, 1)

f (x) = - x (x + 2) (x - 2)

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 5

f (x) = x^{3} + x^{2} - 5 x + 3

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

f (x) = 6 x^{5} - 10 x^{3}

What is the extreme f(x)=(x+1)/(x^2+x+1) ?
The extreme f(x)=(x+1)/(x^2+x+1) is Minimum(-2,-1/3),Maximum(0,1)